Probability That At Least One of Three Players Gets Two Aces When Each Player Is Dealt Cards

Question

Suppose that from a standard deck of 52 cards, three players are each dealt two cards. What is the probability that at least one player receives two Aces? How would I use the Inclusion-Exclusion Formula to answer this question?

Answer 1

There is a systematic direct way to solve it.

Total possibilities are given by $\Large\binom{52}{2}\binom{50}{2}\binom{48}{2}$ which will form the denominator $D$

Denoting aces by A, and non-aces by B, there are only $4$ "good" patterns:

$AA|AA|BB: \;\Large\binom42\binom22\binom{48}2 \times\frac{3!}{2!}$

$AA|AB|AB:\;\Large\binom42\binom21\binom{48}1\binom11\binom{47}1\frac{3!}{2!}$

$AA|AB|BB:\;\Large\binom42\binom21\binom{48}1\binom{47}2\times 3!$

$AA|BB|BB:\;\Large\binom42\binom{48}{2}\binom{46}{2}\times\frac{3!}{2!}$

Adding up the above will give the numerator $N$,

and the answer for the Pr = $\Large\frac{N}{D}$

Answer 2

For player 1:

they get 0 ace, proba is $(48 \times 47)/(52 \times 51)$

they get 1 ace, proba is $(2 \times 4 \times 48)/(52 \times 51)$

they get 2 aces, proba is $(4 \times 3)/(52 \times 51)$

Sum is 1, good.

If player 1 gets 0 ace, Proba for player 2 :

they get 0 ace, proba is $(46 \times 45)/(50 \times 49)$

they get 1 ace, proba is $(2 \times 4 \times 46)/(50 \times 49)$

they get 2 aces, proba is $(4 \times 3)/(50 \times 49)$

If player 1 gets 1 ace, Proba for player 2 :

they get 0 ace, proba is $(47 \times 46)/(50 \times 49)$

they get 1 ace, proba is $(2 \times 3 \times 47)/(50 \times 49)$

they get 2 aces, proba is $(3 \times 2)/(50 \times 49)$

If player 1 gets 0 ace and player2 gets 0 ace, Proba for player 3 :

they get 0 ace, proba is $(44 \times 43)/(48 \times 47)$

they get 1 ace, proba is $(2 \times 4 \times 44)/(48 \times 47)$

they get 2 aces, proba is $(4 \times 3)/(48 \times 47)$

If still 3 aces available, proba for player 3 :

they get 2 aces, proba is $(3 \times 2)/(48 \times 47)$

If still 2 aces available, proba for player 3 :

they get 2 aces, proba is $(2 \times 1)/(48 \times 47)$

Total : $(4 \times 3)/(52 \times 51)$ $+(48 \times 47)/(52 \times 51) \times (4 \times 3)/(50 \times 49) + (2 \times 4 \times 48)/(52 \times 51) \times (3 \times 2)/(50 \times 49) +(48 \times 47)/(52 \times 51) \times (46 \times 45)/(50 \times 49) \times (4 \times 3)/(48 \times 47)$ $+((48 \times 47)/(52 \times 51) \times (2 \times 4 \times 46)/(50 \times 49)$ $ + (2 \times 4 \times 48)/(52 \times 51) \times (47 \times 46)/(50 \times 49)) \times (3 \times 2)/(48 \times 47)$ $+(2 \times 4 \times 48)/(52 \times 51) \times (2 \times 3 \times 47)/(50 \times 49) \times (2 \times 1)/(48 \times 47)$

=1.356%