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?)

This post is only with respect to the "Three of a Kind" brain teaser as interpreted according to poker rules. (Most people disagree with that interpretation).

(01) SK: No, Dan... I hate arguing what should be obvious .... but you MULTIPLIED by (5!). There is no 'this is the same' about it....

Dan: I went haywire when I did that. I believe it would be unethical for me to go back now and correct the original post, so I will either put a heading comment in it, or post a corrected version of it.

Let me break this out. The 52*51*50*49*48=51979200 part is correct if duplicates are not eliminated. 4*3*2 reflects all the ways that three aces can be ordered vis-a-vis each other in a hand of five: (spades - clubs - diamonds), (spades, clubs, hearts), (spades-diamonds-clubs), etc. There really are 4*3*2 = 24 such possibilities. So far, so good. Now each of these orderings will have 2 other cards that are not aces and don't match each other. If you remove all the aces, you're left with 48 cards. If you consider such a card (let's say, a jack), then the other non-ace among the 5 cards can be neither an ace nor a jack, for this to be a "3 of a kind" in the poker sense. There are only 44 cards in a deck of 52 that are neither aces nor jacks. That's where the 48*44= 2112 came from. Where I went wrong here is to multiply by (5!) in a misguided attempt to factor in permutations, since I had aleady considered ace permutations vis-a-vis each other with that 4*3*2 business. What I should have done is considered all the ways 3 aces
of a particual ordering with respect to each other (e.g. Hearts-Clubs-Diamonds = HCD), can lie among 5 cards. There are ten possibilies:
(HCD--), (HC-D-), (HC--D), (H-CD-), (H-C-D),
(H--CD), (-HCD-), (-HC-D), (-H-CD), (--HCD)
So all the 3-ace combinations in a true poker hand of 5 have to be:
(4*3*2)*(48*46)*(10) = 529920.
And the odds are 529920/51979200=0.0101948472
This is a different value than either of us has given so far. I defy anyone to find a flaw in that reasoning.

(02) SK: I tire of your attempt at justifying the wrong answer.
Dan: Yeah, life is rough, isn't it? It's really unfair that you have to suffer so. Maybe Hollywood will turn your life into a movie. I see the role of "SilverKnight" in the movie as being played by John Ritter, Robert Stack, Bob Hope, Art Carney, Gregory Peck or Charles Bronson...

(03)SK: What you wrote DOES imply that *IF* you get 3 of a kind, it will occur on the 5th to 14th card. And this is incorrect. Straights and Flushes have EXACTLY as much bearing on this problem as fullhouses.... Because, otherwise, I can choose WHICH five cards I wish... and get A, A, 1, 1, 2, 2, ... K, K, and wait until I get my 27th card (say an 8) before getting my 3 of a kind. And then I'll choose three 8's, a 9, and a 4, and I have my three of a kind, but I couldn't get my three of a kind before the 27th card.

Dan: This is an interpretation of the puzzle that I hadn't considered. I was just ass-uming that, as soon as you got A A 2 2 3 you now had "two pairs", and it was too late to get a "three-of-a-kind." Interesting..... If this is your interpretatiopn of the poker-verion of the puzzle, then you're right in regarding it as being as dumb as a box of hair.

:-)




Edited on November 20, 2003, 9:06 pm
Edited on November 20, 2003, 9:11 pm

Posted by Dan on 2003-11-20 20:41:35