Wow in just about one week you guys have captioned this video in English, AND it's been translated into Arabic and German! Captioning is a lot of hard work--I know I've done a few--and I'm really appreciative. I read that TH-cam gives some rewards for doing captions, has anyone earned any worthwhile rewards yet? You guys are great, I have so many things to be thankful for this year!
Interesting! Actually, I provided both the english and the german version (the english subs because TH-cam wouldn't provide autogenerated english subs when I wanted to do the german translation... though, they could provide autogenerated once for english ¯\_(ツ)_/¯). I submitted them approximately 2 hours after you uploaded the video and they needed one week to verify them. I would be interested in how they did that? Do they simply ask random people how they find the subtitles? The autogenerated once where actually quite good, and let's be honest: You speak so clearly one does not really need the subtitles for your videos :D
Well you can not start by saying it does not matter the first sock if it is black or white. Which will translate 24/24 to 1, But what matters in the second sock to make a pair and then you have 11/23 chance. So the total probability is 1 times 11/23 which is 11/23. I think this way its easier to go by.
This was so easy I was starting to get paranoid if maybe I'm missing something. By the way who with even the slightest knowledge of how propability works would assume it's 50%... Also I think the better and faster method of considering it is: The first sock you pick doesn't matter which colour is it, but the second one must be matching. Since we've already taken out one of this color sock there are 11 left out of 23. So it's 11/23
It's not that people don't understand how probability works, it's often that people don't realise they should be applying their knowledge of how probability works, instead of going with the easy and intuitive answer. It's similar in nature to the bat and ball question, if you've heard of it.
The solution in the video is theoretically correct. In reality, not only will you always pick unmatched socks, but after the first time doing laundry, you will not have an even number of any single color of socks.
I did it a bit different. We were only asked to get a matching pair. So whatever is picked first, be it black or white, means that on the second pick we have to pick what we picked in the first pick. The odds of getting the second sock to match the first pick is 11 out of 23. That equals 47.8%. After posting this I saw others did it this way too.
This is exactly how I thought about it too. I immediately thought it didn't matter which color you picked first The only probability that mattered was the probability of picking another one of that same color which at that point would be reduced by one therefore 11/23.
Good problem. It's easy for many of us that are used to thinking, but it's good for a lot of viewers to see the easier/basic stuff, too. So thanks for mixing up the difficulty level!
@@commandercaptain4664 No? I don't get the logic. Think about it like this: you must take either a black or a white sock. The odds that you then take a matching sock is 11/23
11/23 is the probability I think. If you pull any one sock, there's a 50/50 chance it will be black or white, but that doesn't matter because no matter which color you pull there'd be 11 of that same color left of 23 remaining socks.
Very smart! Unfortunately this does not work for me since I have multi colored socks. I guess I gotta throw them all out and only buy black and white socks :D
Again, his explanation is far too complex. It should just be this: After you pick the first sock there are 11 more socks of that color, and 12 of the opposite color. So the chances of picking another sock of the same color are 11 out of a total of 23, which equals approximately 47.8%. Done! Why complicate things with the rest of that algebraic notation, confusing verbiage ("sampling without replacement"? shees!), separate consideration of the black sock case vs. the white sock case, etc. Keep it simple, I say.
I agree he tends to overcomplicate explanation but that verbiage is in standard use. I mean obviously from the problem statement it makes more sense without replacement, but since it is never actually specified, you could assume sampling with replacement and achieve an entirely different but also correct answer. But really the simplest way to express it is just as a probability equation. P(black and black or white and white) = 2*P(black and black) which can be simplified to 11/23 in your head in this case because of the simplicity
Yep. Regardless of which color you pick, there will always be 11 of that color left in the pile, of a total of 23 socks. The question is simple. The first grab has no significance on the outcome, only the second grab. Therefore the answer is 11/23.
Everyone is forgetting that this method only works if there is the same quantity of each colour. If there were 11 black socks and 12 white socks your "simple" method would not work
You missed the trick to the question: the socks are identical, so there are no matching pairs. Your black or white socks will either be 2 left socks or 2 right socks, not a pair of 1 left sock and 1 right sock.
Nathan Lang ah, that would be the laundry gnomes. Tricky little buggers make off with one on occasion. The only solution I've found is to sacrifice the other sock in tribute to appease them. Keeps the laundry safe for 2-3 months
This is actually a good point that highlights the flaw in the question. You're still picking the socks at random, you've just changed the distribution. The question doesn't actually say what the distribution is, so hey. Good out of the box answer, now go sort those boxes.
For a matching pair you don't need to add a probability for Pr(bl) and Pr(wh). All you need to do is to calculate the 11/23. Since you can pick any color with your first pick - the question becomes "what is the probability that the next color is the same". Hence 11/23. Result is the same, it's just simpler.
I have a large number of odd socks in my drawer. Getting up before it's light might have led to many odd-sock days! I folded the few matching pairs together when I put them in the drawer to avert this real life problem.
Yeah, amusingly like BlacksmithTWD said, if you had an infinite number of socks, there would actually be a 0% chance of picking any of them, however you’re right in that when you do pick your 0% probably socks, they will always have a 50% chance to be black or white.
At the Railroad we had a similar set of choices. To fix a broken rail at night you need what are called Angle Bars. These bars fit in the web of the rail and hold the bolts and rail to make a complete joint. Usually we would need 2 sets (4 bars) to cut out the defect and add a length of rail. The problem is that the shoulder of the bolts is oval and locks the blot from spinning when you tighten an 1 1/8in bolt. Most bars have the oblong hole to the right end of the bar then alternate oval, round, oval, round, oval, round. There are 6 holes, 3 round and 3 oblong in each bar. Once in a while you find and outlier and it has the oblong/oval hole on the left of the bar. When you put 2 oblong left bars together you make a good joint because every other bolt goes in from the oblong hole and comes out of the other bar through the round hole. So 2 Oblong right bars make a set and if you get an oblong left mixed in you can not make the joint safe because the oblong/oval holes line up with each other and only 3 of the bolts can be held fast. In the dark you always get an odd number of bars just incase you mix the oblong rights and oblong lefts. With an odd number you will always be able to make a joint/pair.
Easy, 11/23. First sock can be anything then 23 socks left. If first white, probability second white 11/23 If first black, probability second black 11/23 (ie 11/23 chance that the second sock is the same as the first)
Nicholas 4321 My exact reasoning. No need to go through the trouble of the 1/2 + 1/2 because the first draw doesn't matter if all you want is any matching pair.
Another way of saying this is: the real problem is the probability of picking a sock with a color equal to the color of 11 socks in the drawer, when there are 12 socks of the other color. But, then, that would be a give-a-way!
I'm trying to deduce if it would be 50% if you picked both at the same time one with each hand given that if you remove 2 at random simultaneously there is never one missing before drawing a second sock?
There are 2 possible scenarios: First, you could unwittingly grab the same sock with both hands. There’s a 1/24 chance of that happening. As this wouldn’t satisfy the “pair” requirement, you’d try again. Second, you grab two distinct socks, one with each hand. It doesn’t matter if the grab is simultaneous or not. The probability of pulling out a pair is still 11/23 as explained in the video. The reason is that your left hand grabs a sock, and since we’re not in the first scenario, there are only 23 socks for your right hand to grab. Of those 23, 11 are matching whatever color you hold in your left hand. Or vice-versa if you grab with your right hand first. Or even if you grab simultaneously. Not grabbing the same sock necessitates there being 23 available socks to see if it’s a match.
