Pulled this one on a girl with a photographic memory. Not sure what the chances of her winning because of that was, but, she did lose because she didn't think to use her memory and count the cards, so even in that scenario it still works (sometimes)
I've been using this as a money closer. But to do that I do it a little different to tip the odds just a little more in my favor. 1) I have them write down their predictions ahead of time as 5 rows of 5 turn the paper over than 5 more rows of 5. I tell them to keep it as random as possible by saying "If you wrote only aces sooner or later an ace is going to turn up" 2) Have them cross out the last number and read the next. Specifically stating "It's what you say not what's written down that counts" 3) When you split it like that and have them write it down it confuses them as to what they wrote down. They do tend to write down A K Q J less but the other cards more so it ends up evening out (I think).
Odds and facts about this challenge: 0.0003694% if you pick the same number until card 49. 0.0044% if the deck is in an order that removes a "full suit" every 13 cards Each round has a (x-n)/x (x is current deck size, n is # cards of your chosen value left) chance of being passed. The actual odds change on each deal. The chance to correctly guess the deck is 100x(((4!)^13) / 52!) % With one suit, if you impose they can't guess something that has occurred gives a 7.69% chance for them to win .
As a note the method of using the most unlikely numbers to come up next isn't going to help you until all cards of one type are gone. (i'm just watching this now as i just found the channel recently) The issue becomes that while your unlikely to hit it on the next card, you can never close based on this. Let's for example say that in the top 10 cards you saw 9, 8, 9, 6, 9, K, J, Q, 3, 2 Now sure you could keep choosing 9 to try and get through as many cards as you can, but without another 9 dropping, your eventually gonna be screwed. and if your doing the method like this, your actually more likely to blow it on the next card assuming it impacts your method as eventually your going to hit a 9, or get to a point where you have 2 cards to switch between, leaving you at a 50% chance your next pick screws you or doesn't impact what your picking. Your best odds is being able to count the cards and going random until you get a set of 4 out, as any method relying on the low chance is actually more likely to work against you in the long term.
So, my theory is that the best strategy is to always guess the card that has come up the most. Once you get three of a kind turned up, you'll be guessing that one repeatedly, in which case you'll lose UNLESS another card value gets to three, and then surpasses the first one. Of course, then you can guess that value that has been turned up all four times through to the end, and win. So, I made a simulator that goes by this strategy, and ran it 1,000,000 times. It won 270,277 times, and lost 729,723 times. This method therefore gives odds of approximately 27.0% winning percentage, or 1 in 3.70. I don't know if this is the best method, but it's the best idea I can think of.
***** You say the strategy is obvious. I feel like it's pretty obvious, but the only thing that makes me wonder if MAYBE something could be better, is this : During the game, you're hoping to "advance through" some stages. The ultimate (winning) stage is having 4 cards of one value drawn. The penultimate stage is having 3 of one or more values, and this must be reached first to reach the ultimate stage. Likewise, reaching the stage of having 2 of one or more values must be reached before the penultimate stage. Basically what I'm getting at is, guessing one from the highest level actually lessens (or in some cases eliminates) your chances of getting to the next stage IMMEDIATELY. Example: If there are 3 aces turned up, but only 0, 1, or 2 of the rest of the values, you are guessing ace every time... but if an ace comes up you lose, whereas you could win more quickly by picking one that has only 2 instances so far. HOWEVER, I do agree with you, in believing the originally mentioned method must be the best... simply because, your odds of survival on each turn are so much greater by choosing from the highest level... and survival here inevitably leads to victory.
Ace Diamond I jumped in on the simulation game too. When making true random guesses, the chance of success was about 1.55%. The 9th card had a higher than 50% chance of a match. When running the Most-Occurring-Value algorithm, chance of success went up to about 27%. This matches your results as well as the odds calculated by UltraMadWizard. Better results, but the 50% chance-of-match mark was only card 13. Four of a kind can be seen as early as card 7. I ran the simulation a handful of times or so (each a million attempts) and only seen this once. Usually it starts at 8 or 9. The latest a four of a kind can be seen is card 40.
I'm a bit late to this party, but here's my thinking about the math side of things. At the start of the guessing, every face value has a 1/13 or 4/52 chance of showing up. After the first card is shown, that card has a 3/51 chance of showing up while all the others have a 4/51 chance, so the mathematically optimal move for the first few moves (assuming you don't bust on the first card) is to guess one of the first cards you've guessed. Depending on how good your memory is, you might be able to keep the first three or four in your mind. On move 2, guessing the card that came up in move 1 will lose just under 6% of the time vs just under 8% of the time for any other card. If it doesn't come up, whatever 2 cards came up in the first 2 rounds will be 6% vs 8% for any of the others. Assuming that you can keep the first 4 cards in your mind, you can track those 4's totals mentally, or if you want by using binary on your fingers by assigning the values of 1 and 2 to a pair of adjacent fingers and and iterating using binary counting on them. Card 1's count can be your left hand's pinky and ring finger, card 2 can be the left hand's index and thumb, card 3 the right index/thumb, and card 4 the right ring/pinky combo. Pinkies and thumbs are 2^0 (1) for their respective cards, and index/ring fingers are 2^1 (2) for their respective cards. So to explain what I mean, I start the game with both fists closed. I say an arbitrary value, say "king" and the card that is pulled for move 1 is a 2. I mentally note that 2 is on my left hand pinky/ring side and stick out my pinky to represent 1. Next move I declare 2, and the card pulled is a 7. I mentally note that 7 is on my left hand index/thumb side and extend my left thumb. Next I declare a 7, and the next card that gets pulled is a 10, so I note 10 is my right index/thumb and stick that thumb out. Next I declare 10 and card 4 is an ace, so I assign ace to my right ring/pinky and stick my right pinky out. I then rotate between those 4 unless I notice 2 of another card come out before any of my 4, then I update my lowest count to the new value on that hand. This should ensure that the mid-game is the most risky time of the game unless you get really lucky or unlucky (which we all know happens). You should get through the first 4 moves at least (12/13) * (48/51) * (47/50) * (46/49) ~= 77% of the time, as opposed to (12/13) * (47/51) * (46/50) * (45/49) ~= 72% chance from pure chance. Not a huge gain in the moment, but over 1000 scams that's 50 beers. Your strategy after this should lead to you weighting your guesses depending on how many of a given card are left in the deck. Unfortunately if you're the guesser, the best scenarios for your odds come infrequently but are ruined by this method: if 2+ of the same card come up in a row in the first 4 draws. If the first 2 cards are, say, King King, then for guesses 3-10 the odds of staying in the game if I guess king every time after that are (48/50) * (47/49) * (46/48) * (45/47) * (44/46) * (43/45) * (42/44) * (41/43) * (40/42) ~= 67% assuming that no other cards show up more than 3 times total as I do this. Comparatively, by 10 cards in if you guess randomly from 3-10 you're at about a 44% chance of staying in (assuming you made it past 1&2), and if it's just picking between 2-4 cards that have each had just 1 copy removed, you're at about a 54% chance. By card 13, if one card hasn't had at least 2 copies eliminated yet, one is about to. So at this point the best strategy you have is to focus on those just a bit more, but not overly so (they still exist somewhere in the deck, after all). A theoretically mathematically perfect machine that tries to guess the next card and can keep track of how many of each card are left could choose at random from the sets with the cards with the lowest count and based on some rough guesstimation, I would expect that to be effective somewhere in the 59-62% range, but I would want to actually run some simulations and analyze the data before actually saying anything definite about that sort of superhuman strategy (or strategy for someone with a photographic memory). But talking this out a bit, let's follow the "keep guessing a single card after it shows up twice" method for a second and extrapolate. Assuming that this happened on turns 1 and 2, where the probability is at the most volatile and evenly distributed, sticking with the king guess gives you 2/(52-n) chances to fail whereas other options first give you 4/(52-n) odds to fail and then as they get eliminated they give you 3/(52-n) odds. But let's look at the most freak scenario with your worst odds: you get past card 14 and every card other than king has come up once. At this point, if you guess "king" for card 15 there is a 2/38 chance you'll be eliminated, versus a 3/38 chance for any other choice (about a 5% chance vs a roughly 8% chance, which is closer than I'd like, but in another way guessing king increases your chances of winning that round by about 60%). The only time you should stick to any single card is when it's out of the deck with you having dodged the 4 bombs of them being pulled, as that reduces your chances of losing to 0% which is the sort of odds we want here. :)
I work in an assisted living home and one of my residents does this when she gets bored of playing solitaire.... her strategy start at ace count through to king if she gets through makes a pile and restarts trying to make 3 more piles (shes very neat so her piles are too and they stay face up).
