I'm trying to calculate the probability of a 'card' being beaten for a certain card game for a school project and require some assistance with my calculations.

So a given card can be beaten by cards of the same suite with higher value, or by cards of the trump suit (determined each round), what I want to calculate is for a given card, what is the probability of someone else drawing, and being able to play a card to beat it, where you play your cards around from player 1 to 4 and then the 'highest' card wins.

So for any card, there is some set of cards that can beat it, say N, and there is a set amount of draws, say M. To calculate the probability of one of the cards in N being drawn, I am using hypergeometric distribution to calculate this, but I believe this is only the first part of what I need to calculate.

So for example: if you are playing first in a round, then you play your card and everyone else plays a card (lets say this round everyone is dealt 4 cards), so the number of cards drawn is 16, so I'm using hypergeometric distribution where population size = 52, number of succeses (cards that can beat the card from your hand) = 25 (in this situation), sample size = 16 and we want X >= x, so this gives the probability of someone around the table having one or more cards that can beat yours. But theres also the information that you may or may not have one of those cards in your hand, which changes the number of successes in the sample, right? The other thing to consider would be that if you were playing second in a round, you only have to consider two people in front of you, so do you use sample size = 8? (as thats how many still to play?)

As you can see im quite unsure with these calculations!

Thanks! Jasper


Let $h$ be the number of cards left per hand, $p$ the number of players playing after you in this round, $c$ the total number of cards in play and $b$ the number of cards in play that can't beat your card. By "in play", I mean cards that the other players could hold, i.e. cards that you haven't seen yet. This excludes the cards that you've already seen played (including in this round) and the cards on your own hand; it includes the cards on the other players' hands and the cards that weren't dealt.

Then the probability that no-one holds a card that can beat yours is

$$ \frac{\binom b{ph}}{\binom c{ph}}=\frac{b!(c-ph)!}{c!(b-ph)!}\;, $$

the number of ways to select $ph$ cards from the cards in play that won't beat yours over the total number of ways to select $ph$ cards from the cards in play.

So for instance, if there are $4$ players and $52$ cards, $27$ of which can't beat yours, each hand has $4$ cards left, you're going second and you've seen $13$ cards ($9$ played and $4$ left on your hand), $6$ of which couldn't have beaten your card, then the probability that no-one can beat you is

$$ \frac{\binom{27-6}{2\cdot4}}{\binom{52-13}{2\cdot4}}=\frac{21!31!}{39!13!}=\frac{35}{10582}\approx0.33\%\;, $$

since $p=2$, $h=4$, $b=27-6$ and $c=52-13$.

  • 1
    $\begingroup$ Awesome. Thanks for your help :) $\endgroup$ – Tehmo3 May 15 '16 at 6:33

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