Qaz Wsx Good catch! Yes, it would be 5/11, I didn’t think that through completely. I should spend more time double checking my work and less time admiring my humor before posting.
Yes, I figured it out. Total time < 2-seconds. I taught AP Statistics for 22-years and am confident that all of my former students would solve this problem, just as quickly.
There's a fundamental flaw in this solution - it makes an assumption (not stated in the question) that the socks will be chosen sequentially, one after the other. The question stated here is "if you pick two socks at random" (ie pick two socks at the same time). There is no inference of time delay.
If you calculate the number of combination : 2 among 12 black sockets + 2 among 12 white sockets And if you divide by 2 among 24 (all 2 sockets combination) You also find 11/23
the flaw is that socks should be assigned to one foot. the ones i buy are and they need to be as my big toe stretches one side. a matching pair need to have same colour and different foot.
Unless you grab the same sock with both hands (a 1/24 probability), then the explanation of the video is correct. Think of it this way. I grab a sock with my left hand, still in the drawer. I grab a distinct sock with my right hand. If I’m not grabbing the same sock, there remain 23 socks available for my right hand to grab. Of those 23, 11 match whatever sock I’m grabbing with my left hand. When I pull my hands out, the probability is 11/23 that they are a match. This works the same way if I grab first right, then left. Also simultaneously. Remember: I’m not grabbing the same sock with both hands.
this is straight forward Probability question, 24 socks in totals, you've picked X color now you can pick from other 11 of that color out of a total of 23 socks. i.e 11/23 = 0.47 I didnt know they had redefined PUZZLE in the dictionary to this
For a brief moment, I was happy that I solved it in a few seconds, but then it hit me, it's really not exactly something to write home about. Good puzzle though, always love the ones with probability, even though it's not my area of expertise.
Another way to think of this: Number all the socks. Black even, white odd. Or black low, white high. Whatever. If you pick two numbers at random, there's a 50/50 chance of getting an odd-even pair. The problem is that the pairs 1:1, 2:2, 3:3 and so on are not pairs you can actually take from the draw. These invalid pairs are all matching pairs, so there are fewer possible matching pairs than 50/50.
if you pull any 1 sock out of the drawer, it will be either black or white, and it doesn't matter which because in either case there'd be 11 of the same color out of 23 remaining socks, so the probability of pulling a matching pair is 11/23. EDIT: I have seen a couple of comments claiming that the probability is different if you choose them one after another vs if you choose two simultaneously, but it isn't any different, and I'd like to try to explain why: Using the combinations formula to find the number of matching pairs of each color (2 in 12) it's 12!/(12-2)! = 12!/10! = 12x11 = 132 (I skipped some steps to save space in the comment), and you'd multiply this by 2 because there are even numbers of both colors. The total number of combinations of two socks (2 in 24) is 24!/(24-2)! = 24!/22! = 24x23 = 552. if you divide 132*2 by 552, the result is ~0.478 or ~47.8%, which is the same as 11/23. I encourage you to go through and do the math yourselves and see if your results are any different.
If you think about it, what is the difference between a) grabbing two socks and b) grabbing a sock then another sock? Either way you've chosen two different socks at random. Who cares how you ended up with 2 socks? End result is the same.
@@wiggles7976 that's exactly the point I was trying to make. It doesn't matter practically speaking nor does it make a difference in the final result mathematically.
You can skip the whole adding the probabilities and dividing by two if you say the probability of picking a sock the same colour as the one you have is 11(12 of each colour minus the one you picked up)/23(total minus one)
Yes, this is a neater way to the solution. The probability that the first sock picked is acceptable is 100%. There are 11 of the same colout left, out of 23 socks remaining in the drawer. 11/23.
You could also do the following..... 1. Total no. of ways of choosing 2 socks = 24 C 2 = 276 2. Ways of choosing 2 white socks or two black socks=12 C 2+12 C 2 = 66+66 = 132 3. Therefore total probability = 132/276 = 0.4782 or 47.82% Note: nCr = n!/ (n-r)! r! A better option !? Lemme know! Love this channel for the standard of questions. Just excellent! 💯
When the problem was proposed, i was thinking about take two socks at the same time. Wouldn't the probabillity be higher if you take 2 instead of 1+1? Because you take WW or BB or BW/WB, so 2/3 of the cases are positive.
1) BW and WB are two different outcomes. 2) You are sampling without replacement; whether you sample at the same time or not, there are slightly more ways of picking a non-matching pair than of picking a matching pair if you take one sock out of 12 pairs.
I can either get hit by a car or not get hit by a car today. I better not leave the house because there is a 50% chance I'll get hit by a car. Just because you grouped BW and WB doesn't make them one "case". I want you to just stop and think for a second... do you actually believe the speed at which you take the socks out of the drawer affects the probability? Like if you quickly snatch two at once, your odds are different than slowly taking one after the other? This is just silly. How you remove the socks doesn't change the math.
The first question I had in mind after the problem was presented was: Am I picking socks one by one, or two of them at the same time? Because that changes the probability outcome.
@@stephenbennett3677 I agree the answer seems wrong, the question was about picking a pair of matching socks, shouldnt it be 50/50 twice ie you have a 25% chance
No it wouldn't. I could argue and say there's no way you'd pick them both at the exact same time, but even if you suppose you do, the answer doesn't change. Name the twelve black socks B1 , B2, B3 ... B12 and the 12 white socks W1, W2 ... W12. Consider all the possibilities there are: You could either pick (B1,W1) ,(B1,W2), (B1, W3) ... (B1 W12), (B1 B2), ... (B1 B12) in total there are 23 different ways you could pick B1. There are 22 different ways you could pick B2 without picking B1 21 ways to pick B3 without picking neither B1 nor B2... Etc, etc... The total amount of pairs you can make is (23+22+21+....+1) = 276. How many of these are matching pairs ? Well there are 11 matching pairs you can make that include B1, 10 matching pairs that include B2 without including B1, etc etc... In total there are (11+10+9+...+1)=66 ways to make a matching pair of black socks (and 66 ways to make a matching pair of white socks). That's a total of 132 matching pairs. The probability to get a matching pair is given by 132/276=11/23 which is the result that was found in the video.
Does waiting a few years in between picks change the probability? Nope. Time doesn't matter. At the same time, or one after the other. It is the same math. If I flip two coins at the same time, odds they match is 50%. If I flip one coin twice, odds it matches is 50%.
Thank you very much, after commenting i figured it had to be the same, but i didn't knew why, but yeah it was as simple as just counting all possibilities and then counting the matching pairs.
@Derek Gooding First of all just want to clarify i knew the answer before he mentioned it in the video, and my reasoning was basically the same as his; take one sock, and then calculate the probability of getting the same color, easy. But then i either had a brainfart or whatever, and i started thinking what if you get them at the same time, that way the idea of getting one and calculating the probability of the next disappears. Kinda like Schrodinger where you dont know what color the first one is so they dont influence each other (still brainfart). Just wanted to let you know because reading back it looks really dumb...