If you have a perfect memory, you will play whichever card has occurred the most times. With that strategy, you win about 27% of the time: simulate_game
cant believe i wasted time on this, but got around 8% chance of winning with strategy: used numerical estimation, 10000 samples Matlab Code: clear all n=10000; win=ones(1,n); suit=1:13; deck=[suit suit suit suit]; guess=1; for j=1:n shuffleddeck=deck(randperm(length(deck))); for i=1:52 if guess==shuffleddeck(i) win(j)=0; break; else burn(i)=shuffleddeck(i); guess=mode(burn); end end j end P=nnz(win)/n
Those odds are still a tad off from reality as every time you draw a card it will be 1 card less in the deck. So when the odds of a randomly picked number being correct is 4/52 or wrong 1 - 4/52 and the 2 drops, you will now not have 1- 4/52 but 1 - 3/51 odds which is a bit higher. Also the odds of the other numbers will increase as 1 card has been drawn and therefor are 4/51, this continues as the deck gets smaller and smaller increasing the odds of picking a right number all the more. Do this for any sequence and think the amount of combinations you can make choosing any card at any given time you will find the odds are actually rediculous.
the math is a little off in this because its not 12/13 chance every flip, its actually a 48/52 for the first shot, 48/51 OR 47/51 on the second depending on if you repeat your guess or not, then third is x/50, x/49, x/48 etc for each card taken out of the deck
Worst case, one number will have come up four times after 40 cards have been shown (13 * 3 + 1). So with perfect memory, you're guaranteed to win after clearing 40 cards. Until then, the best you can do is guess one of the numbers you've seen the most often. In the worst case, that leaves you with about 8% chance to win. Running some random trials, the actual probability seems to be around 1/3.
There are lots of factors like if you will survive the first card. If you have a photographic memory the chances of you getting a win is 42.3% (IF YOU SURVIVE UP TO THAT POINT which is another 17%)= 7.2% 4/52 if the first 4 cards are the same 40/52 until you find a match if they arent the same so average is 22/52 cards that you will find a 4 of a kind and win.
chances of the number that just landed being repeated are slim in a well mixed deck. I'd pic a random number, trying to get it right (laws of probability screwing that up) and then pick the number that was just played, taking mental note on how many of each number was played. When all four of one card has been played, then just pick that card over and over... cause no one will realize that part til the end.
@VHOCambier only works if all four of that number have been drawn already otherwise you WILL run into that number, hence the need to be rainman to track the cards
take two decks of cards and shuffle them individually, take half of each deck and then play the game. Or take two dice and make them say a number 1 to 12 (13 would be correct for a deck of cards) for 52 rolls straight. This factors out the counting. Otoh, the % is variable depending of first cards.. say first 5-10 cards (low number) will contain 4 of a kind =>100% win after that point (so 70-80%). But the percent of such an arrangement is small, but there are 3 runs in total.. so it's doable
I guessed Ace for the first card, and from then on, you name the card you saw. I say "ace", and you flip over a 6, then I'll say "six", you flip 10, I say "ten", etc. This worked for me, and a managed to get through the entire deck. Chances are slim that 2 cards of the same value will be next to each other in the deck.
Manga Boss Scam School lets do the math here (I know a late reply, but still ^^)I now asume you only say the card taht was last shown without counting till you have hit 4 of the same, or when multiple of the same have already passed and you name those. so only by following the system you say. as he said, the first card would be totally random, so 12/13 you get it wrong what is a 92% chance. then you name the card that was shown last time. the chance that you will get another one by following this system is 3/51, instead of 4/52 (what is 1/13 if you say something completely random. this means you have a 48/51 chance to get it wrong (what is what you want) instead of a 48/52 (what is 12/13) so what brian did was do .92x.92 and do this over and over again. the first card is still completely random so still 0.923, the second one and every card following upon that is on its own is 0.941 (around 1.8% better) so do 0.923x0.941=0.868 instead of 0.852. it takes 11 cards to get 50% instead of 8. if you do 26 cards (half of the deck) you still only have 20% chance (totally random it would've been 20 cards for 20%) if you do 52 cards you have 4.15% to get it all correct doing it your way, instead of 1.55% chance. so you now have 2.6 times more chance to win, however it is still a very small chance. if I made any mistakes please tell me so :D (PS: I added you brian because you want to know the number if you did any tricks for this to get it easier right)
EDIT: I think I actually made a mistake, the same one Brian made: the wrongful assumption that the cards don't change, will recalculate and update later The very first card will always be a complete guess and thus be a 12/13, which is the same as 48/52 or 92% chance. After the first card, you have at least 1 value that has only 3 remaining cards of it instead of 4, so by guessing that, you slightly increase your odds to 49/52 or 94% chance per card. In the worst case scenario (for the player, best case for the host), you don't get a a value a second time until the 14th card, because at 13 cards each value could have been in there exactly once. As soon as you get at least one value for a second time, you can use that you increase your odds to 50/52, or 96% chance per card. The next worst case is when you don't get a value for the 3rd time until the 27th card (again, in 26 cards each could be in there exactly twice, 27 must therefore be a value for the 3rd time). From that point onward, you can increase the chances to 51/52, or 98% per card. As soon as you get a value for the 4rth time, which in the worst case would be after the 40th card, you can jjust keep guessing that value and be sure that you get it wrong 100% of the time. So in the worst case scenario (with the strategy), we've got 1 card at 92%, then 13 cards at 94%, 13 cards at 96%, 13 cards at 98%, and the remaining 12 cards at 100%, so we take (48/52)×((49/52)^13)×((50/52)^13)×((51/52)^13)×((52/52)^12)=0.19892 or about 20% (and again, that's absolute worst case for the player), however, the chance of getting such a worst case scenario is only 0.000000001633876%, so practically 0%. So for the player who would have perfect photographic memory, the baseline of the absolute worst chance they're gonna get is still about 20%, and they'll likely have more luck, which would make the payout of $1.50 for $10 already a good bet for the player. Now, what about the best case scenario (using the strategy) for the player? the first card will of course still be a complete guess, so still the 92%. The second guess will be at the 94%, but in order for them to succeed, the card must be a new card as well, as they would be guessing the value that's already been, so the third guess will also still be at 94%. However, after that third guess, there's already the chance that they get the value that they already have one of, but which they didn't guess, so if that happens, they'd already be at 96% for the 4rth card, which must be a value of which there aren't 2 yet, because there's only one value that's already been twice, which it the one that they'd be guessing, so the fifth will be at 96% as well. If the 4rth card was one that we did already have one of, then going into the 5th there is a chance to get a card of the value that's already been twice but which you aren't guessing, so for the 6th guess, you could already be moving on to the 98% chance. Once again, the 6th guess must be one that you don't already have 3 of so the chance for the 7th remains at 98%, and in order to get this absolutely best scenario, the 6th must be the value you've already got 2 of. Then for the 7th guess, you'd need to be lucky enough to hit the card of which you've got 3 but didn't guess, and if you do so, the 8th guess onwards, you can just guess the value that you've already gotten all of, which would result in a 73% chance of winning. Again, however, the chance of this happening is very small, only 0.0001066106% (so again virtually 0%), but despite both the best and worst case scenario having ridiculously low chances of occuring, the best case scenario is still 65250.15 times as likely to happen as the worst case scenario. Most games will be somewhere between absolute best and absolute worst, but given the fact that absolute best is so much more likely to occur than absolute worst, I'm guessing that using the strategy, the majority of the games will be closer to the best case rather than worst case (and in the games that tend more towards the best case, it's also more likely that someone can remember all cards up to that point, increasing their chances of sticking to the strategy)
As they progress, their odds of avoidance actually begin to stabilize. I misstated what I said before as to the percentage chance it is actually close to 1/26. If you would like to know the math behind it please do not hesitate to ask me through email.
