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2007 Alabama ARML TST Problems/Problem 8

Revision as of 16:55, 28 January 2009 by Misof (talk | contribs) (New page: ==Problem== Krugman is a little odd. Really odd actually. He has a collection of Massachusetts state quarters and New Hampshire state quarters. All 29 of his Massachusetts quarters are kep...)
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Problem

Krugman is a little odd. Really odd actually. He has a collection of Massachusetts state quarters and New Hampshire state quarters. All 29 of his Massachusetts quarters are kept face up, while all 11 of his New Hampshire quarters are kept face down on display in his office. Eric, being the prankster that he is, sneaks into Krugman’s office and randomly turns over 20 of Krugman’s quarters. Find the expected number of Krugman’s quarters that are heads up when Eric is finished.

Solution

Solution 1

First thing to note is that once we know the number $j$ of New Hampshire quarters Eric flipped, we know the final number of quarters that are heads up: These will be the $j$ flipped New Hampshire quarters, and the $29-(20-j)=9+j$ not flipped Massachusetts quarters.

We could now solve the task by brute force: for each $j\in\{0,\dots,11\}$ we can compute the probability $p_j$ that Eric flips exactly $j$ of the New Hampshire quarters, and then compute the sum $\sum p_j (9+2j)$.

However, one more observation can help us avoid this tedious computation: The number of quarters Eric flips is in this case exactly one half of the total number of quarters. We can now take all the $40\choose 20$ possibilities which quarters Eric picked, and divide them into complementary pairs.

Pick one such pair. If Eric flipped $j$ of the New Hampshire quarters in one of the possibilities, he flipped the other $11-j$ New Hampshire quarters in the other possibility. In the first case, he would get $9+2j$ quarters heads up, in the second case there would be $9+2(11-j)=31-2j$ such quarters. The average of these two values is $\frac{(9+2j) + (31-2j)}2 = 20$. And as this holds for each pair of possibilities, the expected number of quarters that are heads up is obviously $20$.

Solution 2

As the number of quarters Eric flips is in this case exactly one half of the total number of quarters, each quarter has probability $1/2$ that it will be flipped. Therefore for each quarter the probability that it will be heads up at the end is $1/2$, and the expected number of quarters that are heads up is $40\cdot(1/2)=20$.

See also

2007 Alabama ARML TST (Problems)
Preceded by:
Problem 7 		Followed by:
Problem 9
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