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2007 Alabama ARML TST Problems/Problem 10

Problem

Andrew and Patri each have a deck of cards. Andrew draws two cards from his deck, both of which turn out to be queens. Patri draws two cards from his deck, one of which is an ace, and the other of which is a king. Andrew and Patri then put the remainder of their decks (50 cards left in each deck) down on a table and race to the nearest river. Meanwhile, Chelsea comes along and picks up one of the decks at random. She randomly pulls two more cards out of that deck. It turns out Chelsea’s cards make a pair (both of the same rank, i.e., both are 2’s, both are 3’s, etc.). Find the probability that she drew those two cards from Andrew’s deck.

Solution

We will show how to derive the answer using high school mathematics only.

Let's imagine that Chelsea repeated her experiment $X$ times, where $X$ is very large. In $X/2$ cases she picked Andrew's deck, and in $X/2$ she picked Patri's one.

There are ${50\choose 2} = 1225$ ways to pick two cards from Andrew's deck. Out of these, in $12\cdot {4\choose 2} + {2\choose 2} = 73$ cases we will get a pair. (For queens we have to pick two out of the remaining two, and for any other value we have to pick two of the four cards that have it.)

Thus in $73/1225$ out of the $X/2$ cases where she picked Andrew's deck she will get a pair. Let's call this case "situation A", and let "situation B" be the case where she picked Andrew's deck, but did not get a pair.

Similarly we get that in $72/1225$ out of the $X/2$ cases where she picked Patri's deck she will get a pair ("situation C"), and in the other part of these cases ("situation D") she will pick Patri's deck, but she won't get a pair.

The problem statement gave us the information that Chelsea did get a pair. This means that we are observing one of the experiments that ended either in situation A or in situation C. There were $73X/2450$ cases that resulted in situation A, and $72X/2450$ cases that resulted in situation C. Thus the ratio of situations A to situations C is $73:72$, hence the probability that we just saw situation A is $73 / (72 + 73) = 73/145 \simeq 50.345\%$.

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 9 		Followed by:
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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