No arceus1498 is right if you are using one deck of cards (if an infinite number of cards are used the chance is 1 in 13 if you are using full decks to make the infinite number of cards, confusing and can elaborate if you want). The point is if a three is discarded there are now 3 threes in a 51 card deck so your chance of winning if you guess a 3 the first time is (48/52) 92.31 rounding and a three the second time is (48/51) 94.12 rounding. (# of successes/ # of cards, the total)
did anyone come up with the best way to do this if you have all the cards played so far? because I wrote a program and figured it out I don't have a specific value because it varies but around 25% of getting all the way through if you know the cards
Tried this out a number of times and worked perfectly ... until someone is sober enough to remember how many times they said 1 card value . 1st time its happened, but thankfully I didn't bet anything!
Holy Crap! It's Destin from Screwattack! Never knew he would end up here on ScamSchool. Didn't realize it was him until Brian said his name. I miss him on Hard News.
Love how many people are into the maths of it by their comments but I can tell it doesn't alter its a good trick that most often works in the performers favour. Who cares about maths?
Tracking the picture cards......even with a limited memory talent could help win the bet like the guy did with the Jacks....if you see one of them four times then just call out that card near the end....card counters would easily win this challenge....like suggested in the messages, the fairest way to play would not to show the cards and have a "referee"
if you remember one card and count how many times it showed up, wait for four of them to appear, then keep on saying that number, since there is only 4 of a value
Yeah seems like the best strategy is to keep a count of what cards and keep saying those. Not sure why anyone would take the bet though bet though because if you’re just guessing randomly statistically speaking you should get a match by at least the 13th card, and even if you’re “unlucky” 4 matches in a single deck.
It makes it easier. Get the first card wrong then guess that same first card throughout the entire deck (ex you say six of clubs for the first card but it's a jack of diamonds, then say jack of diamonds for every next card).
I did it on my first two tries. The first time i got all four aces in the first 15 cards, and then went through the rest just saying ace. The second time i got super lucky and went all the way through saying random cards. I then tried ten more times and failed them all
When each card has been revealed twice they have a double half percent chance of making it if they have photographic memory although that still decreases by 13% per new card throughout the entire game
That is not how the probability would work... Since u dont put the revealed cards back in the deck, the probability would be = 1-12!/13! ... non-replacement probability.
+Nidhi Yerr No it depends on your strategy. If your guesses are random you are not factoring your 'history' and therefore the probability of each value on the next turn is equivelant (12/13)
@DJBigUm Wrong. My videos are free, no long commercial sponsors. If I want to avoid advertisements I should stay clear of promoted wed-pages that have them. TH-cam is the last place I should have to deal with it since it's a user based platform. Also; they have the right to have commercials to support their program. I have no issue with that - just as you shouldn't have an issue with me not choosing to watch it or make comments! My choice.
@KianAlexina well technically when you do this to your friends or family and what not, it's basically your game and your rules I assume you can make the rules as you like :-)
It involves a little luck but if you happen to come across a situation where all the 4 suits of a same number card have been gone you can just keep saying the number over and over again and win.
Brian, you are wrong about the odds being 1/13 of them guessing the correct value. That is the probability. The odds for that is 1/12, notation does matter to some people! I should get a deck of cards and an autograph :)
to eliminate "strategy" i told them the rules of the trick and before going through the deck had them write down a number 1-13 52 times had them shuffle the deck and then go through each number they wrote down which so far no one has won out of 20 tries.
when you flip the first card count how many more you need of that card to get rid of then say that card over and over. Example: 9 When all of the 9s are gone say 9 forever. Like if this helped you
in the case of the person with the phtographic meory when they get to the last 2 they are garanteed to win because they wont guess one of the last 2 values left
@pussymstrr yeah kk i see what your doing, i watch the ads to but if every decides to skip them then there is no point them being there and no one visits the site so please respect his decision to put them in there and expect them to be watched. also i don't appreciate your last 4 words, theres no need for it. and i just want to appologise for my bad language i will delete it right away.
Hmm.......... for the probability of random (I don't think it's as simple as (12/13)^52, it only applies if the taken cards can be used again) it's..... err... the first pick is 12/13.... then for the second pick is much much much trickier, because you've taken 1 out... so... the second pick should be... 47/51 if they said the card that is not the one that was taken away, and 48/51 if they said the card that was taken away..... so the probability of being wrong for the second pick is..... 12/13x47/51 + 1/13x48/51, it's.. 612/663, then multiply the result with 12/13 to get the probability of being wrong twice in a row (144/169). The third one is EVEN MORE trickier.... because the second picked card can be the same (number) from the first one (that's... 13x(1/13)x(3/51)=3/51), but can still be different (number) with the first one (48/51). Like from the second scenario, he could say the one that's been taken (11/13 or 12/13, depending on the second card taken away), or the one that hasn't been taken away (or taken away twice) (2/13 or 1/13). So, the probability of being wrong on the third pick is: (48/51)x{(11/13)x(46/50) + (2/13)x(47/50)} + (3/51)x{(12/13)x(46/50) + (1/13)x(48/50)}, that's....... 30600/33150, for guessing the cards wrong 3 times in a row, it's (30600/33150)x(144/169) = 4406400/5602350 = 1728/2197 ..... Sorry, it's the same after all!!!
