Difference between revisions of "1975 USAMO Problems/Problem 5"
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Also, note that when second ace is still the second ace when we draw the deck backwards. (we need to note this, else this argument holds for the first ace and third ace as well) | Also, note that when second ace is still the second ace when we draw the deck backwards. (we need to note this, else this argument holds for the first ace and third ace as well) | ||
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Now, this implies that the expected location of the second ace is in the <math>(k+n+1-k)/2=(n+1)/2</math> place. | Now, this implies that the expected location of the second ace is in the <math>(k+n+1-k)/2=(n+1)/2</math> place. | ||
Revision as of 11:22, 18 May 2014
Problem
A deck of playing cards, which contains three aces, is shuffled at random (it is assumed that all possible card distributions are equally likely). The cards are then turned up one by one from the top until the second ace appears. Prove that the expected (average) number of cards to be turned up is
.
Solutions
Solution 1
We begin by induction. Our base case is naturally when , as there can be no less than
cards in the deck. The only way to turn up the second ace is to turn up the first, and then turn up the second, which requires
moves. This indeed is equal to
. Next, we assume that the expected number of cards required to pull the second ace from a deck of n cards is in fact
. Now, we see that to simulate the case with
cards, we simply must add a card to our deck. By our assumption, we can expect to turn up
cards to be turned up, including our second ace. Thus ahead of our second ace, we expect
cards , and after it
as well. When we add our card, it is ahead of the second ace half the time, and behind if half the time. Therefore we expect to add one card to the number of cards drawn half the time, and add nothing for half the time. Hence we expect to add half a card, and the expected value is now
, or
, the desired outcome for the
case.
Solution 2
Although we can induct as in the situation above, we can use a more direct approach.
Consider a deck. The second ace is in the th place. If we draw the deck backwards (which has equal probability as drawing it forwards), then we will draw it in the
th place, since the
th place will become the first, the
th the second, etc.
Also, note that when second ace is still the second ace when we draw the deck backwards. (we need to note this, else this argument holds for the first ace and third ace as well)
Now, this implies that the expected location of the second ace is in the place.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.