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Three of a Kind (Posted on 2003-11-19) Difficulty: 4 of 5
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?)

No Solution Yet Submitted by Lewis    
Rating: 4.3333 (9 votes)

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re: Additional clarifications for SilverKnight | Comment 26 of 39 |
(In reply to Additional clarifications for SilverKnight by Dan)

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

(1)
I'll quote you AGAIN:
"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"

if you want all PERMUTATIONS of 3 aces and 2 OTHER non-matching-rank cards... then the number should be:
(4*3*2) * 48 * 44.
If you want all COMBINATIONS of 3 aces and 2 OTHER non-matching-rank cards... then the number should be (as I said):
(4*3*2) * 48 * 44 / 3! / 2!

You did NEITHER of these, as can be seen from the quote above.

What's more, the odds you calculated are wrong. As I mentioned in my earlier post the odds that you tried to calculate (odds of 3 Aces and two non-matching cards in a 5 card poker hand) is .001625265. You wrote that it is 0.0195031859. These are not equal. They are not "one way of looking at it... and another way of looking at it".

I tire of your attempt at justifying the wrong answer.

(2)
Please go back and read my post AGAIN. I agree that what you wrote does not imply that you WILL get 3 of a kind. 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.

(3) 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.

You can't have it both ways, Dan. If you take into account Full Houses, then you should also take into account straights and flushes.
Edited on November 20, 2003, 6:23 pm
  Posted by SilverKnight on 2003-11-20 18:21:34

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