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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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Some responses for SilverKnight,,,, | Comment 22 of 39 |
SilverKnight, you wrote:
"...to follow your analysis on the assumption that the intent was a 'poker hand'.
(1) [Dan] wrote:
'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 correct formula is]
52!/(47! * 5!) [for all the 5 card combinations in a deck of 52]"
That you formula is incorrect, Silver Knight, is easily seen by applying it to the case of 2 cards. I think we can agree that there are 52*51 = 2652 possibilites for 2 cards out of 52. But applying your formula:
52!/(50! * 2!) = (52*51)*2 = 5304. Can you really come up with that many different 2-card hands in a deck of 52 ?
Well, SilverKnight, if I am holding 5 cards of a 52 card dack, lets call them A,B,C,D,and E, then there are 52 possibilities for A. For each of these, there are 51 possibilities for B. 52*51= 2652. For each of these 2652 possibilities for A and B, there are 50 possibilities for C. 2652*50= 132600. For each of these 132600 possibilities for A, B and C, there are 49 possibilities for D. 132600*49 = 6497400. For each of these 6497400 posibilities for A,B,C,and D, there are 48 possibilities for E. 6497400*48 = 311875200.
(2) You wrote:
"[Dan] wrote:
'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."
Not inconsistent at all. The explanation for that small figure is that, by the time we get to 15 cards, it is astronomically unlikely that we haven't bumped into a "2 pairs" situation by now. When I said that a poker "3-of-a-kind" could only be achieved between 5 and 14 cards, you misconstrued me to mean that it was a certainty within 5 and 14 cards. That is incorrect. It is only possible within 5 and 14 cards, but may not occur at all. If I were to say "The only way I could knock out Mike Tyson in a 15 round bout, is between rounds 1 and 15". That would hardly mean I am certain of knocking him out within those rounds. My actual chances of knocking out Mike Tyson with 15 rounds are only about 4 in 10.
(3) Lastly, I have taken up your good-natured
challenge to solve the problem given your interpretation, rather than the poker rules one. I will be posting it in a few minutes.
:-)

  Posted by Dan on 2003-11-20 15:37:26
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