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Old 12-12-2002, 12:09 AM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default Simple solution (both problems)

The average number of hands to get all 1326 hands is the average number of hands to get the first hand plus the average number of hands to get the second unique hand, etc. The average number of hands to get a hand we haven't seen is always:

average = 1*P(1st not seen) + 2*P(1st seen)P(2nd not seen) + 3*P(1st seen)P(2nd seen)P(3rd not seen) + ...

If p is the probability of getting a hand we've already seen, then

average = 1*(1-p) + 2*p(1-p) + 3*p^2(1-p) + 4*p^3(1-p) + ...

= 1 + p + p^2 + p^3 + p^4 + ... = 1/(1-p)

Now if n is the number of hands we have seen, then p = n/1326. So the average hands needed to see all the hands is:

sum[n = 0 to 1325] 1/(1-n/1326) = <font color="red">10,300</font color>

I evaluated the above sum in excel.

For the original problem of how many 2 cards hands to see every card in the deck, we can use the same method by pretending that we are just are drawing 1 card at a time and replacing it, instead of 2 card hands, then divide the result by 2. This is a slight approximation since with actual 2 card hands you can't have 2 of the same card in your hand. Since drawing the same card twice in a row has low probability, the difference is small and we save alot of work. The above sum becomes:

sum[n = 0 to 51] 1/(1-n/52)/2 = <font color="red">118</font color>

Here's another approximation you can do to get a quick ballpark figure that doesn't require any sums. In order to have a 50% chance of getting all the cards, each card must have a .5^(1/52) chance of appearing if we treat them as independent. This is an approximation since if one card appears, it changes slightly the required chance of other cards appearing. The chance of any card not appearing in n cards is
(51/52)^n = 1 - .5^(1/52). So n = log[1 - .5^(1/52)]/log(51/52) = 222. This happens in 111 hands.

169 hands is more difficult because these hands don't all have the same probability. Pairs have a different probability than non-pairs. Doesn't look like much fun.
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