Probability if using strategy: The first pick is absolutely 12/13 chance for being wrong. The second, if using strategy, is picking the same number as the previous one, so it's 48/51, in total, it's (12/13)x(48/51) = 576/663 = 192/221 (twice in a row) For the third pick, if using the strategy above, it means player will automatically lose if 2 of the same number occurs twice in a row, so that isn't part of the equation, the next strategy is to pick the number either the first taken away or the second taken away number, so it's 47/50. (192/221)x(47/50) = 9024/11050 = 4512/5525 The fourth one is quite tricky, because the player has a chance to be very lucky and the third taken away number can be any of the first taken away number but not his guess (3/50). If the player was lucky, then it's obvious which card to guess: the one that's been taken twice, it has 47/49 chance of guessing false. The overall probability *to be lucky and to guess false* is (3/50)x(47/49) = 131/2450, but there is still a chance of not being lucky (47/50), and the strategy if the player isn't lucky is to pick either of the taken away cards (46/49 chance of being wrong), overall chance of both not being lucky and guessing false is (47/50)x(46/49) = 2162/2450 Overall chance of guessing wrong on the fourth pick is (2162 + 131)/2450 = 2293/2450, and the chance of guessing wrong four times in a row is (4512/5525)x(2293/2450) = 10346016/13536250 = 5173008/6768125 I will let someonw continue this.... it's already become too complicated for me....
You have a 12/13 chance of guessing wrong for every guess if you randomly choose a number. First: 12/13 (92.3%) Second:12/13 (92.3% x 92.3% = 85.2%) Third: 12/13 (92.3%x92.3%x92.3% = 78.7%) Etc. until you get to the 1.6% cumulative probability. Assuming you aren't memorizing all the cards played, or counting cards (until you know all four of one number has been played), you still CAN increase your odds of not guessing a correct card by saying the previous card shown as your next guess. ---The first guess is still 12/13 chance of course (92.3%) ---The second guess (if you say the previous number) is 48/51 (94.1%) of guessing wrong since there are only three of those numbers left in the 51 card deck. 92.3% x 94.1% = 86.9%. ---So on the second guess you have a 1.7% increased chance of guessing wrong. ---After that, you start getting into some heavy maths because duplicates start showing up so the true probabilities aren't as clean cut as the second guess, but the trend continues with slightly better odds.
Obviously, but that's not what I'm talking about. I meant if, for example, you have you have seen two of that card previously (without knowing it, of course), so three of that number have passed in total, the odds that your next card would be the same as previous drops to 1/X instead of 2/X or 3/X (X being the number of cards left in the deck) like it would be if it were the first or second time you saw that number. To clarify my original continuation, you have to start factoring in the odds that you may or may not have seen two, three or four of the previous cards -- without knowing it. Which gets more complicated than I can represent in a simple manner like the first two draws.
Songbirdo ah, I see. And with a photographic memory, you have to take into account that eventually you'll run into two of the same card. I guess the best way to figure it out is to try it face up so you can see every card and figure out how many times you lose. Or you could try to map all the probabilities... but that seems complicated, and probably takes more time than actually experimenting and seeing what you get.
You can do it mathematically. There's only 52 parts to the equation multiplied together, and I just gave you the first two pieces (12/13 x 48/51). And the last one is theoretically 12/13, too. I'll leave the remaining 49 draws to you! lol The last 10 draws are easy in comparison to the middle 40 as your number of possibilities collapse... Good luck!
Songbirdo Yeah, one can map the probabilities mathematically, probably with some sort of summation, but I'd rather honestly just do a hundred trials. Not to mention I have no idea what the ideal strategy is.
bet someone you can drink 2 pints of beer, faster than they can drink 1 shot glass worth of beer (or any other liquid). only rules are that you can't touch each others glasses, and they can't start until you finish your first beer and set the glass down. when its time for you to set the first glass down, set it upside down over their glass, so they can't drink the shot. now you can drink the second beer at your leisure. impossible to lose
Can i reapet the same number after the four of them arent on the deck. For example the four aces are already out od the deck, and i say Ace all the time :S
There's like only 4 suits, so four of the same type of number. Once you see that one value has all it's suits down, just repeat that number til the end. :)))
Wait but if you have a boat load of cash and you buy a boat then the leftover amount could fit in a space much smaller than a boat so if you were to buy the boat just for the purpose of storing the money it would be a complete waist
@DJBigUm Yes; I got that they have to use advertisement - but that wasn't my original point. My point (the one you chose to discuss) was that I didn't wish to deal with watching it because I had to sit through too many commercials. That's it. I'm on youtube watching videos - I can go all day here without seeing all that; therefore, I choose not to watch this. End of point. I don't give a rat's ass why they do it - they still do it!
![Math for Game Devs [2022, part 10] • Abstract Algebra, Procedural Animation & Splines](http://i.ytimg.com/vi/Pku6WAs1fF8/mqdefault.jpg)
![Math for Game Devs [2022, part 10] • Abstract Algebra, Procedural Animation & Splines](/img/tr.png)
Pulled this one on a girl with a photographic memory. Not sure what the chances of her winning because of that was, but, she did lose because she didn't think to use her memory and count the cards, so even in that scenario it still works (sometimes)
I've been using this as a money closer. But to do that I do it a little different to tip the odds just a little more in my favor.
1) I have them write down their predictions ahead of time as 5 rows of 5 turn the paper over than 5 more rows of 5. I tell them to keep it as random as possible by saying "If you wrote only aces sooner or later an ace is going to turn up"
2) Have them cross out the last number and read the next. Specifically stating "It's what you say not what's written down that counts"
3) When you split it like that and have them write it down it confuses them as to what they wrote down. They do tend to write down A K Q J less but the other cards more so it ends up evening out (I think).
Odds and facts about this challenge:
0.0003694% if you pick the same number until card 49.
0.0044% if the deck is in an order that removes a "full suit" every 13 cards
Each round has a (x-n)/x (x is current deck size, n is # cards of your chosen value left) chance of being passed.
The actual odds change on each deal.
The chance to correctly guess the deck is 100x(((4!)^13) / 52!) %
With one suit, if you impose they can't guess something that has occurred gives a 7.69% chance for them to win .
As a note the method of using the most unlikely numbers to come up next isn't going to help you until all cards of one type are gone. (i'm just watching this now as i just found the channel recently)
The issue becomes that while your unlikely to hit it on the next card, you can never close based on this.
Let's for example say that in the top 10 cards you saw 9, 8, 9, 6, 9, K, J, Q, 3, 2
Now sure you could keep choosing 9 to try and get through as many cards as you can, but without another 9 dropping, your eventually gonna be screwed. and if your doing the method like this, your actually more likely to blow it on the next card assuming it impacts your method as eventually your going to hit a 9, or get to a point where you have 2 cards to switch between, leaving you at a 50% chance your next pick screws you or doesn't impact what your picking.
Your best odds is being able to count the cards and going random until you get a set of 4 out, as any method relying on the low chance is actually more likely to work against you in the long term.
So, my theory is that the best strategy is to always guess the card that has come up the most. Once you get three of a kind turned up, you'll be guessing that one repeatedly, in which case you'll lose UNLESS another card value gets to three, and then surpasses the first one. Of course, then you can guess that value that has been turned up all four times through to the end, and win. So, I made a simulator that goes by this strategy, and ran it 1,000,000 times. It won 270,277 times, and lost 729,723 times. This method therefore gives odds of approximately 27.0% winning percentage, or 1 in 3.70. I don't know if this is the best method, but it's the best idea I can think of.