I believe there is a theorem or proof for the general case that shows that simultaneous is equivalent to serial -- ZP did a nice job on this unique case
No, Frank. Because once you pick a shoe (whether it be left or right, and white or black) there are then only 6 matching socks left in the drawer, instead of 11. Which means the probability is 6/23.
It's given in the original problem. 12 identical black socks and 12 identical white socks. This tells you you don't have to worry about the possibility of two socks of the same color still not being matching pairs. And rightly so - I have never experienced any difference between my left-foot and right-foot socks, they are practically identical.
Changed my mind several times...but if we create two piles of socks in the drawer, 6 black and 6 white in each pile, why shouldn't it be a 50/50 when drawing one sock from each pile? Same principle is valid for random piles. Another way of thinking is (1) grab first sock (2) grab second sock. If second sock has a probability of 11/23 for same color, what will happen with that probability if we release the first sock? Does probability jump up to 12/24?
Before attempting to answer, I would want clarification as to how the socks are arranged. Are they just tossed in at random? There's no reason to assume that would be the case, because most people don't store socks that way. If the socks are organized in some way, it would greatly change the odds of getting a matched pair.
Fair point but the question talks about picking 2 socks "at random". I suppose that is meant to be interpreted as the having a drawer full of "Laplace socks". ;-)
Well if it was an Amazon question for a warehouse position interview... maybe asking about how the socks were stored might be considered a positive inquiry, lol. But this was a written test, so basically you were probably not allowed any more information than what was printed on the page, so you had to interpret it literally as just "random."
This solution (47.8) is wrong. It assumes you "sample without replacement" one sock at a time. But when you PICK PAIRS you are sampling 2 at a time. Close your eyes, open the sock drawer and pick out 2 socks, one in each hand. That is an example of consecutive picking of a pair. The answer is 50 percent. Serial sampling (pick one sock and then pick another) is not explicitly stated in the question. In fact it says "pick a pair" which means 2 socks at the same time.
What if you pick two socks at exactly the same time? If you just sorted the socks blindfolded into 12 random pairs and then picked one of the pairs. Does that change the probability of picking a match?
sween187 still 11/23. As long as you're not using the information of the 1st pick to change the outcome of the second pick, it is equivalent to picking two socks out together.
What if I take one sock out, then wait a few years considering what could happen. Then I take a second sock out. I have better odds because the time between matters right?
The question implied that 2 socks are selceted simultaneously. As you would not know what either is until you have selected both, are the odds still the same, i.e. at the point imediately prior to selecting any, rather than after one is selected?
Is the probability different for grabbing two socks at the same time? As opposed to grabbing them in succession, one at a time? I figured there was three outcomes in that case you'd either grab two whites, two blacks, or one of each, therefore giving you a 2/3s chance of getting a pair
I keep the black socks in the left side of my draw, the white ones on right, even in dark or with sleepy eyes closed I always pick a correctly matching pair if wish with 100% probability. That said, what’s the chance that I can get a paid consultancy job with Amazon to help them optimise order fulfilment?
If the color didn't matter then why would there be a probability of picking either attached to the first sock chosen? Mind you I skipped around in the video and might have missed something.
Kanade Tachibana: The problem states that the socks are identical, except that some are black and some are white. Therefore, they don't get stained; at least, not by the time the pair of socks is selected. It doesn't matter whether there really are such perfect socks in the real world. You solve the problem within the world of the problem as described by the problem.
My guess is 11/23. You choose the first sock. There's 23 socks left: 12 of the other kind, and 11 of the same kind. So the probability of choosing the same kind on the second draw is 11/23.
Are the pairs of socks rolled or placed separately? What time of day am I picking the socks. You have questions. I have questions? Is the corner office taken?
This was a flawed question to begin with. If I was asked this I would respond with the need to know more facts. Am I picking 2 socks at once or one at a time? Are they ankle cut or crew cut socks? Are piled in the drawer all mixed up or are they organized neatly? Are they dress socks or thick wool socks with different texture? Did you fabric softener or starched?......
This was really easy. I instantly figured out that the 2nd sock you would take would have a probability of 11/23 of being the same color as the 1st one...
I think a better question would be: " Why on Earth would you want to work long hours in an un-air-conditioned Amazon sweatshop in India for $2.00 an hour ? "
Yes Presh, I figured it out but rather more simply. The first sock picked is always one half of a potential pair, whatever its colour. So that probability is 100%. Of the remaining 23 socks for the second pick, only 11 will match the first. So the probability of the second pick matching is simply 11/23.
Almost. The *2, though mathematically correct, isn't notated correctly. It indicates that you can have two white socks, or two black socks. The first part indicates the probability of picking one or the other. Therefore, you need a "two choose one" to start, indicating you are selecting either black or white. I.e. (2C1 * 12C2) / (24C2) = 11/23, or 47.8%
Why take 1/2 chance at first though? It doesn't matter wat sock you take first, as long as the second one matches the first one right, so should it not be 1/24*11/23 twice? Why not?
This is a test of your knowledge of probability and of how to analyze word problems. That stuff does matter to Amazon. They perform millions of transactions every few seconds, and the ability to present desired information in a faster timeframe with less computation is literally worth millions of dollars over time.
Stephen Hutchison - Well, no. Amazon computers do not perform MILLIONS of computations every few seconds. Amazon struggles to sell even 600 - 1,000 items per second on its very best days. Alibaba sells and packages FOUR times the volume that Amazon does.
Wow this puzzle is the hardest puzzle I've ever seen proofing the Riemann hypothesis must be so easy compared to this I tried solving it and after five weeks of skipping school, intense thinking and thanks to my 6th grade stochastic skills I came up with 11/23
I figured it would be a 48% chance without doing the math because I had this problem when I was in high school long before you were born and I only vaguely remembered the answer. Good sign I don;t have dementia. (Yet)
I came at the correct answer a different way. Ignore the colour of the first sock. Whatever it is, there are now 11 out of 23 socks of the same colour. Therefore chances are 11/23 you will pick a matching colour.
Gilad I disagree. It was easy, yes, and I got the right answer in less than 10 seconds, but still the problem itself was fair, well-presented, and not too convoluted. Thus, it was not a waste of my time and encouraged one to think critically, if even for just a simple problem
Although I got the correct answer almost instantly, here's a small spin that leads to a different answer. The twist is: your matching pair has to consist of a 'left' sock and a 'right' sock. 6/23
Except that could not be the case, bc it started by saying the socks were identical. That means, "exactly the same"....so, no there's no left or right to it.
Even in your twist example you still said "matching pair" which would still mean no "right" or "left" sock differences, lol. Just figured I'd point that out since it seems you mean socks where there is actually a left/right difference in your hypothetical example ;)
Ringo Garvin None of the probabilities is 25%. The video gives the probability of 2 black socks at 2:05. (1/2)(11/23). 23.91304%. The probability of 2 white is the same.
It would be close to that if they asked specifically for a black or a white pair. Since either is acceptable, add the probability of either outcome to get your answer.
Can you help me with this question? If I enter a contest that can have maximum of 4,790 total entries but I can only enter a maximum of 100 different entries, what are the odds of 1/4 of my entries placing in the top 200 out of 4,790?
