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Jesse and three of his friends are playing Poker with 32 cards, The 32 cards are of every combination of the for patterns with the numbers 1, 7-13. In this game, each player takes five cards randomly.

1) Let X be the number of cards between the five Jesse got, with the number "1" on them. What is the probability function of $X$?

2) What is the probability that Jesse will have at least three cards with the same number?

3) During the evening, the four players played a number of rounds. In every round new cards were given out between the 32. Jesse decided he will stop playing after he will receive at lease three cards with the same number, in two games. What is the probability Jesse will play six games?

WIll someone please help me understand how to solve this question? In part c, is it a geometric probability based on part b ?

Thanks in advance

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  • $\begingroup$ "The 32 cards are of every combination of the for patterns with the numbers 1, 7-13."... what does this mean? Are you trying to say there are $4$ suits of $8$ faces each, numbered $\{1,7,8,9,10,11,12,13\}$ $\endgroup$ – Graham Kemp Sep 15 '14 at 11:49
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Answer:

From what I understand of your problem:

1) P(X = i) where i is the number of ones amongst the five cards Jesse gets in any round.

Thus function of X => $$P(X=i) = \frac{{4\choose i}\times {(28)\choose (5-i)}}{{32\choose 5}} , i = 0,1,2,3,4$$

2) P(three cards being the same in the draw) =$$\frac{{8\choose 1}\left({4\choose 3}{28\choose 2}+{4\choose 4}{28\choose 1}\right)}{{32\choose 5}}$$

3) Part three is an application of negative binomial distribution.

Here In each round Jesse will get all cards in a round the same number $$= \frac{{8\choose 1}\left({4\choose 3}{28\choose 2}+{4\choose 4}{28\choose 1}\right)}{{32\choose 5}} = p$$ In each round Jesse will get all cards different than that in a round $$= 1 - p = q$$

Proabability that she will need two rounds of success in the six game

$${(4+2-1)\choose 4} q^4 p^2 = ({(5)\choose 4}q^4 p^2$$

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  • $\begingroup$ In question 1, did you mean: $ P(X=i) = \frac{{4\choose i}\times {(28)\choose (5-i)}}{{32\choose 5}} , i = 0,1,2,3,4 $? I will always have 28 cards that are not "1", so why should I put $28-i$ there ? Thanks ! $\endgroup$ – CrazyStatistician Sep 15 '14 at 13:25
  • $\begingroup$ Good catch, smart you!!, I will edit it $\endgroup$ – Satish Ramanathan Sep 15 '14 at 13:33
  • $\begingroup$ I made another error in part 2, it says atleast three cards with the same number, I will edit that too $\endgroup$ – Satish Ramanathan Sep 15 '14 at 13:40
  • $\begingroup$ Yeah, I figured it out. I made myself a mistake in part (C), it should be "at least THREE cards" and not "at least FIVE cards" which makes no sense. I am still reading your solution. Will give you an update soon . Thanks! $\endgroup$ – CrazyStatistician Sep 15 '14 at 13:41
  • $\begingroup$ OK, everything is ok now. I solved part (a)+(b) in the same way you suggested and I now have an idea about how to solve part (c). Thanks a lot ! $\endgroup$ – CrazyStatistician Sep 15 '14 at 13:44

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