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Let's say there is a deck of 40 cards with 4 aces (just take any 12 non-ace cards of a regular deck for instance)

We deal three cards for each player (so 40-12=24 remain undealt) what is the probability of exactly each player having one ace?

I was thinking that a player has $\frac{4}{40}$ chance of having one ace on the 1st card then the second player has $\frac{3}{39}$ and so on, totalizing $\frac{4!}{39! - 35!}$, but this doesn't work because I just calculated the possibility of each player getting an ace on the 1st card dealt. It could also be that player 1 gets not an ace on the first card dealt but an ace on the second (remember that each player gets 3 cards)

So should I take into account the way the cards are dealt as well? Because for instance concerning the 1st two cards: say the first is an ace for player 1 $\frac{4}{40}$ then the chance of the second being an ace for player 2 is $\frac{3}{39}$. But if the player 1 first card was not an ace, that probability is changed to $\frac{4}{39}$, since there are still 4 aces left on the deck... How could I take all the possibilities in consideration then? Sounds like a pretty boring task since there are $12 \choose 2$ possible configurations, considering the ways the 12 cards are deal to each player... Is there a better way to compute this probability?

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  • $\begingroup$ NB: $40-12= 28$ $\endgroup$ – Graham Kemp Nov 1 '18 at 1:33
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The first player gets exactly one ace with probability $\dfrac{\binom{4}{1}\binom{36}{2}}{\binom{40}{3}}$ seen by picking which ace followed by picking which two other non-ace cards are used compared to the total number of three-card hands possible.

Given that the first player received exactly one ace, the probability that the second player then receives exactly one ace given what cards remain in the deck occurs with probability $\dfrac{\binom{3}{1}\binom{34}{2}}{\binom{37}{3}}$

Applying multiplication principle, the probability then that both the first and second player receive exactly one ace is $\dfrac{\binom{4}{1}\binom{36}{2}}{\binom{40}{3}}\times\dfrac{\binom{3}{1}\binom{34}{2}}{\binom{37}{3}}$

Continuing in this fashion you get one representation for the probability that every player receives exactly one ace as:

$$\frac{4!\binom{36}{2}\binom{34}{2}\binom{32}{2}\binom{30}{2}}{\binom{40}{3}\binom{37}{3}\binom{34}{3}\binom{31}{3}}$$

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  • $\begingroup$ Note: "We deal three cards to each player" ... $\endgroup$ – Graham Kemp Nov 1 '18 at 1:05
  • $\begingroup$ Good catch. I'll correct once trick or treaters are done and I go upstairs $\endgroup$ – JMoravitz Nov 1 '18 at 1:12

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