This was very easy! If you get a sock what are the probabilities of getting one of the same colour? It is the first video of this channel I solved in less than 10 seconds...
the question isn't asked accurately enough. "Matching pair" might mean socks of same color AND opposite feet, because that's how they "match" if you wanna wear them. In this case, the question should be reformulated simply as "what is the probability of getting two socks of the same color?"
If you pick two at the same time, then you either get two white, two black or one of each. Two of the options are matching, and all three options are equally likely, so, does doing it this way increase your chances to 2/3?
I got the logical part correct... however I didn't understand how the probability of getting a pair of black/white socks is (1/2) x (11/23)... why are we multiplying and not adding?
Nope. Doesn't matter. Imagine if you had only two socks in the drawer, one black, one white. Odds would clearly be zero percent. Now imagine four socks, 2 black, 2 white. Odds are 1 on 3. If the number of socks of white and black are the same, the odds are always (number of socks - 1) in (number of socks * 2 -1). Doesn't matter if you pick them at the same time or some time apart.
I'm from the Czech Republic so I apologise for my English. I solved it different way. My probability was: how many right pairs there are/how many pairs there are overall. If one color has 12 socks then there are 1+2+3+4+5+6+7+8+9+10+11=66 pairs. There were 2 colors so 2*66=132 right pairs. Because there were 24 socks then there are 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23=276 pairs overall. 132/276=11/23
To me it's a paradox, however - I get the 11/23 explanation. But what if you pick blindfolded and simultaneously? Then I see it as: left hand white right hand black left hand white right hand white left hand black white hand white left hand black right hand black which gives 50/50. So the paradox is how could the timing (or for that matter knowing the first color) of the draw change this to 11/23? Edit: Thinking on it more I think I can see that even simultaneously it is NOT 50/50. If I think of it as a volume problem, assuming I do not grab the same sock with both hands, the volume of space in the bin or drawer occupied will be (I believe for both hands simultaneously) larger for the opposite color, in accordance with the probability. So the outcomes I listed above have to be weighted.
On a simultaneous pick, you'll either end up with two black or two white socks (you can summ them up as one "fulfilled" condition) or end up with one black and one white sock (in which case the condition is "unfulfilled"). So your condition could either be "fulfilled" or "unfulfilled". It's like if you pick one sock as your first pick, it will either be black or white. The chance is 50/50 to get one or the other, as long as you pick from a symetrical amount.
Wow in just about one week you guys have captioned this video in English, AND it's been translated into Arabic and German! Captioning is a lot of hard work--I know I've done a few--and I'm really appreciative. I read that TH-cam gives some rewards for doing captions, has anyone earned any worthwhile rewards yet? You guys are great, I have so many things to be thankful for this year!
Interesting! Actually, I provided both the english and the german version (the english subs because TH-cam wouldn't provide autogenerated english subs when I wanted to do the german translation... though, they could provide autogenerated once for english ¯\_(ツ)_/¯). I submitted them approximately 2 hours after you uploaded the video and they needed one week to verify them. I would be interested in how they did that? Do they simply ask random people how they find the subtitles?
The autogenerated once where actually quite good, and let's be honest: You speak so clearly one does not really need the subtitles for your videos :D
If you get me the script of your text. I'm get you Portuguese version of this. :)
Ok. Translated to Portuguese. Just waiting on someone else to review what I wrote. xD
Well you can not start by saying
it does not matter the first sock if it is black or white. Which will translate 24/24 to 1, But what matters in the second sock to make a pair and then you have 11/23 chance. So the total probability is 1 times 11/23 which is 11/23.
I think this way its easier to go by.
What if you pick two socks at once, in the exact same time. Is it still 11/23 ?
Applicants who got this right were put to work in the warehouse fulfilling sock orders.
lol
Applicants who inquired about how the socks were stored upon reading this question were put to work in the warehouse fulfilling sock orders, lol.
Too true. Lol
🤣🤣
Applicants who got this right were sent to China to match the socks before sent to USA.
This was so easy I was starting to get paranoid if maybe I'm missing something. By the way who with even the slightest knowledge of how propability works would assume it's 50%... Also I think the better and faster method of considering it is: The first sock you pick doesn't matter which colour is it, but the second one must be matching. Since we've already taken out one of this color sock there are 11 left out of 23. So it's 11/23
That's the way I did it.
that's how I though of it too
Exactly what I did, even the part about the paranoia.
It's not that people don't understand how probability works, it's often that people don't realise they should be applying their knowledge of how probability works, instead of going with the easy and intuitive answer. It's similar in nature to the bat and ball question, if you've heard of it.
Obiwanek this. I don't know why video tells you about chances on 1st sock as it's always 100%.
Answer is 0. I can tell from experience.
Yes, it's kinda like a USB plug. You have a 50/50 chance of getting it right, but invariably you get it on the third try.
This is the 50/90 law. If you have a 50% chance to get something right, you will get it wrong in 90% of the time.
The solution in the video is theoretically correct. In reality, not only will you always pick unmatched socks, but after the first time doing laundry, you will not have an even number of any single color of socks.
...WRONG WRONG WRONG...
The answer is,..
BUT "ONE" COLOR OF SOCKS. duh lol
MWSin1 12 pairs of socks is good for about 6 months. After that, i need new ones, but will occasionally wear old ones that i hadn't seen in years.
I did it a bit different. We were only asked to get a matching pair. So whatever is picked first, be it black or white, means that on the second pick we have to pick what we picked in the first pick. The odds of getting the second sock to match the first pick is 11 out of 23. That equals 47.8%. After posting this I saw others did it this way too.
This is exactly how I thought about it too. I immediately thought it didn't matter which color you picked first The only probability that mattered was the probability of picking another one of that same color which at that point would be reduced by one therefore 11/23.
I think most people used this method. The explanation in the video seemed to complicate it for no reason.
Good problem. It's easy for many of us that are used to thinking, but it's good for a lot of viewers to see the easier/basic stuff, too. So thanks for mixing up the difficulty level!
Wouldn’t it be easier to have the equation be 23 divided by 24, then divide by 2?
@@commandercaptain4664 No? I don't get the logic. Think about it like this: you must take either a black or a white sock. The odds that you then take a matching sock is 11/23
11/23 is the probability I think. If you pull any one sock, there's a 50/50 chance it will be black or white, but that doesn't matter because no matter which color you pull there'd be 11 of that same color left of 23 remaining socks.
I don,t mind if i wear black or white soks, so i pick 3 then i always have a pair in the same color.
Very smart! Unfortunately this does not work for me since I have multi colored socks. I guess I gotta throw them all out and only buy black and white socks :D
Just pick n+1 socks in which n is how many colours are there
S vdw *grabs mysterious grey sock*
Announcer: "A new challenger has entered the arena!"
pigeon hole!
just wear two different socks
Again, his explanation is far too complex. It should just be this:
After you pick the first sock there are 11 more socks of that color, and 12 of the opposite color. So the chances of picking another sock of the same color are 11 out of a total of 23, which equals approximately 47.8%. Done!
Why complicate things with the rest of that algebraic notation, confusing verbiage ("sampling without replacement"? shees!), separate consideration of the black sock case vs. the white sock case, etc. Keep it simple, I say.
