Three 6s and three 0s could work. The labeled die already goes from 1 to 6 with equal odds, so, if you have a coinflip that adds 0, or 6, it would produce all twelve of the numbers with equal odds.
Yeah the original answer doesn't include one as a possible outcome, only from two up to twelve. I assume this is why one didn't get produced by the formula
Since we need a non zero chance to roll 1 and 12 There is at least one side reserved for 0 and 6 By intuition I try 000666 If you roll a 0 with the blank die: even distribution over 1-6 Roll a 6 with the blank die: even distribution over 7-12
I think at some point you need to identify that each total must have three ways to arrive at, then there must be three sixes and three zeroes, there is no other combination that works.
Your intuition is better than mine. I started at I must have a 0 to get to 1. My intuition went for 000006 Then quick calculation gave me that generates 5 1's and only 1 12. That's no good. Then my intuition went to 000666 That gives me 3 1's and 3 12's but does it give me 3 of 2 through 11? Yes because I have 1 through 6 with 1/6 chance with +6 or 7 through 12 50% of the time :-) Solved and checked :-)
Somebody given that answer in comments, which I find brilliant: Three 6s and three 0s could work. The labeled die already goes from 1 to 6 with equal odds, so, if you have a coinflip that adds 0, or 6, it would produce all twelve of the numbers with equal odds.
Finally a problem that I could figure out without the explanation! Your videos have helped me grow in mathematical skill so much, even though you often cover concepts I don't understand due to being less advanced in classes
Maybe one could just use 0 and 6...and that's it, lol...0 and 6...3 times...3 zeros, 3 sixes, that's 6 sides...0 is necessary to get 1, giving us 1 to 6, and 6 is necessary to get everything from 7 to 12, each of those appears exactly once...and since they don't say repetition is forbidden, we can just use them several times in order to get exactly the same number of occurrences, which is 3 for each possibility, so each of the 12 numbers can appear in 3 ways...
After watching this video I really wanted to try to solve another problem using this method. I know it would be hard, but I suggest you adding an extra problem for us to practice what we've learned (of course not in all videos, as it may be impossible). Keep up the good work! 🙂
Ha...you'll have to have a 0, lol, for starters, because otherwise you won't get 1...and the moment you have 1, you will have 1 to 6 in exactly one way, etc...will it be kept at 1, or do we have to increase it?...I suspect we'll have to increase it, lol...
I looked at the thumbnail, digested it, and then kept scrolling down as a I thought about it. I got the answer in about 15 seconds. I’m actually really happy about myself, usually the puzzles I see on your channel are really difficult for me.
A small mistake I noticed - at around 6:41 you say "so we're going to multiply this by one half, which is also equal to one half of x times zero" - this should be "also equal to one half of x to the power of zero". It's written correctly in the video, and spoken correctly later in the video around 6:55 as "one half multiplied by x to the power of zero" Other than that, great video!
Glad to know someone is paying attention! So this kind of video is always a big risk for me, since it's hard to communicate probability generating functions in a TH-cam video. If the video doesn't resonate quickly with enough people, it naturally won't get recommended through TH-cam's system. Even though my funding is 1% of some education channels, I still take more risks than them to make this kind of video on probability because teaching and education are more important to me than money. It's the ultimate validation when some videos are cited in academic papers, and glad to know they inspire many students to pursue learning these topics.
I'm a Stat student and I studied Generating functions in last sem. Oof. Never thought they had such interesting applications. I should study them once again. Thank you very much.
Andrew Guthrie you need 12 to appear three times and 1 to appear three times. That means three 0’s and three 6’s are needed, otherwise it’s impossible. And like magic, it works for 2-11 too
I love watching your explanations through to the end. Even though I got the answer almost immediately in my head, it was due to simply immediately realizing I needed a 0 and a 6 at the start, and that those two alone gave me all sums equally. I love seeing how you structure the path to the solutions, since I'm usually far too lazy to actually come up with such a plan myself :D
I think this is the second one I actually decided to solve and succeeded! Good thing I'm studying probabilities in my free time. Thank you, Presh Talwalkar!
This can be simply done 1. The only way we get the sum to be one is 0+1 2. So every number should be expressed as sum in only one way to make the probability same 3. 0+1,0+2,0+3,0+4,0+5,0+6,1+6,2+6,3+6,4+6,5+6,6+6
I have been experimenting with dice lately and found really interesting probabilities if you put pips on the corners instead of the faces. I wish there were games that take advantage of weighted randomness like this, so many do not even seem to take account of the Gaussian distribution of two or more d6.
Roll a die (1,..,6), then flip a coin: if heads we are in the original (1,..,6) range, if tails we are in the (7,..,12) range. To get to the (7,..,12) range from the die, we simply offset/shift the die result with +6. No shift, +0, to the die result is needed for the lower range. The coin flip is identical to a die roll with equal probability of a +0 and +6 shift which means that the mystery die should have equal probability between the two shifts (3 sides of each since 3/6=1/2).
You can also solve this problems using convolutions and understanding that the sum of probabilities are the convolution of the probability function. If you don't know how convolution works it would ve quite complex, but knowing how it works it is not so dificult
This truly is the best maths channel on youtube, or the just the best, for that matter Also, probability is pretty cool, so I'd like to see more videos on that
I found this fun because I didn’t know the answer until I drew the blank probability table. I was puzzled because I knew the probability result line for two standard dice shows a “bulge” of density in the middle, and visually, the problem was to spread that out and make it uniform. I then entered the three zeroes, using the same logic as the video (only way to get three chances out of 36 of there being a one). I then recognised that if I put any other numbers in except six there would be be greater than three chances of of a number. Six when combined with any of the numbers on the standard dice would not duplicate any number. That is when I realised the whole “trick” was to avoid the troublesome duplications, and 0 and 6 did that. I smiled at the now “obvious” result. Very similar logic to the video, but in a slightly different order. For me it was a nice illustration how seemingly difficult problems often are simple once you start to lay them out.
