A deck of 52 cards is fairly dealt to 4 hands (with 13 cards each). In this prob

Question

A deck of 52 cards is fairly dealt to 4 hands (with 13 cards each). In this problem we find the probability that one hand gets 3 aces. You may assume that 13 cards are dealt to hand 1, then 13 cards are dealt to hand 2, etc., although the answer does not depend on the order cards are dealt. An outcome is 4 hands of 13 cards, with hands ordered as 1, 2, 3 and 4, but cards within a hand unordered. Define the following events: S = set of all outcomes. A = one hand gets 3 aces, B = hand 1 gets 3 aces and hand 2 gets 1 ace, C = hand 1 gets 3 particular aces and hand 2 gets remaining ace. Let |D| denote the number of outcomes in an event D. (a) Find |S|. (b) Express |A| in terms of |B|. (c) Express |B| in terms of |C|. (d) Find |C|. (e) Combine the above results to determine Pr (A).

Explanation / Answer

(a)

There are 4 outcomes- 4 hands where each hand has 13 cards in any order.

In a standard deck number of cards is 52. Number of ways of selecting 13 cards for first hand is C(52,13). After that number of cards remaining is 39 so number of ways of selecting 13 cards out of 39 is C(39,13). Then number of ways of selecting 13 cards for third hand out of remaining 26 cards is C(26,13). And number of ways of selecting 13 cards out of remaining 13 cards is 1.

So possible number of ways four hands can be selected is

C(52,13)C(39,13)C(26, 13)C(13,13) = 5.3644738e+28

Hence, |S| = C(52,13)C(39,13)C(26, 13)C(13,13) = 5.3644738e+28

(b)

|A| shows the number of ways any one of four hands can get 3 aces.

|B| shows the number of way hand 1 get 3 aces and hand 2 get 1 ace.

Set A will be super set of B.

Number of ways of chossing one hand for three aces out of 4 is C(4,1)= 4 and then one hand for one ace out of 3 is C(3,1)= 3 so

|A| = 4*3*|B| = 12 * |B|

(c)

|B| shows the number of way hand 1 get 3 aces and hand 2 get 1 ace.

|C| shows the number of way hand 1 get 3 particular aces and hand 2 get 1 remaining ace.

Set B will be super set of C.

In set B, we need to chose 3 aces out of 4. That is

|B| = |C| * C(4,3) = 4 *|C|

(d)

Number of ways hand 1 can have 3 parituclar is 1 and number of ways of selecting rest 10 cards out of remaing 48 cards (excluding 4 aces out of 52) is

1 * C(48,10) = 6540715896

and number of ways of selecting 12 cards of second hand from remaning 38 cards is

C(38,12) = 2707475148

Rest two hands can be selected in C(26,13)C(13,13) ways.

So

|C| = C(48,10) C(38,12) C(26,13)C(13,13)

(e)

So

|A| = 12|B| = 12*4 *|C| =  12* 4 * C(48,10) C(38,12) C(26,13)C(13,13)

Thats is

Pr(A) = |A| /|S| = 0.1648

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