Ken Haley
Yup
I agree he tends to overcomplicate explanation but that verbiage is in standard use. I mean obviously from the problem statement it makes more sense without replacement, but since it is never actually specified, you could assume sampling with replacement and achieve an entirely different but also correct answer. But really the simplest way to express it is just as a probability equation.
P(black and black or white and white) = 2*P(black and black) which can be simplified to 11/23 in your head in this case because of the simplicity
Matthew, if you replaced the first sock, then it's certain you'd end up with half a pair of socks of the same colour.
Yep. Regardless of which color you pick, there will always be 11 of that color left in the pile, of a total of 23 socks. The question is simple. The first grab has no significance on the outcome, only the second grab. Therefore the answer is 11/23.
Everyone is forgetting that this method only works if there is the same quantity of each colour. If there were 11 black socks and 12 white socks your "simple" method would not work
The answer is I don't pick any socks, I just buy a new pair on Amazon. "You're hired!"
I buy a used pair from Elon Musk. "You're fired!" 🤣
@@jakubkrcma Ewwww!
Then your socks may not match - "your fired"
You missed the trick to the question: the socks are identical, so there are no matching pairs. Your black or white socks will either be 2 left socks or 2 right socks, not a pair of 1 left sock and 1 right sock.
Probability of picking a matching pair is 100% because I match my socks when I do laundry and roll them together. Too easy >_>
Gregory Painter i put pairs of socks in the laundry and get an odd number back
Nathan Lang ah, that would be the laundry gnomes. Tricky little buggers make off with one on occasion. The only solution I've found is to sacrifice the other sock in tribute to appease them. Keeps the laundry safe for 2-3 months
My probability would be very low because I’ve lost one sock of almost all my pairs.
This is actually a good point that highlights the flaw in the question. You're still picking the socks at random, you've just changed the distribution. The question doesn't actually say what the distribution is, so hey. Good out of the box answer, now go sort those boxes.
For a matching pair you don't need to add a probability for Pr(bl) and Pr(wh). All you need to do is to calculate the 11/23. Since you can pick any color with your first pick - the question becomes "what is the probability that the next color is the same". Hence 11/23. Result is the same, it's just simpler.
good thinking 👍👍
this doesn't generalize to cases where the numbers of socks between different colors differs.
I seem to have an infinite number of socks in my drawer, so the probability is 50%
Rather 0% if they are all different pairs of socks.
As the number of socks approaches infinity, the odds of picking a pair approach 50%
Usually I don't get dressed in the dark, so the probability is 1
I have a large number of odd socks in my drawer. Getting up before it's light might have led to many odd-sock days! I folded the few matching pairs together when I put them in the drawer to avert this real life problem.
Yeah, amusingly like BlacksmithTWD said, if you had an infinite number of socks, there would actually be a 0% chance of picking any of them, however you’re right in that when you do pick your 0% probably socks, they will always have a 50% chance to be black or white.
At the Railroad we had a similar set of choices. To fix a broken rail at night you need what are called Angle Bars. These bars fit in the web of the rail and hold the bolts and rail to make a complete joint. Usually we would need 2 sets (4 bars) to cut out the defect and add a length of rail. The problem is that the shoulder of the bolts is oval and locks the blot from spinning when you tighten an 1 1/8in bolt. Most bars have the oblong hole to the right end of the bar then alternate oval, round, oval, round, oval, round. There are 6 holes, 3 round and 3 oblong in each bar. Once in a while you find and outlier and it has the oblong/oval hole on the left of the bar. When you put 2 oblong left bars together you make a good joint because every other bolt goes in from the oblong hole and comes out of the other bar through the round hole. So 2 Oblong right bars make a set and if you get an oblong left mixed in you can not make the joint safe because the oblong/oval holes line up with each other and only 3 of the bolts can be held fast. In the dark you always get an odd number of bars just incase you mix the oblong rights and oblong lefts. With an odd number you will always be
able to make a joint/pair.
Easy, 11/23. First sock can be anything then 23 socks left.
If first white, probability second white 11/23
If first black, probability second black 11/23
(ie 11/23 chance that the second sock is the same as the first)
Nicholas 4321 My exact reasoning. No need to go through the trouble of the 1/2 + 1/2 because the first draw doesn't matter if all you want is any matching pair.
Yep, same for me: color of first sock doesn't matter, then there are 11 "right" socks left among the remaining 23. Probability 11/23. Done. :-)
Well first sock and then only 11/23 chance of picking correctly on the second pick. Or a 47.8 percent chance. That's my simple ebonics math. :)
In the UK at the moment this would not be allowed as a white sock must be paired with a black, especially if the white ones are female.
the first sock doesn't matter anyway, only the probability for the second sock counts and that's your result.
Well said.
yeah his answer is wrong as the initial colour didn't matter, this isn't conditional probability.
Another way of saying this is: the real problem is the probability of picking a sock with a color equal to the color of 11 socks in the drawer, when there are 12 socks of the other color. But, then, that would be a give-a-way!
11/23 is 47,8% anyways.
daggle77 Exactly!
I'm trying to deduce if it would be 50% if you picked both at the same time one with each hand given that if you remove 2 at random simultaneously there is never one missing before drawing a second sock?
There are 2 possible scenarios:
First, you could unwittingly grab the same sock with both hands. There’s a 1/24 chance of that happening. As this wouldn’t satisfy the “pair” requirement, you’d try again.
Second, you grab two distinct socks, one with each hand. It doesn’t matter if the grab is simultaneous or not. The probability of pulling out a pair is still 11/23 as explained in the video. The reason is that your left hand grabs a sock, and since we’re not in the first scenario, there are only 23 socks for your right hand to grab. Of those 23, 11 are matching whatever color you hold in your left hand. Or vice-versa if you grab with your right hand first. Or even if you grab simultaneously. Not grabbing the same sock necessitates there being 23 available socks to see if it’s a match.
The answer is 10/23 [correction, 5/11, see comments below] because after you pick the first sock, one of the remaining matching socks escapes.
I swear my washing machine contains a portal to the oddsockniverse.
Still that would make the probab = 10/22.
you mean 5/11
Well.. according to the quantum sock entanglement theory anyway. Experimental results may vary.
Qaz Wsx Good catch! Yes, it would be 5/11, I didn’t think that through completely. I should spend more time double checking my work and less time admiring my humor before posting.
Yes, I figured it out. Total time < 2-seconds. I taught AP Statistics for 22-years and am confident that all of my former students would solve this problem, just as quickly.
There's a fundamental flaw in this solution - it makes an assumption (not stated in the question) that the socks will be chosen sequentially, one after the other. The question stated here is "if you pick two socks at random" (ie pick two socks at the same time). There is no inference of time delay.
If you calculate the number of combination : 2 among 12 black sockets + 2 among 12 white sockets
And if you divide by 2 among 24 (all 2 sockets combination)
You also find 11/23
Good lord, the number of people that think this makes a difference is worrying.
the flaw is that socks should be assigned to one foot. the ones i buy are and they need to be as my big toe stretches one side. a matching pair need to have same colour and different foot.
