I am playing poker with three friends and from a well-shuffled deck we have each been dealt five cards. I have a hand consisting of the four kings and the two of hearts. Being a poker wizard I know exactly the probability that I have a winning hand.

But then I suddenly discover that earlier in the day the family’s children were playing with the cards and fed six of them to their pet shark (but I don’t know which). How does this information change the probability with which I believe my hand will beat my opponents? Why?

From http://www.statslab.cam.ac.uk/~rrw1/prob/exprob1.pdf

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    $\begingroup$ The real Questions should be... "Why do Children play with Pet Sharks?" And "Why do Sharks eat Cards?" $\endgroup$
    – Cryol
    Commented Sep 16, 2015 at 10:02
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    $\begingroup$ I'm sunk. In the original exposition it was a goat. I changed it to make the pun. $\endgroup$ Commented Sep 16, 2015 at 13:42
  • $\begingroup$ Question: What variation of poker is played? Like Texas Hold'em with cards open on the table, or all cards held in hands? $\endgroup$ Commented Sep 16, 2015 at 14:15
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    $\begingroup$ @TimCouwelier since he said they have been dealt 5 cards each, and says nothing about cards on the table… $\endgroup$
    – o0'.
    Commented Sep 16, 2015 at 15:59
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    $\begingroup$ Isn't the real question why the children have so many pet sharks that you don't even know which one they fed the cards to? $\endgroup$
    – Hellion
    Commented Sep 17, 2015 at 13:10

7 Answers 7


It changes...

nothing at all. Since you have no information on the cards lost, it's just like putting six random cards at the bottom of the deck, which could be 'absorbed' by the shuffle.

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    $\begingroup$ But don't you think there's a good chance that the children fed ones that looked like people to the shark (I.e jacks and queens) ;-) ? $\endgroup$
    – Gordon K
    Commented Sep 16, 2015 at 11:44
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    $\begingroup$ This is a valid observation; I can verify that it's common knowledge amongst Magic: the Gathering players that "mill" strategies don't work; That is to say, removing cards from the top of the opponent's deck nets you no "gain" on the opponent. This answer is an extension of that tried and true, proven logic. (There are exceptions to the "rule", but we all know that storm is broken) $\endgroup$
    – Kingrames
    Commented Sep 16, 2015 at 11:53
  • $\begingroup$ I've attempted a more elaborate proof.. alot of big numbers, to end up at some conclusion. $\endgroup$ Commented Sep 16, 2015 at 14:49
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    $\begingroup$ @Kingrames To clarify, by "mill strategies don't work," you mean "milling doesn't predictably affect the distribution of the remaining deck" and not "milling isn't a viable strategy," right? $\endgroup$
    – aebabis
    Commented Sep 16, 2015 at 20:49
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    $\begingroup$ @MarkPeters That's a discussion for another site. $\endgroup$
    – Kingrames
    Commented Sep 17, 2015 at 16:48

If your "Statistics intuition" is broken (like mine), here's a simpler example that might make it more clear.

I have a deck of 5 cards, numbered 1-5. My single opponent and I each draw a card, and the higher card wins.

I've draw a 4! My odds of winning, with a normal deck, are 75%, since the only card my opponent can win with is the 5.

Buuuut one of the cards has gone missing, and it could have been any card (other than the 4 I'm holding).

If the 5 is missing (25% chance), then I win automatically! This is a higher probability than a normal deck.

If any other card is missing (75% chance), I have a 66% chance of winning. This is a lower probability than a normal deck.

All told, my overall odds of winning are 25%(1) + 75%(0.66), or 75%. The same.

It turns out that by removing a card, any gains I would get by eliminating winning hands (the 5, or a Royal Flush in the real example) are countered by the fact that I also have a chance to eliminate losing hands (the 1, or a Flush).