Dies are basically modulus operations (which means they cycle every N numbers, for N = number of sides). With that information you can get to the first solution a lot faster than with a table. From 1 to 12 there are two of these 6-sized cycles, so you need two numbers, one to run over the first 6 numbers (that are already in the first die) and one to run over the next 6 numbers. To run over the first 6 you need 0 as base, as they already exist, to shift to the next cycle you need 6 (N) as base. Half 0, half 6. It is very intuitive if you are familiar with the intuitions of modular maths. Works with any size dies as long as both dies are the same size.
I think it was more to showcase the method as a general tool. Which I appreciate. Think you need to do this with over 100 possible inputs and outcomes, even more so if all the other options (there can even be more than two) are convoluted enough. That's when you do it the complicated way, because it's systematic enough you can force a computer to do the work for you quite easily with basically any symbolic computation tool. And such situations do happen (i.e. in video games or probabilistic modelling).
No it's not. It's actually surprisingly simple. Like look at his other videos, it's more complicated.(but simple enough to be understood assuming you know the basic tricks of the problem).
There's another solution to this problem which is to label all 6 sides of the mystery die with a 12 (or any other number >=12). The probabilities of rolling all the numbers 1-12 are 0, and equal. Not as satisfying as the 000666 solution, but I would definitely mention this if I were in the interview because it's clever and super obvious once you realize it. :)
@Manas Ghai, - "The numbers which can be used for labelling have to be whole numbers less than or equal to 6." ... and also must be non-negative. In other words, the only allowable numbers for labeling are 0, 1, 2, 3, 4, 5 and 6. (Just mentioning it for the sake of completeness.) However, it wasn't specified that you can use only one number on each face, and we can label a face with 12 by using the numbers 1 and 2, so it's possible to label a face with "12". (But it's not possible to label a face with, say, "17", because we cannot use a 7.)
I usually solve problems this way: It has to be really complicated with lots of things to remember for me to do otherwise. I took me about 15 seconds, so yeah... great achievement solving that in mind with no pencil nor paper. ( Forgive me if I seem sarcastic ) :D
This is a case where a graphical method -- drawing up and labelling the full table of all possible outcomes, and showing that each value from 1 to 12 occurs with equal frequency -- is quicker than an algebraic method. But what if we were using a pair of icosahedral dice? Now we would have to think about forty from four hundred possible outcomes, rather than a dozen from thirty-six, and a formula begins to look the more attractive proposition. The point is, the formula is really just the instructions for filling in the table.
@bluerizlagirl, The puzzle was relatively simple and quick to solve, because the required probability distribution of the dice pair's total was uniform (each of the outcomes 1 through 12 has equal probability). The "size" of the dice (six-sided or 20-sided) doesn't really matter to its difficulty. Had the required probability distribution been non-uniform, then the "probability generating functions" formula would have been quicker and easier, even for a pair of just six-sided dice.
@@bluerizlagirl thats my point. People say it is easy because they are told a solution exists and second because 6 possibles cases are too few. The PGF approach settles all those issues at at once.
You didn't explain 4:55 well at all. What is x supposed to be and why are you raising it to the power of 2, 3 etc? Also, since Z is a discrete random variable defined by starting at 1, p of i is always equal to 1/y where y would be the max number (in this case 6) so p of i is always 1/6 for each Z.
Making my guess at 1:48! Three sides of the die get 0, the other 3 get a 6. Figured this out by constructing and frequency table - sample space hybrid.
I came up with the three 0's and three 6's solution pretty quickly. Apparently a lot of people did as well. The first thing I noticed was if you are going to get 1 you need a zero and if you are going to get a 12 you need a six. Then I realized that you were only going to hit 12 one in 36 times so you needed three sixes and similar logic worked for one as well and it all seemed to work when I thought about the different numbers. This seems to be the only possible solution. I am going to watch the video now since from the comments it seems like there is another solution? Or another explanation? ETA: There was no trick that I missed. I got the right solution. That was nice and a little boost to my ego after being humbled by watching somebody solving the normal distribution integral. Alas, most of the video involves a complicated way of getting the answer which I didn't understand but I was a lot closer to understanding it than I am to being able to solve normal distribution integral.
Gee, that's a simple puzzle. Why does he need nearly 8 minutes to explain the solution? Let's see his answer... oooh! (Never knew stuff like this could be generalized with polynomials!) On a different note: So, you've reconsidered from using the term "dice" for a singular die, then? :-)
@DeadFish37, - "Me: skips to the end to check I got it right What's he messing with polynomials for?" Presumably to introduce us (or at least those of us with a deeper interest into mathematics) to a more abstract and relatively unknown branch of methods that combines probability functions with algebra. The relatively simple dice riddle is merely the hook to reel us in.
This was surprisingly simple once I realized. There are 36 possible combinations of faces, and since the 12 totals are represented evenly there have to be three combinations for each total. The only way to roll a total of 1 is with 1 on the normal die and 0 on the blank one, so there have to be three faces labeled 0 on the blank die. Likewise, the only way to total 12 is 6 on the normal die and 6 on the blank, so the other three faces on the blank all must be 6. With that you’ve used up all the blank faces, and from there it’s trivial to show that the probabilities are all equal.
Well, there are 36 possibilities using 2 6-sided dice. This means each number from 1 through 12 needs to be summed to 3 times. This requires the second die to have 3 zeros and 3 sixes. This doesn't require much thinking.
I solved it in the head, no need of equations. Just think about it: you already have 1-6 in one die, and to get 7-12 you just need to add 6. So to make it equal chance, have 3 of the sides 0 and the other 3 as 6.
Oh my... After understanding the requested outcome, I had the answer in 5 seconds. I never would've guessed the mathematical approach would be so much harder
If you let the other die stay as a normal die you can take that die's outcome, subtract one and multiply it by six (number of outcomes on the other die) and get an even distribution for 1-36.
I did this in my head though~ (To get 1,2,3,4,5,6 we add a 0 to the new die. To get 7,8,9,10,11,12 we add a 6 to the new die. This gives the P(X=1,2..12) as 1/6. If we fill the new die with 3 0s and 3 6s we get P(X=1,2..12) as 1/3. This satisfies our requirement)
Wow, this relates directly to coding much more than most of their questions. Except perhaps biased toward D&D players, who understand this kind of modulo intuitively. Sometimes you want to get a random hour of the day in a TTRPG, and make a d24 this way.
