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?)
(In reply to A non-poker solution (computer program not used)
In the dialog manner,
There are 52*51*50=132600 ways to draw the first 3 cards from a 52 card deck. (SilverKnight says you need to divide by 3! to eliminate the duplicates. But as I conclusively demonstrated in "Some additional clarificatioins for SilverKnight", this is like "improving" an equation by dividing both sides by 1. You get the same result either way.) If these 3 cards are all aces, then there are 4*3*2=24 possibilities for these cards (e.g. spades/clubs/diamonds, hearts/clubs/spades...). So the odds of getting 3 aces in 3 draws are: 24/132600 = 0.0001809955
There are 52*51*50*49=6497400 ways to draw the first 4 cards from a deck of 52. If the 4rth card is an ace, there are 2+1=3 places 3 aces can be situated in 4 cards. For each of these, there are 48 possibilities for the non-ace. 24*3*48 = 3456 4-card draws containing 3 aces. The odds of this are 0.0005319051.
So far so good, so long as we are going to multiply by 13 to get the probability of any three of a kind, not just three aces.
There are 52*51*50*49*48=311875200 ways to draw the first 5 cards from a 52 card deck. If the 5rth card is an ace, there are 3+2+1=6 places for the aces cards to be located among the 5 drawn cards. For each of these, there are 48*47 =2256 possibilities for the two non-aces. 24*5*2256 = 270720 5-card draws containing 3 aces. The odds of this are: 0.0008680395
Conceptually, with that 13x caveat, this is still good; however there's a bookkeeping mistake: the 6 ways of placing the other 2 aces among the previous 4 cards is carried down into the final multiplication as a 5 rather than a 6. It should indeed be 6, as 2C4, or 4*3/2. I don't know how it was derived as 3+2+1, but that's beside the point.
Once that correction is made, as with the previously defined numbers, once they're multiplied by 13 and by 1,000,000 they agree within the expected statistical variations with the simulation results.
Similarly, the odds...
Just when they easy ones go by the wayside, the description ends. Six is the first time we come to the situation where the numbers other than the newly completed triplet also themselves could contain a triplet and must be excluded from the consideration that this was the first triplet. It's irrelevant when considering ace-only triplets, but kills the possibility of merely multiplying by 13 to get the probability of any triplet first being completed at the given stage. Since the others were capable of being multiplied by 13 to get the latter probability, the numbers are not even proportional to the given probabilities.
I also note that the probabilities for the cases of 3, 4 and 5 were individual probabilities of completing the first (ace) triplet at that number, rather than a cumulative probability. The numbers you post are monotonically increasing and add up to more than 1; are these intended as cumulative probabilities, by adding the "similar" calculations? At what point to you switch over from individual to cumulative probabilities?
The procedure you describe, correct for finding what is the probability of completing a set of 3 aces first at draw n, can be stated mathematically as
which can be verified for the first few from the following table generated with that formula:
The values for 3 and 4 agree with yours. The value for 5 is 6/5 what you give, consistent with your mistakenly bringing down 5 instead of 6 as a factor. The values for 6 and 7 agree with your table. Then your table goes off into something only you know how was calculated--not "similarly" to the other calculations. You haven't told us at all how these were arrived at.
Posted by Charlie
on 2003-11-20 22:36:38