What follows below assumes a variation of poker is played with no cards visible on the table

Hands to which you lose given a full deck:

  • 4 aces + random kicker
  • any straight flush

Given you have 5 cards, there's 47 left.
To get 4 aces + any random, chances are: 4/47 * 3/46 * 2/45 * 1/44 * 43/43 or 1.032/184.072.680
To get one particular straight flush, chances are: 5/47 * 4/46 * 3/45 * 2/44 * 1/45 = 120 / 184.072.680
There's 8 possible straight ranges (A-5 up to 8-Q) in 4 suits , so 32 possible straight flushes. We also need to scratch A-5 of hearts and 2-6 of hearts, so 30 options left. Odds are now 3.600 / 184.072.680 you lose to a straight flush.

Total odds of losing to one particular friend: 4632 / 184.072.680 (approx 1/37739), odds of losing to either of the three = 13.896 / 184.072.680 (approx 1/12580)

Now assume you have 5 cards, but there's only 42 left
Chances for each option remaining are calculated in a similar way, but with 42 cards rather then 47:
4 Aces + random kicker = 4/42 * 3/41 * 2/40 * 1/39 * 38/38 = 912 / 102.080.160
Any of the 30 straight flushes = 30 * 5/42 * 4/41 * 3/40 * 2/39 * 1/38 = 3.600 / 102.080.160

This however assumes that all the required cards are left remaining. For each of the above options, the occurence rate should be multiplied with the chances that set is actually possible with the remaining cards, or: the odds none of the required cards have been taken out.

For the aces, we need the odds that none of the 4 aces were in the 5 cards taken out of the set of 47. For the straight flushes, we need the odds that none of the 5 cards for a given straight flush were taken out of the set of 47.

To select 5 cards from a deck of 47, there are 47*46*45*44*43 = 184.072.680 combinations. The amount of combinations that does NOT hold at least one of the 5 cards is 42 * 41 * 40 * 39 * 38 = 102.080.160 To select 5 cards from a deck of 47, the amount of combinations without a single ace being in them is 43*42*41*40*39 = 115.511.760

Now if we multiply odds of hand beating us and odds of the hand being possible, and sum that over all hands:

912 / 102.080.160 * 115.511.760 / 184.072.680 = 1032 / 184.072.600 (same as normal)
Straight flushes
3.600 / 102.080.160 * 102.080.160 / 184.072.680 =3600 / 184.072.680 (same as normal)

Total odds of losing to one particular friend: 4632 / 184.072.680 (approx 1/37739), odds of losing to either of the three = 13.896 / 184.072.680 (approx 1/12580)

LONG ANSWER SHORT: It's the exact same

  • $\begingroup$ The math is a lot easier actually. if we assume that there is a 1 in 47 chance that you lose (just to establish an arbitrary number, very high, but whatever), the odds of that one card being drawn against you does not change, whether it was eaten or not, because the odds of it being left in the deck are not subject to any non-random treatment. $\endgroup$
    – Kingrames
    Commented Sep 16, 2015 at 20:58
  • $\begingroup$ I just wanted it written out without any intuïtive assumptions. While I agree the logic is 'right', I find it rather abstract. $\endgroup$ Commented Sep 17, 2015 at 7:12

Instead of feeding the cards to the shark, imagine you deal it in to the game. This doesn't change whether your hand is best among the human players.

Alternatively play in a casino but when it comes to burn a card, feed it to a shark. Obviously this makes no difference to the outcome of the game.

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I don't know for real the math behind it, but take in consideration a few things:

  • Not many hands in poker win from 'KKKK2': I don't play 5 Cards Poker, but i guess , if it goes by the same winning principle that Holdem have , the only hands that could win against that are: 'AAAAX'(let X be a random card) , 'AKQJT'(all of the same kind, or Royal Straight Flush ) and Straight Flushes . That said, if you have the four kings in hand, nobody can make the 'AKQJT' combination . With that in mind , only 'AAAAX' and Straight Flushes < K can beat you. In Texas Hold'em ( or 2 cards poker ) , the chance of making 'Four of a kind' is '4,164 : 1' or 0.0240% , and for Straight Flushes there are 0.0014% or 72,192 : 1 odds , which is even worse.