Having a 50% chance to add 6 to your roll gives you equal chances for everything from 1 to 12, dungeons and dragons actually being useful for something, wow
It's exactly "magic" in that you need to "know" that this approach is going to work to consider trying it. If the table didn't have such complete clues, he would've just wasted time or ended up with a partial solution.
for 2 6-sides dice there are a total of 36 outcomes of equal probability, if we want to map 36 outcomes to 12 outcomes with equal probabilities then it must be that 3 of the 36 outcomes map to 1 of the 12 outcomes. one of the dice is given to be a normal dice. what outcomes map to a sum of 1? well we must roll a 1 on the normal dice and a 0 on the other, since this must be able to happen in 3 of the 36 outcomes, there must be 3 0s on the non-normal die. what outcomes map to a sum of 12? well we must roll a 6 on the normal dice and a 6 on the other, since this must be able to happen in 3 of the 36 outcomes, there must be 3 6s on the non-normal die. simple checking reveals that having the second dice have 3 0s and 3 6s does indeed solve the problem.
thought process: 1. 2 dice = 36 combinations, for 1 to 12 to be equally likely, I need 3 of the combinations to get 1, so that's 3 sides with 0, similarly, I need 3 sides with 6 to get 12 with 1/12 chance, so 3 0's and 3 6's are the only combination that can work. 2. Yes it does work. if I roll a 0 on the blank die, it is equally likely to be 1-6, if I roll a 6 on the blank die, it is equally likely to be 7-12
The problem is easy because you are told such a solution exists. The right way to settle the issue is via the probability generating function. Imagine you had a dice with 100 faces, how would you go about it with the former method?
We could also use a six sided die and a coin, valuing heads as six and tails as zero. One side of the coin gives all the low numbers (with equal probability) and the other gives all the high numbers. Having spotted this intuitively, if you now insist on a die with six faces as the second die, this is synthesised by numbering three faces as six and three as zero.
I really couldn't understand what the question wanted. I was wondering how it would be possible for the sum of the 5 faces showing to possibly do 1-12. I took the drawing of the dice as a literal point of view for the problem. And by roll, I thought they meant rotating on the same axis for some reason.
Im 14 so let me explain how I solved this: So, since its 1 through 12, we know we need at least one 0 so that we can have a sum of 1 (1+0). This zero, if added to 1 through 6 gives you 1 through 6, but we still need 7 through 12. To get this, we can add something else to 1 through 6 to get this (remember the probability to roll all numbers has to be the same) so we would need a 6 to add to 1 through 6 to get 7 through 12. Now we know that, if we have a 0 and a 6, or the same amount of 0's and 6's, we will have the same probability from 1 through 12. This means, to fill up the die, we need three 0's and three 6's. Edit: note that 0 and 6 are being added to 1 through 6 because those are the numbers on the other die.
It's not clear from the problem description that you can label a face with zero. Normal dice don't have a "zero" face, and a "zero" face on a normal die would appear blank, which would seem to violate the requirement that we need to label all the faces. The example (1, 2, 3, 4, 4, 5) should have included a zero.
I just drew a 6x6 state-space matrix. The fixed die (1..6) labels the columns, and the variable die-faces label the rows. If every result (1..12) is equiprobable, then each sum must have 36/12 = 3 cells in the matrix. The first problem is "How to get 3 cells having the row-col sum = 1?" This implies that 3 faces must be labeled with 0s. This takes care of filling the top 3 rows. The bottom 3 rows? You can see that these need a 6 on the die face to obtain 3 each of the (7..12) cells. Answer: 0,0,0,6,6,6
A single die has a uniform distribution from 1 to 6. You can create a uniform distribution from 1 to 12 by either adding 0 or 6 with equal probability. That's how I reasoned through this problem.
36 different two dice rolls, so each of twelve numbers must come up three times. The only way to roll a one is with a one and a zero, so the second die has three zeros. Then each other number on the first dice is rolled three ways with the zeros, so no number less than six is on the remaining spaces of the second die. That leaves six to be filled in on those three spaces, and that works, seven through twelve can be rolled three ways.
I think it is 3 0's and 3 6's. You have an equal chance of getting a 0 or a 6 on the custom die. This gives you an equal chance of getting on the low side (1-6) or highs (7-12). Then the natural die has an equal chance on any result.
This was actually fairly trivial. In order to get 1, you need a face of 0 (only 1+0 can equal 0). For 12, you need a 6 (only 6+6 can equal 12). No other numbers can be added, since they would not cover either 1 or 12, and the outcomes have to be equally likely - ergo the mystery die must be 0, 0, 0, 6, 6, 6
3 0s and 3 6s. There are 36 possible arrangements with 2 die => 36/12 means each number from 1 to 12 needs to show up 3 times. If we choose a 0 on the blank, the 6 possible results of both dies with hat 0 face become 1 to 6. If we choose 6, the 6 results go from 7 to 12. With 2 numbers the whole line from 0 to 12 is shown. Since each result should show up 2 mroe times, the other 4 numbers are 2 0s and 2 6s Edit I wrote this before the solution was presented
My idea was that all chance events are 36, and since there has to be equal chance of 12 numbers occurring, it has to be 1/12=3/36 Therefore 3 sums from 2 dices must add up to 1 and 3 sums must add up to 12. The only way we can achieve both effects is 0+1, 0+1, 0+1 and 6+6, 6+6, 6+6 Since we have one normal die, the other must have 3 0s and 3 6s to give these sums
Could you use the same technique in the following situation ? You are given two blank dice. Label the two dice using positive integers 1, 2, 3, ... (0 is not allowed) so that when you roll the two dice the sum shows each whole number from 2 to 12 with exactly the same probability as two regular dice. Of course, none of the two dice will be labeled the same way as a regular die.