Riff
You may want to think a little longer and a little harder about this one…
Unless you grab the same sock with both hands (a 1/24 probability), then the explanation of the video is correct.
Think of it this way. I grab a sock with my left hand, still in the drawer. I grab a distinct sock with my right hand. If I’m not grabbing the same sock, there remain 23 socks available for my right hand to grab. Of those 23, 11 match whatever sock I’m grabbing with my left hand. When I pull my hands out, the probability is 11/23 that they are a match.
This works the same way if I grab first right, then left. Also simultaneously. Remember: I’m not grabbing the same sock with both hands.
this is straight forward Probability question, 24 socks in totals, you've picked X color now you can pick from other 11 of that color out of a total of 23 socks.
i.e 11/23 = 0.47
I didnt know they had redefined PUZZLE in the dictionary to this
The probability of picking a matched pair is 100% if you take 3 socks.
For a brief moment, I was happy that I solved it in a few seconds, but then it hit me, it's really not exactly something to write home about.
Good puzzle though, always love the ones with probability, even though it's not my area of expertise.
it may be an easy problem but it still helps you learn about calculating probabilities, which is great fun in my opinion
Another way to think of this:
Number all the socks. Black even, white odd. Or black low, white high. Whatever. If you pick two numbers at random, there's a 50/50 chance of getting an odd-even pair. The problem is that the pairs 1:1, 2:2, 3:3 and so on are not pairs you can actually take from the draw. These invalid pairs are all matching pairs, so there are fewer possible matching pairs than 50/50.
one of your easiest problems so far, 95% of people should get this right
if you pull any 1 sock out of the drawer, it will be either black or white, and it doesn't matter which because in either case there'd be 11 of the same color out of 23 remaining socks, so the probability of pulling a matching pair is 11/23.
EDIT: I have seen a couple of comments claiming that the probability is different if you choose them one after another vs if you choose two simultaneously, but it isn't any different, and I'd like to try to explain why:
Using the combinations formula to find the number of matching pairs of each color (2 in 12) it's 12!/(12-2)! = 12!/10! = 12x11 = 132 (I skipped some steps to save space in the comment), and you'd multiply this by 2 because there are even numbers of both colors.
The total number of combinations of two socks (2 in 24) is 24!/(24-2)! = 24!/22! = 24x23 = 552.
if you divide 132*2 by 552, the result is ~0.478 or ~47.8%, which is the same as 11/23. I encourage you to go through and do the math yourselves and see if your results are any different.
If you think about it, what is the difference between a) grabbing two socks and b) grabbing a sock then another sock? Either way you've chosen two different socks at random. Who cares how you ended up with 2 socks? End result is the same.
@@wiggles7976 that's exactly the point I was trying to make. It doesn't matter practically speaking nor does it make a difference in the final result mathematically.
100%. Because the drawer is mine and I bundle my pairs.
You can skip the whole adding the probabilities and dividing by two if you say the probability of picking a sock the same colour as the one you have is 11(12 of each colour minus the one you picked up)/23(total minus one)
Yes, this is a neater way to the solution. The probability that the first sock picked is acceptable is 100%. There are 11 of the same colout left, out of 23 socks remaining in the drawer. 11/23.
You could also do the following.....
1. Total no. of ways of choosing 2 socks = 24 C 2 = 276
2. Ways of choosing 2 white socks or two black socks=12 C 2+12 C 2
= 66+66 = 132
3. Therefore total probability = 132/276 = 0.4782 or 47.82%
Note: nCr = n!/ (n-r)! r!
A better option !? Lemme know! Love this channel for the standard of questions. Just excellent! 💯
When the problem was proposed, i was thinking about take two socks at the same time. Wouldn't the probabillity be higher if you take 2 instead of 1+1? Because you take WW or BB or BW/WB, so 2/3 of the cases are positive.
1) BW and WB are two different outcomes. 2) You are sampling without replacement; whether you sample at the same time or not, there are slightly more ways of picking a non-matching pair than of picking a matching pair if you take one sock out of 12 pairs.
I can either get hit by a car or not get hit by a car today. I better not leave the house because there is a 50% chance I'll get hit by a car.
Just because you grouped BW and WB doesn't make them one "case".
I want you to just stop and think for a second... do you actually believe the speed at which you take the socks out of the drawer affects the probability? Like if you quickly snatch two at once, your odds are different than slowly taking one after the other? This is just silly. How you remove the socks doesn't change the math.
I've come across your videos now & then, this is the first time I ever got the right answer (and for the right reason even!)
The first question I had in mind after the problem was presented was: Am I picking socks one by one, or two of them at the same time? Because that changes the probability outcome.
Exactly , I thought the same, how would you solve if its that way? i dread to think the solution , it will suck me into chaos.
I think it would be 50% exactly if picked simultaneously. Presh???
@@stephenbennett3677 I agree the answer seems wrong, the question was about picking a pair of matching socks, shouldnt it be 50/50 twice ie you have a 25% chance
@@ambrosiad1588 no since it doesnt matter if you get two white or two black
it doesnt change the probability
Thank goodness that somebody's testing this important problem
I`m seriously wondering if the answer would change if we picked both the socks at the same time
No it wouldn't. I could argue and say there's no way you'd pick them both at the exact same time, but even if you suppose you do, the answer doesn't change. Name the twelve black socks B1 , B2, B3 ... B12 and the 12 white socks W1, W2 ... W12. Consider all the possibilities there are:
You could either pick (B1,W1) ,(B1,W2), (B1, W3) ... (B1 W12), (B1 B2), ... (B1 B12) in total there are 23 different ways you could pick B1.
There are 22 different ways you could pick B2 without picking B1
21 ways to pick B3 without picking neither B1 nor B2...
Etc, etc... The total amount of pairs you can make is (23+22+21+....+1) = 276.
How many of these are matching pairs ?
Well there are 11 matching pairs you can make that include B1, 10 matching pairs that include B2 without including B1, etc etc...
In total there are (11+10+9+...+1)=66 ways to make a matching pair of black socks (and 66 ways to make a matching pair of white socks).
That's a total of 132 matching pairs.
The probability to get a matching pair is given by 132/276=11/23 which is the result that was found in the video.
Does waiting a few years in between picks change the probability? Nope. Time doesn't matter. At the same time, or one after the other. It is the same math.
If I flip two coins at the same time, odds they match is 50%. If I flip one coin twice, odds it matches is 50%.
Thank you very much, after commenting i figured it had to be the same, but i didn't knew why, but yeah it was as simple as just counting all possibilities and then counting the matching pairs.
@Derek Gooding
First of all just want to clarify i knew the answer before he mentioned it in the video, and my reasoning was basically the same as his; take one sock, and then calculate the probability of getting the same color, easy.
But then i either had a brainfart or whatever, and i started thinking what if you get them at the same time, that way the idea of getting one and calculating the probability of the next disappears. Kinda like Schrodinger where you dont know what color the first one is so they dont influence each other (still brainfart). Just wanted to let you know because reading back it looks really dumb...
I believe there is a theorem or proof for the general case that shows that simultaneous is equivalent to serial -- ZP did a nice job on this unique case
YAY! Another puzzle in dark mode! And a fun little explanation of probability too!
And how do you know that there is not two left foot socks
I suggested this alternate version where you have to get a matching pair that consists of a left and a right sock.
