Thread: Probability Question View Single Post
#3
12-07-2005, 01:06 PM
 pzhon Member Join Date: Mar 2004 Posts: 66
Re: Probability Question

a. 106/5 = 21.2.
b. 53/2 = 26.5.

A brute-force method:

For the 2nd ace to be in the nth position, there must be 1 ace in the first n-1 positions, and 2 aces in the last 52-n. There are (n-1)C1*(52-n)C2 ways to arrange these, so the probability that the 2nd ace is in position n is (n-1)C1*(52-n)C2/(52C4). Sum n*(n-1)C1*(52-n)C2/(52C4) to get 106/5.

The same idea works for the 7th club. However, it is possible to read off the answers above with no complicated calculation by using a symmetry argument.

Imagine adding a 5th ace to the deck. Shuffle, then cut to a random ace and remove it. The average distance from the cut ace to the second ace remaining must be 53 * 2/5 since the 5 possible aces give you 5 such distances and these intervals wrap around the 53-card deck twice, so they add up to 53*2.

Example: Suppose the 5 aces are in positions 4, 10, 20, 23, and 45 in the 53-card deck.

If you cut to the ace in position 4, the second ace will be in position 16.
If you cut to the ace in position 10, the second ace will be in position 13.
If you cut to the ace in position 20, the second ace will be in position 25.
If you cut to the ace in position 23, the second ace will be in position 34.
If you cut to the ace in position 45, the second ace will be in position 18.

16+13+25+34+18=106, so the average among these is 106/5.