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 evanski 12-07-2005 07:37 AM

Probability Question

You shuffle a deck of 52 cards and start dealing off the top.

a. How many cards should you expect to turn over before you get to the second ace?

b. The 7th club?

-Evan

 LetYouDown 12-07-2005 11:53 AM

Re: Probability Question

Intuitively, 25 and 27 respectively. I'm trying to determine the flaw in my logic as we speak.

 pzhon 12-07-2005 01:06 PM

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.

 12-07-2005 01:29 PM

Re: Probability Question

That's really cool.

 evanski 12-07-2005 11:35 PM

Re: Probability Question

Thanks.

 BruceZ 12-08-2005 03:20 AM

Re: Probability Question

[ QUOTE ]
You shuffle a deck of 52 cards and start dealing off the top.

a. How many cards should you expect to turn over before you get to the second ace?

[/ QUOTE ]

Consider the 4 aces plus 1 non-ace. In order for the non-ace to occur before the 2nd ace, the non-ace must occur in positions 1 or 2 of these 5 cards, and this has a probability of 2/5. This is true for each of the 48 non-aces, so adding these 48 probabilities of 2/5 gives the expected value of the number of non-aces before the 2nd ace, which is 48*2/5. Adding the 2 aces, the expected value of the number of cards occurring up to and including the 2nd ace is 48*2/5 + 2 = 21.2.

[ QUOTE ]
b. The 7th club?

[/ QUOTE ]

Consider the 13 clubs plus 1 non-club. In order for the non-club to occur before the 7th club, the non-club must occur in positions 1-7 of these 14 cards, and this has a probability of 7/14 or 1/2. This is true for each of the 39 non-clubs, so adding these 39 probabilities of 1/2 gives the expected value of the number of non-clubs before the 7th club, which is 39*1/2. Adding the 7 clubs, the expected value of the number of cards occurring up to and including the 7th club is 39*1/2 + 7 = 26.5.

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