The answer is YES, though it is not no easy to see it. The generating function for a regular die is D(x)=(1/6)⋅(x¹+x²+x³+x⁴+x⁵+x⁶) and we want to find two other generating functions P(x) and Q(x) for non-regular dice such that P(x)⋅Q(x) = D(x)². First, note that x+x²+x³+x⁴+x⁵+x⁶ = x⋅(1+x)⋅(1+x+x²)⋅(1-x+x²) − I will explain in another response how we can find this. Hence we shall have P(x)⋅Q(x) = (1/6)²⋅x²⋅(1+x)²⋅(1+x+x²)²⋅(1-x+x²)². It should be obvious that there will be a 1/6 in each of P(x) and Q(x). Also, there will be a x in each of P(x) and Q(x) because of the condition "0 is not allowed" which means that there will be no x⁰ in the generating functions. Now... I don't know how to choose the repartition of the other factors, except to test every possible choice. Eventually, we find : P(x) = (1/6)⋅x⋅(1+x)⋅(1+x+x²) = (1/6)⋅(x¹+2x²+2x³+x⁴) Q(x) = (1/6)⋅x⋅(1+x)⋅(1+x+x²)⋅(1-x+x²)² = (1/6)⋅(x¹+x³+x⁴+x⁵+x⁶+x⁸) And the dice are (1,2,2,3,3,4) and (1,3,4,5,6,8).
@@rontyson6118 The question was not to have equal probabilities between the 11 possibilities, but to have the same probabilities as two normal dice. I.e. 1/36 for the 2 and the 12, 2/36 for the 3 and the 11, and so on up to 6/36 for the 7.
You want to generate a number between 1 and 12 with equal probability. So half the time, you want to generate a number between 1 and 6 with equal probability (which is exactly the same as a single die throw); and the other half of the time, you want to generate a number between 7 and 12 with equal probability (which is equivalent to a single die throw plus a constant 6). So the second die must generate either 0 or 6 with equal probability: that is to say, half the faces must be sixes, and the other half blank. Instead of two dice you could use a die and a coin, adding 6 for a head.
I started by noticing that you're going to have issues if there are two different things you could roll to get some of the totals, because you're not going to be able to do that for all of the totals, and there's no way the boundary could work. Then, if you use anything from 1 to 5, you can't also use any other number. So it's got to be using only 0 and 6, which are far enough apart.
I solved it in seconds. 1) there has to be possibility for 12, so there has to be at least one 6 2) like wise there has to be at least one 0 3) the 6 that we have generates 7-12 with equal chances. and the 0 that we have generates 1-6 with equal chances. so if we triple the result all numbers are covered with equal chances 4) after we solve, it is worth showing that it is the only possible solution. that can be achieved by showing that any other number created asymmetry that can not be compensated with another number or numbers
I don't understand what is going on with all these equations, but I did the task in two minutes. Just wrote down 1 to 12 and the numbers that sums up to them. Then you notice what has bigger and what has smaller probability, then you cross out numbers that cause big probability and done!
I figured it out before the solution is given, do I get the job now? I figured it out by knowing that the chance for each number is 1/12 so that means each number has 3 possible combinations to get to it and since with 0 you can get the first six numbers and with 6 you can get the other six the answer is 3 0s and 3 6s
Hey Presh !! Greetings . I am from India and would like to thank you for these videos.
Three 6s and three 0s could work. The labeled die already goes from 1 to 6 with equal odds, so, if you have a coinflip that adds 0, or 6, it would produce all twelve of the numbers with equal odds.
Yeah the original answer doesn't include one as a possible outcome, only from two up to twelve. I assume this is why one didn't get produced by the formula
I thought of that one quickly
Lol exactly man i did come up with that solution too
Yeah this is way better. I did this in my head just thinking that all you need is 3 different ways of making each number (36/12)
That's what I came up with.
When you actually solve it before he reveals the solution and you think you're 300 iq but everyone else says it's ez in the comments
@@pauliusaleknavicius4518 oh lol
Yeah i solved and felt really good, after 1 minute I read about this guy who solved in 10 secs.... I got depressed....
There’s probably some who have seen it before, or study probably and stats
I thought it was 1,1,3,3,6,6 / 1,1,3,4,6,6 . I guess I was kinda close...
I solved it in 1-2 mins
Since we need a non zero chance to roll 1 and 12
There is at least one side reserved for 0 and 6
By intuition I try 000666
If you roll a 0 with the blank die: even distribution over 1-6
Roll a 6 with the blank die: even distribution over 7-12
That's how I solved it
I think at some point you need to identify that each total must have three ways to arrive at, then there must be three sixes and three zeroes, there is no other combination that works.
Your intuition is better than mine.
I started at I must have a 0 to get to 1.
My intuition went for 000006
Then quick calculation gave me that generates 5 1's and only 1 12. That's no good.
Then my intuition went to 000666
That gives me 3 1's and 3 12's but does it give me 3 of 2 through 11?
Yes because I have 1 through 6 with 1/6 chance with +6 or 7 through 12 50% of the time :-)
Solved and checked :-)
Somebody given that answer in comments, which I find brilliant: Three 6s and three 0s could work. The labeled die already goes from 1 to 6 with equal odds, so, if you have a coinflip that adds 0, or 6, it would produce all twelve of the numbers with equal odds.
So i drew out a diagram of numbers 1 thru 6 and 1 thru 12, aknowledged that they dont over lap and went oh i need 3 of each
Finally a problem that I could figure out without the explanation! Your videos have helped me grow in mathematical skill so much, even though you often cover concepts I don't understand due to being less advanced in classes
Maybe one could just use 0 and 6...and that's it, lol...0 and 6...3 times...3 zeros, 3 sixes, that's 6 sides...0 is necessary to get 1, giving us 1 to 6, and 6 is necessary to get everything from 7 to 12, each of those appears exactly once...and since they don't say repetition is forbidden, we can just use them several times in order to get exactly the same number of occurrences, which is 3 for each possibility, so each of the 12 numbers can appear in 3 ways...
After watching this video I really wanted to try to solve another problem using this method. I know it would be hard, but I suggest you adding an extra problem for us to practice what we've learned (of course not in all videos, as it may be impossible). Keep up the good work! 🙂
Ha...you'll have to have a 0, lol, for starters, because otherwise you won't get 1...and the moment you have 1, you will have 1 to 6 in exactly one way, etc...will it be kept at 1, or do we have to increase it?...I suspect we'll have to increase it, lol...
