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Three players are each dealt, in a random manner, five cards from a deck containing 52 cards. Four of the 52 cards are aces. Find the probability that at least one person receives exactly two aces in their five cards.

Effort: I think the probability of one person getting two aces would be 3*(4 choose 2)*(48 choose 3)/(52 choose 5), but assuming that the first person gets two aces, to find the probability of the second person, I think this might require some sort of multinomial coefficient, I am just unsure of what that is!

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  • $\begingroup$ What are your thoughts on the problem. On this forum you are expected to show your efforts $\endgroup$ – Shailesh Sep 29 '15 at 2:28
  • $\begingroup$ Hint: Call the players A, B, and C. The probability that A gets exactly $2$ Aces is $\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$. If we multiply by $3$, we will double-count, for example, the case where each of A and B gets $2$ Aces. $\endgroup$ – André Nicolas Sep 29 '15 at 2:32
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The total space is all the ways to select three hands of five cards. $$\begin{align}\dbinom{52}{5,5,5,37} & = \dbinom{52}{5}\dbinom{47}{5}\dbinom{42}{5} \\ & = \dfrac{52!}{{5!}^3\,47!}\end{align}$$ Now, for exactly one person receives has two aces, the favoured space would be the way to draw one of the three hands with exactly two of the four aces (and three other cards), and to draw the other two hands so that neither one holds both of the remaining aces.

On the other hand, for two people receive two aces, the favoured space would be the way to select those hands, the way those hands can be selected, and the way the third hand could be selected.

Effort: I think the probability of one person getting two aces would be $3{4 \choose 2}{48 \choose 3}\big/{52 \choose 5}$, but assuming that the first person gets two aces, to find the probability of the second person, I think this might require some sort of multinomial coefficient, I am just unsure of what that is!

Yes, that's a good start.   Can you complete? $$\dfrac{3\binom{4}{2}\!\binom{48}{3}\left(\Box-2\Box\right)+3 \binom{4}{2,2}\binom{48}{3,3,5,37}}{\binom{52}{5}\binom{47}{5}\binom{42}{5}}$$

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  • $\begingroup$ $\binom{52}{5}\binom{48}{5}\binom{44}{5}$ ?? $\endgroup$ – true blue anil Sep 29 '15 at 17:56
  • $\begingroup$ A typo, yes. Thanks for pointing it out. $\endgroup$ – Graham Kemp Sep 29 '15 at 21:49
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Here's another way, taking the aces to be prize winning #s in a lottery.

Imagine the 52 numbers are divided into 4 groups $A-B-C-D\;$ of $\;5-5-5-37$,
and we want to count either A alone has exactly $2$ aces, or $A-B$ have 2 aces each.

There aren't too many patterns to count, we can just add up.

$2-1-0-1\text{ and } 2-0-1-1 : 2{5\choose2}{5\choose 1}{37\choose1}$

$2-1-1-0 : {5\choose 2}{5\choose1}{5\choose1}$

$2-0-0-2 : {5\choose 2}{37\choose 2}$

$2-2-0-0 : {5\choose 2}{5\choose 2}$

We counted only $A$ and $AB$, so we need to multiply by $3$

$Pr = \dfrac{3\left[2\dbinom{5}{2}\dbinom{5}{1}\dbinom{37}{1}+ \dbinom{5}{2}\dbinom{5}{1}\dbinom{5}{1}+\dbinom{5}{2}\dbinom{37}{2}+\dbinom{5}{ 2}\dbinom{5}{2}\right]}{\dbinom{52}{4}}$

You might find this more straightforward.

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Use inclusion-exclusion. Call the players A, B, and C. The probability of only one player getting two aces will be 3 times the probability of any one of them getting two aces minus the instances where two of the players get two aces: P(A)+P(B)+P(C)-{P(AB)+P(AC)+P(BC)} So,3*(4 choose 2)(48 choose 3)/(52 choose 5)-3*(4 choose 2)(48 choose 3,3, and 42)/(52 choose 5,5, and 42).

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  • $\begingroup$ Please use MathJax and format your answer in $\LaTeX.$ $\endgroup$ – Adrian Keister Jul 30 '18 at 13:03

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