***** You say the strategy is obvious. I feel like it's pretty obvious, but the only thing that makes me wonder if MAYBE something could be better, is this :
During the game, you're hoping to "advance through" some stages. The ultimate (winning) stage is having 4 cards of one value drawn. The penultimate stage is having 3 of one or more values, and this must be reached first to reach the ultimate stage. Likewise, reaching the stage of having 2 of one or more values must be reached before the penultimate stage.
Basically what I'm getting at is, guessing one from the highest level actually lessens (or in some cases eliminates) your chances of getting to the next stage IMMEDIATELY.
Example: If there are 3 aces turned up, but only 0, 1, or 2 of the rest of the values, you are guessing ace every time... but if an ace comes up you lose, whereas you could win more quickly by picking one that has only 2 instances so far.
HOWEVER, I do agree with you, in believing the originally mentioned method must be the best... simply because, your odds of survival on each turn are so much greater by choosing from the highest level... and survival here inevitably leads to victory.
Ace Diamond I jumped in on the simulation game too. When making true random guesses, the chance of success was about 1.55%. The 9th card had a higher than 50% chance of a match.
When running the Most-Occurring-Value algorithm, chance of success went up to about 27%. This matches your results as well as the odds calculated by UltraMadWizard. Better results, but the 50% chance-of-match mark was only card 13.
Four of a kind can be seen as early as card 7. I ran the simulation a handful of times or so (each a million attempts) and only seen this once. Usually it starts at 8 or 9. The latest a four of a kind can be seen is card 40.
I tried this and I guessed 2...the first card was a 2
Lol, RIP
happened to me wif 7
I'm a bit late to this party, but here's my thinking about the math side of things.
At the start of the guessing, every face value has a 1/13 or 4/52 chance of showing up. After the first card is shown, that card has a 3/51 chance of showing up while all the others have a 4/51 chance, so the mathematically optimal move for the first few moves (assuming you don't bust on the first card) is to guess one of the first cards you've guessed. Depending on how good your memory is, you might be able to keep the first three or four in your mind.
On move 2, guessing the card that came up in move 1 will lose just under 6% of the time vs just under 8% of the time for any other card. If it doesn't come up, whatever 2 cards came up in the first 2 rounds will be 6% vs 8% for any of the others.
Assuming that you can keep the first 4 cards in your mind, you can track those 4's totals mentally, or if you want by using binary on your fingers by assigning the values of 1 and 2 to a pair of adjacent fingers and and iterating using binary counting on them. Card 1's count can be your left hand's pinky and ring finger, card 2 can be the left hand's index and thumb, card 3 the right index/thumb, and card 4 the right ring/pinky combo. Pinkies and thumbs are 2^0 (1) for their respective cards, and index/ring fingers are 2^1 (2) for their respective cards.
So to explain what I mean, I start the game with both fists closed. I say an arbitrary value, say "king" and the card that is pulled for move 1 is a 2. I mentally note that 2 is on my left hand pinky/ring side and stick out my pinky to represent 1. Next move I declare 2, and the card pulled is a 7. I mentally note that 7 is on my left hand index/thumb side and extend my left thumb. Next I declare a 7, and the next card that gets pulled is a 10, so I note 10 is my right index/thumb and stick that thumb out. Next I declare 10 and card 4 is an ace, so I assign ace to my right ring/pinky and stick my right pinky out. I then rotate between those 4 unless I notice 2 of another card come out before any of my 4, then I update my lowest count to the new value on that hand.
This should ensure that the mid-game is the most risky time of the game unless you get really lucky or unlucky (which we all know happens). You should get through the first 4 moves at least (12/13) * (48/51) * (47/50) * (46/49) ~= 77% of the time, as opposed to (12/13) * (47/51) * (46/50) * (45/49) ~= 72% chance from pure chance. Not a huge gain in the moment, but over 1000 scams that's 50 beers.
Your strategy after this should lead to you weighting your guesses depending on how many of a given card are left in the deck. Unfortunately if you're the guesser, the best scenarios for your odds come infrequently but are ruined by this method: if 2+ of the same card come up in a row in the first 4 draws. If the first 2 cards are, say, King King, then for guesses 3-10 the odds of staying in the game if I guess king every time after that are (48/50) * (47/49) * (46/48) * (45/47) * (44/46) * (43/45) * (42/44) * (41/43) * (40/42) ~= 67% assuming that no other cards show up more than 3 times total as I do this. Comparatively, by 10 cards in if you guess randomly from 3-10 you're at about a 44% chance of staying in (assuming you made it past 1&2), and if it's just picking between 2-4 cards that have each had just 1 copy removed, you're at about a 54% chance.
By card 13, if one card hasn't had at least 2 copies eliminated yet, one is about to. So at this point the best strategy you have is to focus on those just a bit more, but not overly so (they still exist somewhere in the deck, after all). A theoretically mathematically perfect machine that tries to guess the next card and can keep track of how many of each card are left could choose at random from the sets with the cards with the lowest count and based on some rough guesstimation, I would expect that to be effective somewhere in the 59-62% range, but I would want to actually run some simulations and analyze the data before actually saying anything definite about that sort of superhuman strategy (or strategy for someone with a photographic memory). But talking this out a bit, let's follow the "keep guessing a single card after it shows up twice" method for a second and extrapolate.
Assuming that this happened on turns 1 and 2, where the probability is at the most volatile and evenly distributed, sticking with the king guess gives you 2/(52-n) chances to fail whereas other options first give you 4/(52-n) odds to fail and then as they get eliminated they give you 3/(52-n) odds. But let's look at the most freak scenario with your worst odds: you get past card 14 and every card other than king has come up once. At this point, if you guess "king" for card 15 there is a 2/38 chance you'll be eliminated, versus a 3/38 chance for any other choice (about a 5% chance vs a roughly 8% chance, which is closer than I'd like, but in another way guessing king increases your chances of winning that round by about 60%).
The only time you should stick to any single card is when it's out of the deck with you having dodged the 4 bombs of them being pulled, as that reduces your chances of losing to 0% which is the sort of odds we want here. :)
I work in an assisted living home and one of my residents does this when she gets bored of playing solitaire.... her strategy start at ace count through to king if she gets through makes a pile and restarts trying to make 3 more piles (shes very neat so her piles are too and they stay face up).
I did it on the first try at home without seeing the trick yet. Thank you Derren Brown for teaching me how to count cards lol.
Chances are 32.7 for people with photographic memory
If you have a perfect memory, you will play whichever card has occurred the most times. With that strategy, you win about 27% of the time:
simulate_game
cant believe i wasted time on this, but got around 8% chance of winning with strategy: used numerical estimation, 10000 samples
Matlab Code:
clear all
n=10000;
win=ones(1,n);
suit=1:13;
deck=[suit suit suit suit];
guess=1;
for j=1:n
shuffleddeck=deck(randperm(length(deck)));
for i=1:52
if guess==shuffleddeck(i)
win(j)=0;
break;
else
burn(i)=shuffleddeck(i);
guess=mode(burn);
end
end
j
end
P=nnz(win)/n
hahaha the intro is just amazing!