I looked at the thumbnail, digested it, and then kept scrolling down as a I thought about it. I got the answer in about 15 seconds. I’m actually really happy about myself, usually the puzzles I see on your channel are really difficult for me.
Can we all work at Amazon now?
A small mistake I noticed - at around 6:41 you say "so we're going to multiply this by one half, which is also equal to one half of x times zero" - this should be "also equal to one half of x to the power of zero". It's written correctly in the video, and spoken correctly later in the video around 6:55 as "one half multiplied by x to the power of zero"
Other than that, great video!
Glad to know someone is paying attention! So this kind of video is always a big risk for me, since it's hard to communicate probability generating functions in a TH-cam video. If the video doesn't resonate quickly with enough people, it naturally won't get recommended through TH-cam's system. Even though my funding is 1% of some education channels, I still take more risks than them to make this kind of video on probability because teaching and education are more important to me than money. It's the ultimate validation when some videos are cited in academic papers, and glad to know they inspire many students to pursue learning these topics.
MindYourDecisions Thank you for teaching us :D
@@MindYourDecisions True Presh! Continue the MISSION sans CO.
I'm a Stat student and I studied Generating functions in last sem. Oof. Never thought they had such interesting applications. I should study them once again. Thank you very much.
first thought: the only way to get 1 is 0+1, 12 is 6+6
solved
Same here.
Btw pressure inducing tailgater
So that's two of your die's faces, what about the other four?
Andrew Guthrie you need 12 to appear three times and 1 to appear three times. That means three 0’s and three 6’s are needed, otherwise it’s impossible. And like magic, it works for 2-11 too
The only problem in your channel that I solved under 10 seconds
I also got the answer in less than 10 sec. However, I am here to understand how the question is formulated based on the puzzle.
I got it under a sec bitch
@@mandirasaha9033 i got it before i even read it
@@erez2417 i got it before i was born 😎😎
U r preposterous
I love watching your explanations through to the end. Even though I got the answer almost immediately in my head, it was due to simply immediately realizing I needed a 0 and a 6 at the start, and that those two alone gave me all sums equally. I love seeing how you structure the path to the solutions, since I'm usually far too lazy to actually come up with such a plan myself :D
I think this is the second one I actually decided to solve and succeeded! Good thing I'm studying probabilities in my free time. Thank you, Presh Talwalkar!
1:26
Presh: Give it a try...
Me: (Skips 10 seconds)
Wow, I always sucked at probability questions and I got this one in like a minute... Those videos REALLY make you understand maths better...
There is something in this video that is far more important then solution of problem
This can be simply done
1. The only way we get the sum to be one is 0+1
2. So every number should be expressed as sum in only one way to make the probability same
3. 0+1,0+2,0+3,0+4,0+5,0+6,1+6,2+6,3+6,4+6,5+6,6+6
This is the way I too thought the solution will be revealed. I also had a vague idea of solving it like the first method shown here.
yep thats how I did it.
I have been experimenting with dice lately and found really interesting probabilities if you put pips on the corners instead of the faces. I wish there were games that take advantage of weighted randomness like this, so many do not even seem to take account of the Gaussian distribution of two or more d6.
Great video. I felt smart getting that "Aha!" Moment getting the answer
same here
Roll a die (1,..,6), then flip a coin: if heads we are in the original (1,..,6) range, if tails we are in the (7,..,12) range. To get to the (7,..,12) range from the die, we simply offset/shift the die result with +6. No shift, +0, to the die result is needed for the lower range. The coin flip is identical to a die roll with equal probability of a +0 and +6 shift which means that the mystery die should have equal probability between the two shifts (3 sides of each since 3/6=1/2).
You can also solve this problems using convolutions and understanding that the sum of probabilities are the convolution of the probability function. If you don't know how convolution works it would ve quite complex, but knowing how it works it is not so dificult
Good problem, I think this is the first one I got right on the first try! Keep it up, Presh!
This truly is the best maths channel on youtube, or the just the best, for that matter
Also, probability is pretty cool, so I'd like to see more videos on that
You can also use the sum of geometric sequences to perform the division of the polynomials here
So I said to Presh, What's the singular word for _dice,_ he said _Die,_ so I said, Don't be like that.
Jack Sainthill dice and die both work as singular and plural
I found this fun because I didn’t know the answer until I drew the blank probability table. I was puzzled because I knew the probability result line for two standard dice shows a “bulge” of density in the middle, and visually, the problem was to spread that out and make it uniform. I then entered the three zeroes, using the same logic as the video (only way to get three chances out of 36 of there being a one). I then recognised that if I put any other numbers in except six there would be be greater than three chances of of a number. Six when combined with any of the numbers on the standard dice would not duplicate any number. That is when I realised the whole “trick” was to avoid the troublesome duplications, and 0 and 6 did that. I smiled at the now “obvious” result. Very similar logic to the video, but in a slightly different order. For me it was a nice illustration how seemingly difficult problems often are simple once you start to lay them out.
Dies are basically modulus operations (which means they cycle every N numbers, for N = number of sides). With that information you can get to the first solution a lot faster than with a table. From 1 to 12 there are two of these 6-sized cycles, so you need two numbers, one to run over the first 6 numbers (that are already in the first die) and one to run over the next 6 numbers. To run over the first 6 you need 0 as base, as they already exist, to shift to the next cycle you need 6 (N) as base. Half 0, half 6. It is very intuitive if you are familiar with the intuitions of modular maths. Works with any size dies as long as both dies are the same size.
What's the song please this is like my 15th times asking that
I've never known Presh to respond to comments.
Presh didn't gave this problem a try
@@AdityaKumar-ij5ok
Did you figure it out? ;)
That's an unnecessarily complicated way to solve it. I got it in my head in a few seconds.
I think it was more to showcase the method as a general tool. Which I appreciate.
