Correction to Homework - Part 2

In my June 11th article I addressed two questions about the situation where all 52 cards in a deck of cards are dealt out to four people (such as in a game of Bridge). One of the questions asked was to calculate the probability that one player's hand contained all four Aces. My solution to this problem was incorrect. This was brought to my attention by my good friend and colleague Stewart Ethier, who is a Professor of Statistics at the University of Utah. My thanks to Stewart. Interestingly enough, Stewart was the only one of the many readers who caught the error and flagged me concerning it. Here are the details.

I correctly pointed out that an Ace is as likely to be in one hand as in another. Then I looked at possible pattern types for the Ace distributions and created the following table:

Then I wrote the following. "Since each of these patterns is equally likely, the probability that the four Aces are all in one hand is 4/35 or approximately 11.4286%." This is not true. I was attempting to use a calculation that avoided the huge numbers that arise in problems such as this and in doing so I carelessly inferred that since Aces are as likely to be in one hand as in another that the above patterns are also equally likely. Not so.

What is the correct solution? Recall the notation we frequently use here for n things chosen k at a time: C(n, k). If we put the four Aces in the first position and none in the second, third, and fourth positions there are

C(4,4) C(0,0) C(0,0) C(0,0)

ways of doing this and this number is just 1. This leaves us with 48 cards. If we choose 9 of these for the first position, there are C(48,9) ways of doing this. This done, there are C(39,13) ways of choosing the 13 cards for the second position, C(26,13) ways of choosing the 13 cards for the third position, and C(13,13) or 1 way of choosing the 13 cards for the fourth position. Altogether then there are

C(48,9) C(39,13) C(26,13) C(13,13)

ways of choosing four 13-card hands with 4 Aces in the first hand.

Continuing in this fashion there are C(52,13) C(39,13) C(26,13) C(13,13) ways of choosing four 13-card hands, period. The ratio of these two large numbers is the probability of dealing four 13-card hands with all four Aces in hand 1. This looks hard to calculate and it is if you do it the wrong way. But the ratio of these two numbers provide for a lot of cancellations and if you reduce the fraction to lowest terms the answer is

11/(17 x 5 x 49)

So my table above should have had another column labeled probability and the above number should have been opposite the (4, 0, 0, 0) entry. Since the four Aces can go in either of the 4 available hands, the number we are looking for is

44/(17 x 5 x 49)

And this number expressed as a percentage is just 1.0564225%. Sorry for my error but this should set things straight. Oddly enough the calculation in part b) is essentially what we did here and I got that right. Strange! Thanks again, Stewart.