Those odds are still a tad off from reality as every time you draw a card it will be 1 card less in the deck. So when the odds of a randomly picked number being correct is 4/52 or wrong 1 - 4/52 and the 2 drops, you will now not have 1- 4/52 but 1 - 3/51 odds which is a bit higher. Also the odds of the other numbers will increase as 1 card has been drawn and therefor are 4/51, this continues as the deck gets smaller and smaller increasing the odds of picking a right number all the more. Do this for any sequence and think the amount of combinations you can make choosing any card at any given time you will find the odds are actually rediculous.
Just say when you make the rules you can't say the same card more than 4 times in a row
the math is a little off in this because its not 12/13 chance every flip, its actually a 48/52 for the first shot, 48/51 OR 47/51 on the second depending on if you repeat your guess or not, then third is x/50, x/49, x/48 etc for each card taken out of the deck
o god.. brother i got, so many tricks to empress girls from your channel..I love you man.. episodes all r so excellent
Worst case, one number will have come up four times after 40 cards have been shown (13 * 3 + 1). So with perfect memory, you're guaranteed to win after clearing 40 cards. Until then, the best you can do is guess one of the numbers you've seen the most often. In the worst case, that leaves you with about 8% chance to win. Running some random trials, the actual probability seems to be around 1/3.
There are lots of factors like if you will survive the first card.
If you have a photographic memory the chances of you getting a win is 42.3% (IF YOU SURVIVE UP TO THAT POINT which is another 17%)= 7.2%
4/52 if the first 4 cards are the same
40/52 until you find a match if they arent the same
so average is 22/52 cards that you will find a 4 of a kind and win.
I'm trying my hardest, but I'm not following you. Can you explain this in a different/more clear way?
chances of the number that just landed being repeated are slim in a well mixed deck. I'd pic a random number, trying to get it right (laws of probability screwing that up) and then pick the number that was just played, taking mental note on how many of each number was played.
When all four of one card has been played, then just pick that card over and over... cause no one will realize that part til the end.
@VHOCambier only works if all four of that number have been drawn already otherwise you WILL run into that number, hence the need to be rainman to track the cards
take two decks of cards and shuffle them individually, take half of each deck and then play the game.
Or take two dice and make them say a number 1 to 12 (13 would be correct for a deck of cards) for 52 rolls straight.
This factors out the counting.
Otoh, the % is variable depending of first cards.. say first 5-10 cards (low number) will contain 4 of a kind =>100% win after that point (so 70-80%). But the percent of such an arrangement is small, but there are 3 runs in total.. so it's doable
Sometimes the theme song sounds like "Stab School..."
I suppose that works, too.
Lol!
I hear "Scom School, Meow Meow. Scom School, Meow Meow."
Nathan Griffing Stab School, yum yum! Stab School, yum yum!
Nathan Griffing Stab School, yum yum! Stab School, yum yum!
I guessed Ace for the first card, and from then on, you name the card you saw. I say "ace", and you flip over a 6, then I'll say "six", you flip 10, I say "ten", etc. This worked for me, and a managed to get through the entire deck. Chances are slim that 2 cards of the same value will be next to each other in the deck.
Manga Boss Scam School
lets do the math here (I know a late reply, but still ^^)I now asume you only say the card taht was last shown without counting till you have hit 4 of the same, or when multiple of the same have already passed and you name those. so only by following the system you say.
as he said, the first card would be totally random, so 12/13 you get it wrong what is a 92% chance.
then you name the card that was shown last time. the chance that you will get another one by following this system is 3/51, instead of 4/52 (what is 1/13 if you say something completely random. this means you have a 48/51 chance to get it wrong (what is what you want) instead of a 48/52 (what is 12/13)
so what brian did was do .92x.92 and do this over and over again.
the first card is still completely random so still 0.923, the second one and every card following upon that is on its own is 0.941 (around 1.8% better)
so do 0.923x0.941=0.868 instead of 0.852.
it takes 11 cards to get 50% instead of 8.
if you do 26 cards (half of the deck) you still only have 20% chance (totally random it would've been 20 cards for 20%)
if you do 52 cards you have 4.15% to get it all correct doing it your way, instead of 1.55% chance.
so you now have 2.6 times more chance to win, however it is still a very small chance.
if I made any mistakes please tell me so :D
(PS: I added you brian because you want to know the number if you did any tricks for this to get it easier right)
Manga Boss it isnt actually that smalli went through with no strategy and i saw a lot of pairs together
Manga Boss it isnt actually that smalli went through with no strategy and i saw a lot of pairs together
EDIT: I think I actually made a mistake, the same one Brian made: the wrongful assumption that the cards don't change, will recalculate and update later
The very first card will always be a complete guess and thus be a 12/13, which is the same as 48/52 or 92% chance.
After the first card, you have at least 1 value that has only 3 remaining cards of it instead of 4, so by guessing that, you slightly increase your odds to 49/52 or 94% chance per card.
In the worst case scenario (for the player, best case for the host), you don't get a a value a second time until the 14th card, because at 13 cards each value could have been in there exactly once. As soon as you get at least one value for a second time, you can use that you increase your odds to 50/52, or 96% chance per card.
The next worst case is when you don't get a value for the 3rd time until the 27th card (again, in 26 cards each could be in there exactly twice, 27 must therefore be a value for the 3rd time). From that point onward, you can increase the chances to 51/52, or 98% per card.
As soon as you get a value for the 4rth time, which in the worst case would be after the 40th card, you can jjust keep guessing that value and be sure that you get it wrong 100% of the time.
So in the worst case scenario (with the strategy), we've got 1 card at 92%, then 13 cards at 94%, 13 cards at 96%, 13 cards at 98%, and the remaining 12 cards at 100%, so we take (48/52)×((49/52)^13)×((50/52)^13)×((51/52)^13)×((52/52)^12)=0.19892 or about 20% (and again, that's absolute worst case for the player), however, the chance of getting such a worst case scenario is only 0.000000001633876%, so practically 0%. So for the player who would have perfect photographic memory, the baseline of the absolute worst chance they're gonna get is still about 20%, and they'll likely have more luck, which would make the payout of $1.50 for $10 already a good bet for the player.
Now, what about the best case scenario (using the strategy) for the player?
the first card will of course still be a complete guess, so still the 92%.
The second guess will be at the 94%, but in order for them to succeed, the card must be a new card as well, as they would be guessing the value that's already been, so the third guess will also still be at 94%. However, after that third guess, there's already the chance that they get the value that they already have one of, but which they didn't guess, so if that happens, they'd already be at 96% for the 4rth card, which must be a value of which there aren't 2 yet, because there's only one value that's already been twice, which it the one that they'd be guessing, so the fifth will be at 96% as well. If the 4rth card was one that we did already have one of, then going into the 5th there is a chance to get a card of the value that's already been twice but which you aren't guessing, so for the 6th guess, you could already be moving on to the 98% chance. Once again, the 6th guess must be one that you don't already have 3 of so the chance for the 7th remains at 98%, and in order to get this absolutely best scenario, the 6th must be the value you've already got 2 of. Then for the 7th guess, you'd need to be lucky enough to hit the card of which you've got 3 but didn't guess, and if you do so, the 8th guess onwards, you can just guess the value that you've already gotten all of, which would result in a 73% chance of winning. Again, however, the chance of this happening is very small, only 0.0001066106%
(so again virtually 0%), but despite both the best and worst case scenario having ridiculously low chances of occuring, the best case scenario is still 65250.15 times as likely to happen as the worst case scenario.