Think you need to do this with over 100 possible inputs and outcomes, even more so if all the other options (there can even be more than two) are convoluted enough. That's when you do it the complicated way, because it's systematic enough you can force a computer to do the work for you quite easily with basically any symbolic computation tool.
And such situations do happen (i.e. in video games or probabilistic modelling).
No it's not. It's actually surprisingly simple. Like look at his other videos, it's more complicated.(but simple enough to be understood assuming you know the basic tricks of the problem).
There's another solution to this problem which is to label all 6 sides of the mystery die with a 12 (or any other number >=12).
The probabilities of rolling all the numbers 1-12 are 0, and equal.
Not as satisfying as the 000666 solution, but I would definitely mention this if I were in the interview because it's clever and super obvious once you realize it. :)
The numbers which can be used for labelling have to be whole numbers less than or equal to 6.
@Manas Ghai,
- "The numbers which can be used for labelling have to be whole numbers less than or equal to 6."
... and also must be non-negative. In other words, the only allowable numbers for labeling are 0, 1, 2, 3, 4, 5 and 6. (Just mentioning it for the sake of completeness.)
However, it wasn't specified that you can use only one number on each face, and we can label a face with 12 by using the numbers 1 and 2, so it's possible to label a face with "12". (But it's not possible to label a face with, say, "17", because we cannot use a 7.)
What a marvelous method to solve this problem!
Another title for this video: The complexity of dice :-0
Woo hoo
Got it correct
And i solved it in my mind
No pencil ✏️ paper or anything
I usually solve problems this way:
It has to be really complicated with lots of things to remember for me to do otherwise.
I took me about 15 seconds, so yeah... great achievement solving that in mind with no pencil nor paper. ( Forgive me if I seem sarcastic ) :D
same, but i used the simple method and not the systemetic one
@@Mihau_desu You must be so much fun a parties
@@tibbsazoid are you german
You have put some complicated stuff at the end of a very simple question. Awesome
This is a case where a graphical method -- drawing up and labelling the full table of all possible outcomes, and showing that each value from 1 to 12 occurs with equal frequency -- is quicker than an algebraic method. But what if we were using a pair of icosahedral dice? Now we would have to think about forty from four hundred possible outcomes, rather than a dozen from thirty-six, and a formula begins to look the more attractive proposition.
The point is, the formula is really just the instructions for filling in the table.
@bluerizlagirl,
The puzzle was relatively simple and quick to solve, because the required probability distribution of the dice pair's total was uniform (each of the outcomes 1 through 12 has equal probability). The "size" of the dice (six-sided or 20-sided) doesn't really matter to its difficulty.
Had the required probability distribution been non-uniform, then the "probability generating functions" formula would have been quicker and easier, even for a pair of just six-sided dice.
@@bluerizlagirl thats my point. People say it is easy because they are told a solution exists and second because 6 possibles cases are too few. The PGF approach settles all those issues at at once.
@@yurenchu yes it does if you are not told a solution exists.
I did it!! I finally answered one of your questions correctly!!
You didn't explain 4:55 well at all. What is x supposed to be and why are you raising it to the power of 2, 3 etc? Also, since Z is a discrete random variable defined by starting at 1, p of i is always equal to 1/y where y would be the max number (in this case 6) so p of i is always 1/6 for each Z.
Love probability videos! Thanks for much for all the great work
Making my guess at 1:48! Three sides of the die get 0, the other 3 get a 6. Figured this out by constructing and frequency table - sample space hybrid.
Excellent explaination ! Practical and mathematical.. both !!
I came up with the three 0's and three 6's solution pretty quickly. Apparently a lot of people did as well. The first thing I noticed was if you are going to get 1 you need a zero and if you are going to get a 12 you need a six. Then I realized that you were only going to hit 12 one in 36 times so you needed three sixes and similar logic worked for one as well and it all seemed to work when I thought about the different numbers.
This seems to be the only possible solution. I am going to watch the video now since from the comments it seems like there is another solution? Or another explanation?
ETA: There was no trick that I missed. I got the right solution. That was nice and a little boost to my ego after being humbled by watching somebody solving the normal distribution integral. Alas, most of the video involves a complicated way of getting the answer which I didn't understand but I was a lot closer to understanding it than I am to being able to solve normal distribution integral.
Wow you are so smart.
Gee, that's a simple puzzle. Why does he need nearly 8 minutes to explain the solution? Let's see his answer... oooh!
(Never knew stuff like this could be generalized with polynomials!)
On a different note: So, you've reconsidered from using the term "dice" for a singular die, then? :-)
Me: *skips to the end to check I got it right* What's he messing with polynomials for?
@DeadFish37,
- "Me: skips to the end to check I got it right What's he messing with polynomials for?"
Presumably to introduce us (or at least those of us with a deeper interest into mathematics) to a more abstract and relatively unknown branch of methods that combines probability functions with algebra. The relatively simple dice riddle is merely the hook to reel us in.
Both "die" and "dice" are accepted singular terms.
This was surprisingly simple once I realized. There are 36 possible combinations of faces, and since the 12 totals are represented evenly there have to be three combinations for each total. The only way to roll a total of 1 is with 1 on the normal die and 0 on the blank one, so there have to be three faces labeled 0 on the blank die. Likewise, the only way to total 12 is 6 on the normal die and 6 on the blank, so the other three faces on the blank all must be 6. With that you’ve used up all the blank faces, and from there it’s trivial to show that the probabilities are all equal.
Well, there are 36 possibilities using 2 6-sided dice. This means each number from 1 through 12 needs to be summed to 3 times. This requires the second die to have 3 zeros and 3 sixes. This doesn't require much thinking.
Please post few more videos on Probability. The above video was excellent
I labeled all of them with 12 dots. Now the sum of the 2 dice will always shows each whole number from 1 to 12 with equal chance.
0.
You can only use 0-6, can't use 12. I like that thought though
Oh yes Presh, more Probability videos pls!!!!! I do have good time by just reading the problem. Lol
I solved it in the head, no need of equations. Just think about it: you already have 1-6 in one die, and to get 7-12 you just need to add 6. So to make it equal chance, have 3 of the sides 0 and the other 3 as 6.
