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?)

SilverKnight wrote: "(1) [Dan] wrote that:
52!/(50! * 2!) = (52*51)*2 = 5304
No, this is incorrect.
52!/(50! * 2!) = (52*51)/2 = 1326
And yes, Dan, I *do* think I can come up with that many different 2-card hands in a deck of 52."

SilverKnight, you divided by 2! to elimiate duplicates, and I didn't. We're both correct.
To demontrate this, consider a 6-card deck:
Ah, Ad, Ac, Ks, Kh, Kd. (i.e Ace of hearts, Ace of diamonds, etc.) If you draw 2 cards, what are your chances of drawing 2 aces ? All the ways to draw 2 consecutive cards from such a deck are:
(Ah,Ad), (Ah,Ac), (Ah,Ks), (Ah,Kh), (Ah,Kd),
(Ad,Ah), (Ad,Ac), (Ad,Ks), (Ad,Kh), (Ad,Kd),
(Ac,Ah), (Ac,Ad), (Ac,Ks), (Ac,Kh), (Ac,Kd),
(Ks,Ah), (Ks,Ad), (Ks,Ac), (Ks,Kh), (Ks,Kd),
(Kh,Ah), (Kh,Ad), (Kh,Ac), (Kh,Ks), (Kh,Kd),
(Kd,Ah), (Kd,Ad), (Kd,Ac), (Kd,Ks), Kd,Kh)
30 total possibilities, with 6 ace pairs. Odds are 1 in 5. But if you now insist on dividing by 2!, they become:
(Ah,Ad), (Ah,Ac), (Ah,Ks), (Ah,Kh), (Ah,Kd),
(Ad,Ac), (Ad,Ks), (Ad,Kh), (Ad,Kd),
(Ac,Ks), (Ac,Kh), (Ac,Kd), (Ks,Kh), (Ks,Kd),
(Kh,Kd).
15 total possibilites, 3 ace pairs. Same result - 1 in 5. The elimination of duplicate ace pairs is counterbalanced by the elimination of all duplicates. The division by 2! buys you nothing.

SilverKnight wrote:
"(2) Please forgive me for taking what you wrote LITERALLY. I simply thought that when you used the word 'only' that you meant 'only'. Silly me."

I still cannot fathom what point you are trying to make here, SilverKnight. If I were to say, correctly, that "It will only be possible for me to win the Nobel Prize in Chemistry within the next 30 years" I really WOULD mean only. And I would not be implying that I was certain to win it. This is Logic 101, sir.

SilverKnight wrote:
"(3) Did you take into account (I'm betting that you didn't...) the notion that you could get a straight or flush before you get your three of a kind?"

Smart bet, SilverKnight. Of course I didn't,
because straights and flushes are completely irrrelevant to the problem at hand. A straight would mean that there were no matches at all, as would a flush. For a match, the two cards MUST be of equal "face value" (2 aces, 2 jacks etc) AND of different suit. Neither a straight nor a flush have any two cards that meet both those requirements.
:-)
Edited on November 20, 2003, 5:01 pm

Posted by Dan on 2003-11-20 16:58:08