Most games will be somewhere between absolute best and absolute worst, but given the fact that absolute best is so much more likely to occur than absolute worst, I'm guessing that using the strategy, the majority of the games will be closer to the best case rather than worst case (and in the games that tend more towards the best case, it's also more likely that someone can remember all cards up to that point, increasing their chances of sticking to the strategy)
using the counting system when counting cards can give a person an idea on what number to choose depending on what the current count is
As they progress, their odds of avoidance actually begin to stabilize. I misstated what I said before as to the percentage chance it is actually close to 1/26. If you would like to know the math behind it please do not hesitate to ask me through email.
No arceus1498 is right if you are using one deck of cards (if an infinite number of cards are used the chance is 1 in 13 if you are using full decks to make the infinite number of cards, confusing and can elaborate if you want). The point is if a three is discarded there are now 3 threes in a 51 card deck so your chance of winning if you guess a 3 the first time is (48/52) 92.31 rounding and a three the second time is (48/51) 94.12 rounding. (# of successes/ # of cards, the total)
did anyone come up with the best way to do this if you have all the cards played so far? because I wrote a program and figured it out I don't have a specific value because it varies but around 25% of getting all the way through if you know the cards
You have a 32% chance with random and strategy I did mupitle test and came up with numbers varying from 30%-34%
Tried this out a number of times and worked perfectly ... until someone is sober enough to remember how many times they said 1 card value . 1st time its happened, but thankfully I didn't bet anything!
@Conshabi no its a 6 pause at the right time and youll see it
Holy Crap! It's Destin from Screwattack! Never knew he would end up here on ScamSchool. Didn't realize it was him until Brian said his name. I miss him on Hard News.
You should get Richard Turner on Scam School! He would Definitely teach you some wizardry!! He knows martial arts too!! 💪🏽😼
0:19 The background music is from Auction Hunters. They play it at the part where they total their profits.
Love how many people are into the maths of it by their comments but I can tell it doesn't alter its a good trick that most often works in the performers favour. Who cares about maths?
Done on family and they raged
nice!
Good to see Destin still hangin' on after leaving Screwattack!
Tracking the picture cards......even with a limited memory talent could help win the bet like the guy did with the Jacks....if you see one of them four times then just call out that card near the end....card counters would easily win this challenge....like suggested in the messages, the fairest way to play would not to show the cards and have a "referee"
@SuperSoylent2
If you go to their site there are marks where the add's end...
what are those cards called? i just know the german ones but they dont have 52 cards and start at the number 7.
they are regular bycicle rider back cards. You probibly saw cards to play bridge with
they are regular bycicle rider back cards. You probibly saw cards to play bridge with
+garget kumet I know Euchre decks only have 9-K of each suit.
if you remember one card and count how many times it showed up, wait for four of them to appear, then keep on saying that number, since there is only 4 of a value
Yeah seems like the best strategy is to keep a count of what cards and keep saying those. Not sure why anyone would take the bet though bet though because if you’re just guessing randomly statistically speaking you should get a match by at least the 13th card, and even if you’re “unlucky” 4 matches in a single deck.
It makes it easier. Get the first card wrong then guess that same first card throughout the entire deck (ex you say six of clubs for the first card but it's a jack of diamonds, then say jack of diamonds for every next card).
@crazyguy980 if you play TF2, it was a pun on when the spy says "rigt behind you."
Your way of calculation would only be true if you would put the card back after picking it.
Thomas Bezençon Nope. as long as you don't see the previous cards and can't memorize them, calculation is correct.
Holy shit you're right
I did it on my first two tries. The first time i got all four aces in the first 15 cards, and then went through the rest just saying ace. The second time i got super lucky and went all the way through saying random cards. I then tried ten more times and failed them all
When each card has been revealed twice they have a double half percent chance of making it if they have photographic memory although that still decreases by 13% per new card throughout the entire game
what if we see there are 4 cards down could we just say the numbers/letters of that card to win
That is not how the probability would work... Since u dont put the revealed cards back in the deck, the probability would be = 1-12!/13! ... non-replacement probability.
Brushwood did you once do a trick with 1 knife 1 turntable and 4 foam cups?????? Cuz if you did please show your scar on the next video please
It would only be (12/13)^52 if there were replacements. It would be 12/13 * 47/51* 23/25* 45/49 etc. At least that's what I understood.
+Nidhi Yerr No it depends on your strategy. If your guesses are random you are not factoring your 'history' and therefore the probability of each value on the next turn is equivelant (12/13)
@DJBigUm Thanks but have no reason to go anywhere. I'm here on youtube.
@DJBigUm Wrong. My videos are free, no long commercial sponsors. If I want to avoid advertisements I should stay clear of promoted wed-pages that have them. TH-cam is the last place I should have to deal with it since it's a user based platform. Also; they have the right to have commercials to support their program. I have no issue with that - just as you shouldn't have an issue with me not choosing to watch it or make comments! My choice.
@KianAlexina well technically when you do this to your friends or family and what not, it's basically your game and your rules I assume you can make the rules as you like :-)
It involves a little luck but if you happen to come across a situation where all the 4 suits of a same number card have been gone you can just keep saying the number over and over again and win.
@TshirtTimee ooohh lol yaa ur right thanks man
@8bitmagic He went to work at Frogster Interactive. Then IGN.
in what video do they teach the illegal trick? I cant seem to find it, was it even uploaded?
lolll Netflix "10 million users" :D . That escalated quickly
haha
Brian, you are wrong about the odds being 1/13 of them guessing the correct value. That is the probability. The odds for that is 1/12, notation does matter to some people! I should get a deck of cards and an autograph :)
to eliminate "strategy" i told them the rules of the trick and before going through the deck had them write down a number 1-13 52 times had them shuffle the deck and then go through each number they wrote down which so far no one has won out of 20 tries.
really nice dude..awsum video
when you flip the first card count how many more you need of that card to get rid of then say that card over and over. Example: 9 When all of the 9s are gone say 9 forever. Like if this helped you
@ChrisSosa it's not a s ire thing cause ur going to come up to a "7" at some point in the deck
in the case of the person with the phtographic meory when they get to the last 2 they are garanteed to win because they wont guess one of the last 2 values left
@pussymstrr yeah kk i see what your doing, i watch the ads to but if every decides to skip them then there is no point them being there and no one visits the site so please respect his decision to put them in there and expect them to be watched. also i don't appreciate your last 4 words, theres no need for it. and i just want to appologise for my bad language i will delete it right away.
@TheIrishboy13 Yep.