It's not that much harder though.
Isn't it the exact same then?
No, Frank.
Because once you pick a shoe (whether it be left or right, and white or black) there are then only 6 matching socks left in the drawer, instead of 11. Which means the probability is 6/23.
It's given in the original problem. 12 identical black socks and 12 identical white socks. This tells you you don't have to worry about the possibility of two socks of the same color still not being matching pairs. And rightly so - I have never experienced any difference between my left-foot and right-foot socks, they are practically identical.
Patrik Szabó - That is not true for everybody. For me, my toes angle pretty sharply from big toe down, so the ends of my socks have diagonals in them.
Changed my mind several times...but if we create two piles of socks in the drawer, 6 black and 6 white in each pile, why shouldn't it be a 50/50 when drawing one sock from each pile? Same principle is valid for random piles. Another way of thinking is (1) grab first sock (2) grab second sock. If second sock has a probability of 11/23 for same color, what will happen with that probability if we release the first sock? Does probability jump up to 12/24?
Before attempting to answer, I would want clarification as to how the socks are arranged. Are they just tossed in at random? There's no reason to assume that would be the case, because most people don't store socks that way. If the socks are organized in some way, it would greatly change the odds of getting a matched pair.
Fair point but the question talks about picking 2 socks "at random". I suppose that is meant to be interpreted as the having a drawer full of "Laplace socks". ;-)
Well if it was an Amazon question for a warehouse position interview... maybe asking about how the socks were stored might be considered a positive inquiry, lol. But this was a written test, so basically you were probably not allowed any more information than what was printed on the page, so you had to interpret it literally as just "random."
Sock death leaping from the basket is so dreadful.
You clearly don't get the concept of 'abstraction'
My first thought was that he didn't specify whether we were grabbing 2 at a time or one at a time. Totally changes everything. :)
This solution (47.8) is wrong. It assumes you "sample without replacement" one sock at a time. But when you PICK PAIRS you are sampling 2 at a time. Close your eyes, open the sock drawer and pick out 2 socks, one in each hand. That is an example of consecutive picking of a pair. The answer is 50 percent. Serial sampling (pick one sock and then pick another) is not explicitly stated in the question. In fact it says "pick a pair" which means 2 socks at the same time.
jorgensenmj I hope you were trying to troll
You are wrong . Try this for real with 2 pair of sock and you will be amazed.
I pick two socks at random and get two pairs 100% of the time, since I fold matching pairs together in my drawer.
What if you pick two socks at exactly the same time? If you just sorted the socks blindfolded into 12 random pairs and then picked one of the pairs. Does that change the probability of picking a match?
what if you draw the socks out at the same time, one with your left hand and one with your right?
sween187 still 11/23. As long as you're not using the information of the 1st pick to change the outcome of the second pick, it is equivalent to picking two socks out together.
Shrey Rupani you're making it too complicated. You're picking 2/24 socks. Not 1/24, then 1/23.
You're only making one selection of 2 socks in this case.
What if I take one sock out, then wait a few years considering what could happen. Then I take a second sock out. I have better odds because the time between matters right?
Derek Gooding the fact that time passes matters. The length of time doesn't. 1/24, then 1/23 Is different than picking 2/24
Also, how many socks would you have to take out of the drawer before you get a matching pair?
11/23 cuz, first sock choose next sock color, then its 11 of same left of a total of 23
markus dahle
Somehow after reading your comment I don't understand it anymore
The question implied that 2 socks are selceted simultaneously. As you would not know what either is until you have selected both, are the odds still the same, i.e. at the point imediately prior to selecting any, rather than after one is selected?
This was a very easy questions. I'm 14 years old and was able to solve this problem in just 20 seconds using permutations.
If you'd just thought "11 same 12 different" you'd probably have solved it in 3 seconds. It's good to be smart but it's smart to find an easier way.
Oh ya I'm only 7 years old and was able to solve the problem in just about 4.28 seconds using permutations.
Permutations? Fail
Smart kid
Is the probability different for grabbing two socks at the same time? As opposed to grabbing them in succession, one at a time? I figured there was three outcomes in that case you'd either grab two whites, two blacks, or one of each, therefore giving you a 2/3s chance of getting a pair
The insightful part of the discussion is that you are guaranteed to get a matching pair of socks if you pick 3.
I keep the black socks in the left side of my draw, the white ones on right, even in dark or with sleepy eyes closed I always pick a correctly matching pair if wish with 100% probability. That said, what’s the chance that I can get a paid consultancy job with Amazon to help them optimise order fulfilment?
i was gonna say 11/23, but out of respect for today's date, i'm gonna say 11/26
This is a true thinker here. Amazon will now offer you a job.
Good man.
This was one of the easiest questions posted, I want to see more problems like the avg distant between two dots in a unit square
What kind of a Savage fills their drawer with mis-matched socks?
I like the way you think, you're hired.
If the color didn't matter then why would there be a probability of picking either attached to the first sock chosen? Mind you I skipped around in the video and might have missed something.
What about uneven number of holes in both sets???
Again, the problem stated the socks are identical except for color.
I just don't get it, how about stains ? They indeed change color.
Kanade Tachibana: The problem states that the socks are identical, except that some are black and some are white. Therefore, they don't get stained; at least, not by the time the pair of socks is selected. It doesn't matter whether there really are such perfect socks in the real world. You solve the problem within the world of the problem as described by the problem.
Very easy...you just lowered the bars of questions asked in Amazon...I thought it would be a challenging one
My guess is 11/23. You choose the first sock. There's 23 socks left: 12 of the other kind, and 11 of the same kind. So the probability of choosing the same kind on the second draw is 11/23.
Kyle Amoroso its not a guess, its based on your calculation
Are the pairs of socks rolled or placed separately? What time of day am I picking the socks. You have questions. I have questions? Is the corner office taken?
Very ill phrased. First "picking" then "drawing" I could pick two at a time, yet we hear of the first and second draw.
This is why mathematics and linguistics don’t mix.
Where does it say you have to pick one at a time? If I use both hands and take out two at the same time then it's 50% yes?
This was a flawed question to begin with. If I was asked this I would respond with the need to know more facts. Am I picking 2 socks at once or one at a time? Are they ankle cut or crew cut socks? Are piled in the drawer all mixed up or are they organized neatly? Are they dress socks or thick wool socks with different texture? Did you fabric softener or starched?......
This was really easy. I instantly figured out that the 2nd sock you would take would have a probability of 11/23 of being the same color as the 1st one...
I think a better question would be: " Why on Earth would you want to work long hours in an un-air-conditioned Amazon sweatshop in India for $2.00 an hour ? "
Easy, if the only alternatives are to work even more hours in less favorable conditions for even lower pay per hour.
If I got $2 an hour in india for that job, I'd quit my job and join Amazon...
Yes Presh, I figured it out but rather more simply. The first sock picked is always one half of a potential pair, whatever its colour. So that probability is 100%. Of the remaining 23 socks for the second pick, only 11 will match the first. So the probability of the second pick matching is simply 11/23.
(12C2)/(24C2) *2=47.8%
Almost. The *2, though mathematically correct, isn't notated correctly. It indicates that you can have two white socks, or two black socks. The first part indicates the probability of picking one or the other. Therefore, you need a "two choose one" to start, indicating you are selecting either black or white.