Oh my... After understanding the requested outcome, I had the answer in 5 seconds. I never would've guessed the mathematical approach would be so much harder
Whats the name of your theme music? Thanks
Thanks, always beautiful and inspiring videos!
You are a true hero
If you let the other die stay as a normal die you can take that die's outcome, subtract one and multiply it by six (number of outcomes on the other die) and get an even distribution for 1-36.
Best channel 😍😍
I solved it in just seconds, when I get started?
I did this in my head though~
(To get 1,2,3,4,5,6 we add a 0 to the new die. To get 7,8,9,10,11,12 we add a 6 to the new die. This gives the P(X=1,2..12) as 1/6. If we fill the new die with 3 0s and 3 6s we get P(X=1,2..12) as 1/3. This satisfies our requirement)
mathematical interpretation was very impressive. I watch this channel for brain fitness.
Wow, this relates directly to coding much more than most of their questions. Except perhaps biased toward D&D players, who understand this kind of modulo intuitively. Sometimes you want to get a random hour of the day in a TTRPG, and make a d24 this way.
Did Dr Yakubov feature in a Numberphile video?
Would u plz give a project result for secondary level
How about all being zeros??
I solved it in no time. But it seemed tough as your other questions. But the answer came to me abruptly.
As a quick logical guess from me when you showed the problem, I suspected three 0’s and three 6’s. Was surprised that was the answer
Hey, I just came across your channel. Do you have any advice on how I can internalize theory like this?
Having a 50% chance to add 6 to your roll gives you equal chances for everything from 1 to 12, dungeons and dragons actually being useful for something, wow
"By magic we find the answer", I think you meant "By logic"
"Magic", as in "witchcraft", as in "the work of Satan".
Notice that the answer is: 0, 0, 0, *6, 6, 6* !
It's exactly "magic" in that you need to "know" that this approach is going to work to consider trying it. If the table didn't have such complete clues, he would've just wasted time or ended up with a partial solution.
for 2 6-sides dice there are a total of 36 outcomes of equal probability, if we want to map 36 outcomes to 12 outcomes with equal probabilities then it must be that 3 of the 36 outcomes map to 1 of the 12 outcomes.
one of the dice is given to be a normal dice.
what outcomes map to a sum of 1? well we must roll a 1 on the normal dice and a 0 on the other, since this must be able to happen in 3 of the 36 outcomes, there must be 3 0s on the non-normal die.
what outcomes map to a sum of 12? well we must roll a 6 on the normal dice and a 6 on the other, since this must be able to happen in 3 of the 36 outcomes, there must be 3 6s on the non-normal die.
simple checking reveals that having the second dice have 3 0s and 3 6s does indeed solve the problem.
Superb stimulating question ..
thought process:
1. 2 dice = 36 combinations, for 1 to 12 to be equally likely, I need 3 of the combinations to get 1, so that's 3 sides with 0, similarly, I need 3 sides with 6 to get 12 with 1/12 chance, so 3 0's and 3 6's are the only combination that can work.
2. Yes it does work. if I roll a 0 on the blank die, it is equally likely to be 1-6, if I roll a 6 on the blank die, it is equally likely to be 7-12
The problem is easy because you are told such a solution exists. The right way to settle the issue is via the probability generating function. Imagine you had a dice with 100 faces, how would you go about it with the former method?
I solved this in less than 20 sec and thought that maybe didnt understand well the problem...
Same. Brute forced it in my mind
We could also use a six sided die and a coin, valuing heads as six and tails as zero. One side of the coin gives all the low numbers (with equal probability) and the other gives all the high numbers.
Having spotted this intuitively, if you now insist on a die with six faces as the second die, this is synthesised by numbering three faces as six and three as zero.
I really couldn't understand what the question wanted. I was wondering how it would be possible for the sum of the 5 faces showing to possibly do 1-12. I took the drawing of the dice as a literal point of view for the problem. And by roll, I thought they meant rotating on the same axis for some reason.
The music at the end of your videos... What song is it from?
Im 14 so let me explain how I solved this:
So, since its 1 through 12, we know we need at least one 0 so that we can have a sum of 1 (1+0). This zero, if added to 1 through 6 gives you 1 through 6, but we still need 7 through 12. To get this, we can add something else to 1 through 6 to get this (remember the probability to roll all numbers has to be the same) so we would need a 6 to add to 1 through 6 to get 7 through 12. Now we know that, if we have a 0 and a 6, or the same amount of 0's and 6's, we will have the same probability from 1 through 12. This means, to fill up the die, we need three 0's and three 6's.
Edit: note that 0 and 6 are being added to 1 through 6 because those are the numbers on the other die.
Where you collect all hard or intelligent question?
It's not clear from the problem description that you can label a face with zero. Normal dice don't have a "zero" face, and a "zero" face on a normal die would appear blank, which would seem to violate the requirement that we need to label all the faces.
The example (1, 2, 3, 4, 4, 5) should have included a zero.
It does say 0 to 6 in the question. It would be inconsistent to exclude 0 but include 6
The answer is so surprising! Keep it up!
More amazon interview questions please!
I just drew a 6x6 state-space matrix. The fixed die (1..6) labels the columns, and the variable die-faces label the rows. If every result (1..12) is equiprobable, then each sum must have 36/12 = 3 cells in the matrix. The first problem is "How to get 3 cells having the row-col sum = 1?" This implies that 3 faces must be labeled with 0s. This takes care of filling the top 3 rows. The bottom 3 rows? You can see that these need a 6 on the die face to obtain 3 each of the (7..12) cells. Answer: 0,0,0,6,6,6
I was able to realise that 0 and 6 should be there but I couldn't find out that each answer should be there 3 times. Nice puzzle though
A single die has a uniform distribution from 1 to 6. You can create a uniform distribution from 1 to 12 by either adding 0 or 6 with equal probability. That's how I reasoned through this problem.
36 different two dice rolls, so each of twelve numbers must come up three times. The only way to roll a one is with a one and a zero, so the second die has three zeros. Then each other number on the first dice is rolled three ways with the zeros, so no number less than six is on the remaining spaces of the second die. That leaves six to be filled in on those three spaces, and that works, seven through twelve can be rolled three ways.