Hmm.......... for the probability of random (I don't think it's as simple as (12/13)^52, it only applies if the taken cards can be used again) it's..... err... the first pick is 12/13.... then for the second pick is much much much trickier, because you've taken 1 out... so... the second pick should be... 47/51 if they said the card that is not the one that was taken away, and 48/51 if they said the card that was taken away..... so the probability of being wrong for the second pick is..... 12/13x47/51 + 1/13x48/51, it's.. 612/663, then multiply the result with 12/13 to get the probability of being wrong twice in a row (144/169). The third one is EVEN MORE trickier.... because the second picked card can be the same (number) from the first one (that's... 13x(1/13)x(3/51)=3/51), but can still be different (number) with the first one (48/51). Like from the second scenario, he could say the one that's been taken (11/13 or 12/13, depending on the second card taken away), or the one that hasn't been taken away (or taken away twice) (2/13 or 1/13). So, the probability of being wrong on the third pick is: (48/51)x{(11/13)x(46/50) + (2/13)x(47/50)} + (3/51)x{(12/13)x(46/50) + (1/13)x(48/50)}, that's....... 30600/33150, for guessing the cards wrong 3 times in a row, it's (30600/33150)x(144/169) = 4406400/5602350 = 1728/2197
..... Sorry, it's the same after all!!!
For people who didn't read it because it was too long: At last, I found out that it's (12/13)^52 after all...
Probability if using strategy:
The first pick is absolutely 12/13 chance for being wrong. The second, if using strategy, is picking the same number as the previous one, so it's 48/51, in total, it's (12/13)x(48/51) = 576/663 = 192/221 (twice in a row)
For the third pick, if using the strategy above, it means player will automatically lose if 2 of the same number occurs twice in a row, so that isn't part of the equation, the next strategy is to pick the number either the first taken away or the second taken away number, so it's 47/50. (192/221)x(47/50) = 9024/11050 = 4512/5525
The fourth one is quite tricky, because the player has a chance to be very lucky and the third taken away number can be any of the first taken away number but not his guess (3/50). If the player was lucky, then it's obvious which card to guess: the one that's been taken twice, it has 47/49 chance of guessing false. The overall probability *to be lucky and to guess false* is (3/50)x(47/49) = 131/2450, but there is still a chance of not being lucky (47/50), and the strategy if the player isn't lucky is to pick either of the taken away cards (46/49 chance of being wrong), overall chance of both not being lucky and guessing false is (47/50)x(46/49) = 2162/2450
Overall chance of guessing wrong on the fourth pick is (2162 + 131)/2450 = 2293/2450, and the chance of guessing wrong four times in a row is (4512/5525)x(2293/2450) = 10346016/13536250 = 5173008/6768125
I will let someonw continue this.... it's already become too complicated for me....
Tried this on myself. The first 2 trys i got the first card right. Only works when you don't want it to.
I just wanted u to get a scam that I can impress my school do u recomed a video
hm.... we only have... 300 of them so far...
@zheranthan Yes it is
@BalliSkate When I looked at the card again I saw that there are 6 clubs, so yes you're right it's a 6.
27% chance of winning as a math geek with perfect memory.
Hey it's Destin from IGN!
You have a 12/13 chance of guessing wrong for every guess if you randomly choose a number.
First: 12/13 (92.3%)
Second:12/13 (92.3% x 92.3% = 85.2%)
Third: 12/13 (92.3%x92.3%x92.3% = 78.7%)
Etc. until you get to the 1.6% cumulative probability.
Assuming you aren't memorizing all the cards played, or counting cards (until you know all four of one number has been played), you still CAN increase your odds of not guessing a correct card by saying the previous card shown as your next guess.
---The first guess is still 12/13 chance of course (92.3%)
---The second guess (if you say the previous number) is 48/51 (94.1%) of guessing wrong since there are only three of those numbers left in the 51 card deck. 92.3% x 94.1% = 86.9%.
---So on the second guess you have a 1.7% increased chance of guessing wrong.
---After that, you start getting into some heavy maths because duplicates start showing up so the true probabilities aren't as clean cut as the second guess, but the trend continues with slightly better odds.
But if duplicates show up, you lose
Obviously, but that's not what I'm talking about.
I meant if, for example, you have you have seen two of that card previously (without knowing it, of course), so three of that number have passed in total, the odds that your next card would be the same as previous drops to 1/X instead of 2/X or 3/X (X being the number of cards left in the deck) like it would be if it were the first or second time you saw that number.
To clarify my original continuation, you have to start factoring in the odds that you may or may not have seen two, three or four of the previous cards -- without knowing it. Which gets more complicated than I can represent in a simple manner like the first two draws.
Songbirdo ah, I see.
And with a photographic memory, you have to take into account that eventually you'll run into two of the same card.
I guess the best way to figure it out is to try it face up so you can see every card and figure out how many times you lose. Or you could try to map all the probabilities... but that seems complicated, and probably takes more time than actually experimenting and seeing what you get.
You can do it mathematically. There's only 52 parts to the equation multiplied together, and I just gave you the first two pieces (12/13 x 48/51). And the last one is theoretically 12/13, too.
I'll leave the remaining 49 draws to you! lol
The last 10 draws are easy in comparison to the middle 40 as your number of possibilities collapse... Good luck!
Songbirdo Yeah, one can map the probabilities mathematically, probably with some sort of summation, but I'd rather honestly just do a hundred trials. Not to mention I have no idea what the ideal strategy is.
For your next epsiode you should teach how to always win tonk in one move
i want to learn how i can move the hole on the card!!!
do a remake on this trick using a marked deck
bet someone you can drink 2 pints of beer, faster than they can drink 1 shot glass worth of beer (or any other liquid). only rules are that you can't touch each others glasses, and they can't start until you finish your first beer and set the glass down.
when its time for you to set the first glass down, set it upside down over their glass, so they can't drink the shot. now you can drink the second beer at your leisure. impossible to lose
0.92^52
Guesses the same card the whole game after already seeing all four queens.😎
@SuperSoylent2
Good thing there's a fast forward option (; Don't cry.
Is that dailydestin from screwattack?
Damn. I got through 75% of the deck then I said king and it Poped up. That was with the ace, two, three, ext.
Made it through without getting one right on my second try wooooooo why am i happy about this lol
Can i reapet the same number after the four of them arent on the deck. For example the four aces are already out od the deck, and i say Ace all the time :S
but cuba!? well played, brian.
There's like only 4 suits, so four of the same type of number. Once you see that one value has all it's suits down, just repeat that number til the end. :)))
you're brilliant
@champjb25 can u tell me how to do this because the video for me froze so how do u do the trick
Wait but if you have a boat load of cash and you buy a boat then the leftover amount could fit in a space much smaller than a boat so if you were to buy the boat just for the purpose of storing the money it would be a complete waist
One of my friends did this to me and my friend about a ear ago, he got to the 41st card, and i lost on the first!
On my first try, I got the third card right. Same thing happened the second time. On my third try I guessed wrong all the way :)
PS: in a deck there are 4 different kinds then your done with that number forever
@l96sniper25 not just .92^52 ?
@thomaste33 your asking right now
.92^2^52 is the possibility do getting through without a strategy
@Russer24561 can u tell me how to do this because the video for me froze so how do u do the trick
@DJBigUm Yes; I got that they have to use advertisement - but that wasn't my original point. My point (the one you chose to discuss) was that I didn't wish to deal with watching it because I had to sit through too many commercials. That's it. I'm on youtube watching videos - I can go all day here without seeing all that; therefore, I choose not to watch this. End of point. I don't give a rat's ass why they do it - they still do it!
Is that Destin from screwattack/ign????