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Two players each draw a single card, in turn, from a standard deck of 52 cards, without returning it to the deck. The winner is the player with the highest value on their card. If the value on both cards is equal then all cards are returned to the deck, the deck is shuffled and both players draw again with the same rules.

Given that the second player is drawing from a deck that has been modified by the first player removing their card, I'm wondering if either player is more likely to win than the other?

Does this change as the number of players increases?

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up vote 4 down vote accepted

No, each player wins with probability $\frac 12$. If there are $n$ players, each wins with probability $\frac 1n$. Maybe the easiest way to see this is to ask what is the chance the second player draws a particular card, say the ace of spades. The chance is $\frac {51}{52}$ (that the first player didn't take it) times $\frac 1{51}$ (that the second player does take it), making $\frac 1{52}$. So the chance the second player gets any given card is the same as the first player.

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Suppose I'm the first player, and I rig the deck in such a way that you always get the same card as I do, in the same suit, but one rank lower (except that if I get a $2$, you get an ace). Then I'll beat you $12$ times out of $13$, even though your chance of getting any particular card is $1/52$, just like mine. The two players having the same probability distribution is neither necessary nor sufficient for a fair game. (With independence, it would be; but here the draws are not independent.) – mjqxxxx Nov 8 '12 at 17:23
@mjqxxxx: True, but it is implied that the deck is shuffled. – Ross Millikan Nov 8 '12 at 17:26
Of course. And the game is fair. My point is that your argument doesn't actually prove that. You could make the same argument in the rigged case I describe, but in that case the game is unfair. – mjqxxxx Nov 8 '12 at 17:29

Another easy way of seeing that the probability must be $\frac{1}{2}$ for each of the two players (and in general, $\frac{1}{n}$ for each of $n$ players): we can build a one-to-one map from configurations where player A wins to configurations where player B wins by simply swapping the two cards drawn - that is, the configuration where A wins by drawing the 9 of Spades while B draws the 4 of Clubs maps to a configuration where A loses by drawing the 4 of Clubs while B draws the 9 of Spades. This means there must be exactly as many drawings where B wins as where A wins, and so each player has a 50-50 chance of winning.

One caveat: don't confuse this a priori fairness of the game with the a posteriori odds after A has drawn - once A has drawn then the probability that he wins the game changes. For instance, if A draws an Ace then he has a $\frac{48}{51}$ chance of winning immediately and a $\frac{3}{51}$ chance of having to shuffle and redraw (after which, as we've just shown, his chances of winning are $\frac{1}{2}$), meaning that his overall odds of winning are $\frac{48}{51}\cdot 1+\frac{3}{51}\cdot\frac{1}{2} = \frac{99}{102}$, roughly 97%.

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If the second player were drawing from a full deck, he would draw each of the $13$ ranks with equal probability. The only change when he draws from the $51$-card deck that remains after the first player’s draw is that the rank of the first player’s card becomes less probable; the other $12$ ranks remain equally likely. Thus, given that the game is decided in this round, the second player’s probability of winning is the same for both decks, namely, $\frac12$. The only effect of not replacing the first player’s card is to decrease the expected number of tied rounds before the game is won or lost.

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All orderings of the cards are equally likely. It follows that the probability the first player wins is equal to the probability the second player wins.

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