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Three of a Kind (Posted on 2003-11-19) Difficulty: 4 of 5
You have a standard pack of 52 playing cards. You then shuffle them and begin to draw out cards until you have three of a kind. What is the most likely number of cards drawn when this happens?

You then shuffle another pack of 52 playing cards into the pile. What happens to the expected number of cards now? (i.e. does it double / halve / stay the same?)

No Solution Yet Submitted by Lewis    
Rating: 4.3333 (9 votes)

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Solution Described for One Case | Comment 36 of 39 |
The following is an example of the calculation involved for a particular value of n, in the 1-deck case. We'll do the case of n=16--completion of the first triplet at the 16th draw.

There are 52 possible choices for the triplet-completing card. Then that must be multiplied by the number of combinations of cards that could precede that card and still have that card be the first completion of a triplet. The computation of those combinations of the first 15 without a triplet follows:

In the case of n=16, of the first 15 cards drawn are two of the same denomination as the 16th. As there are three others of that denomination in the deck, there are C(3,2)=3 possible ways of choosing these two cards.

That leaves 13 cards still to be accounted for, but only 12 denominations to choose from, so there must be at least one pair; that is, the number of pairs, p, cannot be zero.

If there is 1 pair, there are 12 choices for the denomination of the pair. As there are C(4,2)=6 possible pairs of individual cards of this denomination, we must also multiply by 6. In this case there is no choice for which singlets we use--we have to use all the 11 remaining denominations. (This is equivalent to C(11,11)=1 way of doing it.) There are 4 choices (of suit) for each of the 11 singlets as well so we have to multiply by 4^11=4194304.

Then the next term:
If there are 2 pairs, there are C(12,2)=66 choices for the denomination of the pair. As there are C(4,2)=6 possible pairs of individual cards of each of these denominations, we must also multiply by 6^2=36. Now there is a choice for which 9 singlets we use out of the remaining 10 denominatins--C(10,9)=10. There are 4 choices (of suit) for each of the 9 singlets as well so we have to multiply by 4^9=262144.

Then the next term:
If there are 3 pairs, there are C(12,3)=220 choices for the denomination of the pair. As there are still C(4,2)=6 possible pairs of individual cards of each of these denominations, we must also multiply by 6^3=216. Now there is a choice for which 7 singlets we use out of the remaining 9 denominatins--C(9,7)=36. There are 4 choices (of suit) for each of the 7 singlets as well so we have to multiply by 4^7=16384.

Then the next term:
If there are 4 pairs, there are C(12,4)=495 choices for the denomination of the pair. As there are still 6 possible pairs of individual cards of each of these denominations, we must also multiply by 6^4=1296. Now there is a choice for which 5 singlets we use out of the remaining 8 denominatins--C(8,5)=56. There are 4 choices (of suit) for each of the 5 singlets as well so we have to multiply by 4^5=1024.

Then the next term:
If there are 5 pairs, there are C(12,5)=792 choices for the denomination of the pair. As there are still 6 possible pairs of individual cards of each of these denominations, we must also multiply by 6^5=7776. Now there is a choice for which 3 singlets we use out of the remaining 7 denominatins--C(7,3)=35. There are 4 choices (of suit) for each of the 3 singlets as well so we have to multiply by 4^3=64.

Then the next term:
If there are 6 pairs, there are C(12,6)=924 choices for the denomination of the pair. As there are still 6 possible pairs of individual cards of each of these denominations, we must also multiply by 6^6=46656. Now there is a choice for which singlet we use out of the remaining 6 denominatins--C(6,1)=6. There are 4 choices (of suit) for each of the 3 singlets as well so we have to multiply by 4^1=4.

So the terms are added together and the sum muliplied by that 52.

The terms as described above are 905969664, 18685624320, 84085309440, 110361968640, 41385738240 and 3103930368, adding up to 258528540672. We have to multiply this by 52, and then divide by 52 times the total number of combinations that the first 15 cards chosen from 15 could have. Note that the 52's in the numerator and denominator actually cancel. C(51,15)=3188675231420. Doing the division comes up with the 0.0810771 that's shown in the table for the row labeled 16 in the analytic solution.

In the program, the 52 possible choices for the triplet-completing card are distributed over the individual terms we just summed, but it doesn't make any difference if the sum is multiplied by 52 or each individual term is. Also, the factor of 52 is included both in the dividend and the divisor. It could have been left out.
  Posted by Charlie on 2003-11-23 22:25:55
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