Fyi I like probability questions more than geometry so please do more of them.
Omg I finally solved a mind your decisions problem
I think it is 3 0's and 3 6's. You have an equal chance of getting a 0 or a 6 on the custom die. This gives you an equal chance of getting on the low side (1-6) or highs (7-12). Then the natural die has an equal chance on any result.
Why did you put the x’s to the power of the index ? Like x6 is x1, 6 times ?
- "Like x6 is x1, 6 times ?"
Well, if you roll a die six times, and it lands six times on "1", then the total of your six rolls is equal to "6".
This was actually fairly trivial. In order to get 1, you need a face of 0 (only 1+0 can equal 0). For 12, you need a 6 (only 6+6 can equal 12). No other numbers can be added, since they would not cover either 1 or 12, and the outcomes have to be equally likely - ergo the mystery die must be 0, 0, 0, 6, 6, 6
3 0s and 3 6s. There are 36 possible arrangements with 2 die => 36/12 means each number from 1 to 12 needs to show up 3 times. If we choose a 0 on the blank, the 6 possible results of both dies with hat 0 face become 1 to 6. If we choose 6, the 6 results go from 7 to 12. With 2 numbers the whole line from 0 to 12 is shown. Since each result should show up 2 mroe times, the other 4 numbers are 2 0s and 2 6s
Edit I wrote this before the solution was presented
Another super easy one. Especially if grew up playing weird dice games.
My idea was that all chance events are 36, and since there has to be equal chance of 12 numbers occurring, it has to be 1/12=3/36
Therefore 3 sums from 2 dices must add up to 1 and 3 sums must add up to 12. The only way we can achieve both effects is 0+1, 0+1, 0+1 and 6+6, 6+6, 6+6
Since we have one normal die, the other must have 3 0s and 3 6s to give these sums
I'm afraid it's spelled "jacta" (not "iacta", although it was pronounced like that), from where we get the modern word "jet" (not "iet").
"By magic we have solved the answer!"
Me: "Wow, I have graduated from Hogwarts!"
Could you use the same technique in the following situation ?
You are given two blank dice.
Label the two dice using positive integers 1, 2, 3, ... (0 is not allowed) so that when you roll the two dice the sum shows each whole number from 2 to 12 with exactly the same probability as two regular dice.
Of course, none of the two dice will be labeled the same way as a regular die.
The answer is YES, though it is not no easy to see it.
The generating function for a regular die is D(x)=(1/6)⋅(x¹+x²+x³+x⁴+x⁵+x⁶) and we want to find two other generating functions P(x) and Q(x) for non-regular dice such that P(x)⋅Q(x) = D(x)².
First, note that x+x²+x³+x⁴+x⁵+x⁶ = x⋅(1+x)⋅(1+x+x²)⋅(1-x+x²) − I will explain in another response how we can find this.
Hence we shall have P(x)⋅Q(x) = (1/6)²⋅x²⋅(1+x)²⋅(1+x+x²)²⋅(1-x+x²)². It should be obvious that there will be a 1/6 in each of P(x) and Q(x). Also, there will be a x in each of P(x) and Q(x) because of the condition "0 is not allowed" which means that there will be no x⁰ in the generating functions. Now... I don't know how to choose the repartition of the other factors, except to test every possible choice.
Eventually, we find :
P(x) = (1/6)⋅x⋅(1+x)⋅(1+x+x²) = (1/6)⋅(x¹+2x²+2x³+x⁴)
Q(x) = (1/6)⋅x⋅(1+x)⋅(1+x+x²)⋅(1-x+x²)² = (1/6)⋅(x¹+x³+x⁴+x⁵+x⁶+x⁸)
And the dice are (1,2,2,3,3,4) and (1,3,4,5,6,8).
Those probabilities are not equal. It's impossible because 11 outcomes will not go into 36 possible outcomes equally.
@@rontyson6118 The question was not to have equal probabilities between the 11 possibilities, but to have the same probabilities as two normal dice. I.e. 1/36 for the 2 and the 12, 2/36 for the 3 and the 11, and so on up to 6/36 for the 7.
@@oliviermiakinen197 Oh, I misread. Sorry. That's a nice solution.
You want to generate a number between 1 and 12 with equal probability. So half the time, you want to generate a number between 1 and 6 with equal probability (which is exactly the same as a single die throw); and the other half of the time, you want to generate a number between 7 and 12 with equal probability (which is equivalent to a single die throw plus a constant 6). So the second die must generate either 0 or 6 with equal probability: that is to say, half the faces must be sixes, and the other half blank.
Instead of two dice you could use a die and a coin, adding 6 for a head.
I started by noticing that you're going to have issues if there are two different things you could roll to get some of the totals, because you're not going to be able to do that for all of the totals, and there's no way the boundary could work. Then, if you use anything from 1 to 5, you can't also use any other number. So it's got to be using only 0 and 6, which are far enough apart.
I solved it in seconds.
1) there has to be possibility for 12, so there has to be at least one 6
2) like wise there has to be at least one 0
3) the 6 that we have generates 7-12 with equal chances. and the 0 that we have generates 1-6 with equal chances.
so if we triple the result all numbers are covered with equal chances
4) after we solve, it is worth showing that it is the only possible solution.
that can be achieved by showing that any other number created asymmetry that can not be compensated with another number or numbers
Is it like dynamic programming
I don't understand what is going on with all these equations, but I did the task in two minutes. Just wrote down 1 to 12 and the numbers that sums up to them. Then you notice what has bigger and what has smaller probability, then you cross out numbers that cause big probability and done!
What if I label all the faces with1
Oh thanks i forgit this
I thought that i have to make only equal prob. for all outcomes
I figured it out before the solution is given, do I get the job now?
I figured it out by knowing that the chance for each number is 1/12 so that means each number has 3 possible combinations to get to it and since with 0 you can get the first six numbers and with 6 you can get the other six the answer is 3 0s and 3 6s
Sir I want to know that which software you use to arrange the words.....so I request you sir please tell me...