    With that in mind, we can go to the second point:

  • We need to know what is the probability of one of those 6 cards being an ACE , which would make the 'AAAAX' combination impossible . If you have a 52 cards deck, and remove 5( 'KKKK2' ) , there is 47 cards left. On those 47, what's the chance on the removal of 6 that one will be an ACE? Pretty low chance, lower than 0.0240% .

  • Going by the same principle that you can't make a combination if you don't have that card on the deck , we don't even need to recalculate the probability for straight flushes , because they're going to be even harder to get , so less than 0.0014% of chance.

Well , I don't know the probability for any of those cases , but at least it would make me confortable enough to go 'all-in' on that hand.

Thanks Tim Couwelier for the remind, i completely forgot about straight flushes.

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    $\begingroup$ Straight flush trumps four of a kind, not only a 'royal' flush. $\endgroup$ Commented Sep 16, 2015 at 14:17
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    $\begingroup$ The only question is whether it changes your probability of winning. Since it's equivalent to moving the six cards to the bottom, it doesn't. $\endgroup$
    – Deusovi
    Commented Sep 16, 2015 at 14:58

This is a complicated (almost unanswerable) question as there is no known implementation of optimal strategy in Poker.

By seeing the cards that you know are not eaten (your cards + cards opened on the table), you get a list of cards of which you know for sure the 6 cards will be in. You will have to account for the probability of each card being missing in your strategy. As you keep playing many hands, you will eventually get to know which cards are missing.

Until then, you will even have to account for information that will be revealed in the future, and how this benefits you and the opponent right now and later on. For example, you will have a probability distribution of all cards and the probability of them being missing. You will also have a probability distribution of how each card opened is going to affect your and your opponent's position currently. You will then need to compute your and your opponent's future possible states with probabilities. In each of these possible scenarios (which number in billions and trillions), you will have to compute how new information that is assumed to be revealed and that which is yet to be revealed affect your position in the game.

What is most interesting is that you may even have to take into account future hands, when playing a given hand. An extremely rare circumstance is that you would voluntarily give up the current hand and fold, so that your cards remain secret, and this will benefit you in all future hands.

As of today, I think this is all beyond our computation. Most humans will probably assume the deck is complete (until they know the missing cards), and computers will fare only slightly better.

  • $\begingroup$ The only question is whether it changes your probability of winning. Since it's equivalent to moving the six cards to the bottom, it doesn't. $\endgroup$
    – Deusovi
    Commented Sep 16, 2015 at 14:58
  • $\begingroup$ @Deusovi Yes, it will slightly change your probability if you continue to employ the same strategy because you are not taking into account future hands, and your opponent is. $\endgroup$ Commented Sep 16, 2015 at 18:03
  • $\begingroup$ @Deusovi It might even improve your probability of winning the current hand at the cost of the probability of winning later hands. $\endgroup$ Commented Sep 16, 2015 at 18:04
  • $\begingroup$ As long as you don't get to the last 6, it makes no difference whether the last six cards are eaten or sitting on the bottom. And since you don't know which six cards they are, then it doesn't matter - the distribution is even, so it's just like a normal shuffle. Once you reshuffle though, I agree that it changes the probability - you can use information from the previous hands to figure out what cards might be missing. The question only asks about the current hand, though, $\endgroup$
    – Deusovi
    Commented Sep 16, 2015 at 18:09
  • $\begingroup$ @Deusovi 1. They have not mentioned whether this is the only hand to be played. Hence I assumed it is not. 2. The cards may sit at the bottom but we don't know that. We will also consider the possibility that they do get missed out. 3. For one hand only, your answer is perfectly fine. If there is more than one hand to be played, the opponent will be willing to sacrifice the current hand marginally, if this improves his position in later hands. $\endgroup$ Commented Sep 16, 2015 at 18:32

If the children fed random cards to the shark, and neither you nor your opponents know which ones, then the probabilities haven't changed.

If either you or one of your opponents know which cards were fed to the shark, that player would have an advantage; they could calculate more precisely what is the probability of winning or losing with the hand they have.

Same if the cards are not picked randomly: Whoever knows how to eaten cards were picked has the same advantage as before.


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