I.e. (2C1 * 12C2) / (24C2) = 11/23, or 47.8%
Why take 1/2 chance at first though? It doesn't matter wat sock you take first, as long as the second one matches the first one right, so should it not be 1/24*11/23 twice? Why not?
How could this possibly matter in real life? This is your hiring question? Not like they are gonna be mixing senseitive chemicals.
This is a test of your knowledge of probability and of how to analyze word problems. That stuff does matter to Amazon. They perform millions of transactions every few seconds, and the ability to present desired information in a faster timeframe with less computation is literally worth millions of dollars over time.
Stephen Hutchison - Well, no. Amazon computers do not perform MILLIONS of computations every few seconds. Amazon struggles to sell even 600 - 1,000 items per second on its very best days. Alibaba sells and packages FOUR times the volume that Amazon does.
Doesn't this assume taking 1 sock at at time? If you pull both socks at the same time, simultaneously, does the probability shift?
No.
Wow this puzzle is the hardest puzzle I've ever seen proofing the Riemann hypothesis must be so easy compared to this I tried solving it and after five weeks of skipping school, intense thinking and thanks to my 6th grade stochastic skills I came up with 11/23
I figured it would be a 48% chance without doing the math because I had this problem when I was in high school long before you were born and I only vaguely remembered the answer. Good sign I don;t have dementia. (Yet)
I came at the correct answer a different way. Ignore the colour of the first sock. Whatever it is, there are now 11 out of 23 socks of the same colour. Therefore chances are 11/23 you will pick a matching colour.
what if you grab a pair at the same time, ie both socks in one grab, does it alter the result?
Too easy 🙄
Gilad I disagree. It was easy, yes, and I got the right answer in less than 10 seconds, but still the problem itself was fair, well-presented, and not too convoluted. Thus, it was not a waste of my time and encouraged one to think critically, if even for just a simple problem
I don’t understand why you add the probability of a second matching pair when the question only asks for the first probability. Can you clarify?
Although I got the correct answer almost instantly, here's a small spin that leads to a different answer. The twist is: your matching pair has to consist of a 'left' sock and a 'right' sock.
6/23
mrBorkD
I think that would be the pair-of-gloves problem or perhaps the pair-of-shoes problem.
That'd be a nice things to make it more intuitive, thanks!
Except that could not be the case, bc it started by saying the socks were identical. That means, "exactly the same"....so, no there's no left or right to it.
Even in your twist example you still said "matching pair" which would still mean no "right" or "left" sock differences, lol. Just figured I'd point that out since it seems you mean socks where there is actually a left/right difference in your hypothetical example ;)
Lol, I thought it was 25%
Ringo Garvin
None of the probabilities is 25%. The video gives the probability of 2 black socks at 2:05. (1/2)(11/23). 23.91304%. The probability of 2 white is the same.
It would be close to that if they asked specifically for a black or a white pair. Since either is acceptable, add the probability of either outcome to get your answer.
in a drawer of infinite socks where the first pick matters you are correct.
Given my gf's sock drawer that isn't a crazy assumption
Can you help me with this question?
If I enter a contest that can have maximum of 4,790 total entries but I can only enter a maximum of 100 different entries, what are the odds of 1/4 of my entries placing in the top
200 out of 4,790?
Yep - sampling without replacement - the answer makes sense. Good video!
This was very easy! If you get a sock what are the probabilities of getting one of the same colour? It is the first video of this channel I solved in less than 10 seconds...
Would there be a difference if the socks are picked from the drawer at the same time? Would the probability be 50%?
What if you were to pick them both SIMULTANEOUSLY- without stopping to recalculate the odds in between? Would it not revert to 50/50 again?
What :Dd
What if you grab both socks at the same time?
As soon as you started talking it became really obvious.
Thanks for sharing this.
The question did not state that you take one at a time. If you take both at the same time, would it not still be 1:2?
Applying Hypergeometric method gives the answer 12/23. Can you please explain why?
the question isn't asked accurately enough. "Matching pair" might mean socks of same color AND opposite feet, because that's how they "match" if you wanna wear them.
In this case, the question should be reformulated simply as "what is the probability of getting two socks of the same color?"
5/23
I was already a follower of Presh back when he still made videos on math *and* game theory :D
Surely the probability of choosing a matching pair (if the color does not matter) is 11/23, or how have I got wrong?
If you pick two at the same time, then you either get two white, two black or one of each. Two of the options are matching, and all three options are equally likely, so, does doing it this way increase your chances to 2/3?
Mario Morales
The probabilities aren't equally likely. They are 11/46, 11/46 and 12/23.
I got the logical part correct... however I didn't understand how the probability of getting a pair of black/white socks is (1/2) x (11/23)... why are we multiplying and not adding?
If you draw both sock simultaneously it changes the outcome right? It would go back to 50%
What if you grab two socks at one time instead of one then the other? Would it still be considered 11\23???
What about if you didnt pick one sock at a time ? And you picked one sock with each hand at the exact same time ?
would it change anything for the math if you had to pick two socks at once?
What if I stick my hands in the draw and select two random socks at the same time?
I FIGURED IT OUT IN SECONDS. SUPER EASY.
So does this apply if we pick 2 socks directly from the drawer instead of getting one at a time ?
Nope. Doesn't matter. Imagine if you had only two socks in the drawer, one black, one white. Odds would clearly be zero percent. Now imagine four socks, 2 black, 2 white. Odds are 1 on 3. If the number of socks of white and black are the same, the odds are always (number of socks - 1) in (number of socks * 2 -1). Doesn't matter if you pick them at the same time or some time apart.
I'm from the Czech Republic so I apologise for my English. I solved it different way. My probability was: how many right pairs there are/how many pairs there are overall. If one color has 12 socks then there are 1+2+3+4+5+6+7+8+9+10+11=66 pairs. There were 2 colors so 2*66=132 right pairs. Because there were 24 socks then there are 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23=276 pairs overall. 132/276=11/23
To me it's a paradox, however -
I get the 11/23 explanation. But what if you pick blindfolded and simultaneously? Then I see it as: left hand white right hand black
left hand white right hand white
left hand black white hand white
left hand black right hand black
which gives 50/50.
So the paradox is how could the timing (or for that matter knowing the first color) of the draw change this to 11/23?
Edit: Thinking on it more I think I can see that even simultaneously it is NOT 50/50. If I think of it as a volume problem, assuming I do not grab the same sock with both hands, the volume of space in the bin or drawer occupied will be (I believe for both hands simultaneously) larger for the opposite color, in accordance with the probability. So the outcomes I listed above have to be weighted.
On a simultaneous pick, you'll either end up with two black or two white socks (you can summ them up as one "fulfilled" condition) or end up with one black and one white sock (in which case the condition is "unfulfilled"). So your condition could either be "fulfilled" or "unfulfilled".
It's like if you pick one sock as your first pick, it will either be black or white. The chance is 50/50 to get one or the other, as long as you pick from a symetrical amount.
They made that way harder than needed. Doesn’t matter what color sock is picked first, you have a 11/23 chance of matching the color. That’s 47.8%