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November 5th, 2011 at 10:15:57 AM
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Hi,

I have an urn with 35 balls of which 3 are red, 7 blue and 25 green.

I will pick balls from the urn without putting them back, and as soon as I have 3 of a kind the game is over.

My questions is, what is the probability to end the game with get three of a kind in red balls?

I realised this is much more difficult than I thought from the start, so it thought I would turn to you professionals.

BR

Mutch

I have an urn with 35 balls of which 3 are red, 7 blue and 25 green.

I will pick balls from the urn without putting them back, and as soon as I have 3 of a kind the game is over.

My questions is, what is the probability to end the game with get three of a kind in red balls?

I realised this is much more difficult than I thought from the start, so it thought I would turn to you professionals.

BR

Mutch

November 5th, 2011 at 10:33:06 AM
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Read up on the multivariate hypergeometric distribution.

Then use it to find the probability of all ways to get 2 red and 0..2 of blue and 0..2 of green (you can't get 3 of either because then you'd end the game).

Then for each of those, multiply by the chance of picking the last red ball next (to end the game with 3 reds).

Then add it all up for the total probability.

Then use it to find the probability of all ways to get 2 red and 0..2 of blue and 0..2 of green (you can't get 3 of either because then you'd end the game).

Then for each of those, multiply by the chance of picking the last red ball next (to end the game with 3 reds).

Then add it all up for the total probability.

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563

April 20th, 2012 at 8:06:09 AM
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So, I'll make a try:

Using the multivariate hypergeometric distribution to find the probabilities of the allowed combinations before drawing a third red ball:

P(RR) = [C(3,2)/C(35,2)]

P(RRB) = [(C(3,2)*C(7,1))/C(35,3)]

P(RRG) = [(C(3,2)*C(25,1))/C(35,3)]

P(RRBB) = [(C(3,2)*C(7,2))/C(35,4)]

P(RRGG) = [(C(3,2)*C(25,2))/C(35,4)]

P(RRBG) = [(C(3,2)*C(7,1)*C(25,1))/C(35,4)]

P(RRBBG) = [(C(3,2)*C(7,2)*C(25,1))/C(35,5)]

P(RRBGG) = [(C(3,2)*C(7,1)*C(25,2))/C(35,5)]

P(RRBBGG) = [(C(3,2)*C(7,2)*C(25,2))/C(35,6)]

And then we draw the third red ball. The 1st probability is multiplied by 1/33, the 2nd and 3rd probs. are multiplied by 1/32, the 4th, 5th and 6th probs. are multiplied by 1/31, the 7th and 8th probs. are multiplied by 1/30 and the 9th prob. is multiplied by 1/29. Summing it all up gives me a total probability of 0.0027380074

Comments?

Using the multivariate hypergeometric distribution to find the probabilities of the allowed combinations before drawing a third red ball:

P(RR) = [C(3,2)/C(35,2)]

P(RRB) = [(C(3,2)*C(7,1))/C(35,3)]

P(RRG) = [(C(3,2)*C(25,1))/C(35,3)]

P(RRBB) = [(C(3,2)*C(7,2))/C(35,4)]

P(RRGG) = [(C(3,2)*C(25,2))/C(35,4)]

P(RRBG) = [(C(3,2)*C(7,1)*C(25,1))/C(35,4)]

P(RRBBG) = [(C(3,2)*C(7,2)*C(25,1))/C(35,5)]

P(RRBGG) = [(C(3,2)*C(7,1)*C(25,2))/C(35,5)]

P(RRBBGG) = [(C(3,2)*C(7,2)*C(25,2))/C(35,6)]

And then we draw the third red ball. The 1st probability is multiplied by 1/33, the 2nd and 3rd probs. are multiplied by 1/32, the 4th, 5th and 6th probs. are multiplied by 1/31, the 7th and 8th probs. are multiplied by 1/30 and the 9th prob. is multiplied by 1/29. Summing it all up gives me a total probability of 0.0027380074

Comments?