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 I think I've got it. (No computer program used).
Again, you can stand by your previous analysis or anything else you want.
To be very clear: I think the intent of this problem is to have THREE CARDS of the same rank (ignoring poker hands).
But I decided to follow your analysis on the assumption that the intent was a "poker hand".
I assumed that your analysis of combinatorial probability was correct (and I didn't previously check your figures). It is not correct.
"There are 52*51*50*49*48=311875200 ways to draw the first 5 cards, of which there are (4*3*2)*(48*44)*(5*4*3*2*1)=2880*(48*44)=6082560 ways to get 3 aces and 2 nonmatching cards. Odds=0.0195031859"
This is definitely wrong.
The proper analysis of the likelihood of getting 3 Aces and 2 non-matching OTHER cards is:
Total combinations (not permutations) of getting ANY 5 cards is:
52!/(47! * 5!) which is 52*51*50*49*58/(5*4*3*2*1)
which is 2,598,960
So far, this is similar to what you did, except you were working with permutations not combinations (so you didn't divide by 5!).
To continue, to identify the number of combinations that have Three Aces and two non-matching cards:
4*3*2*48*44/(3! * 2!) = 4*24*44 = 4224
Therefore, the likelihood of getting three aces is:
4224/2598960 = .001625265
(This is less than one tenth of the figure you came up with.)
By symmetry, multiply this by 13 (not just aces) and you get:
~.02112845 is the probability of being dealt a three of a kind poker hand (not including a full house).
This is definitely not equal to what you came up with.
I will allow you to extend this logic, if you wish, to 6 cards, 7 cards, etc.
You can only achieve a true "three of a kind" hand by drawing between 5 and 14 cards.
You still haven't addressed why you have a probability for getting this on the 15th draw:
"(15 draws) = 0.0000002468"
This is inconsistent.