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10-08-2016 , 09:36 AM

RJ91

enthusiast

Join Date: Jan 2012 Posts: 83

potentially a simple solution i'm missing here but i'm now very lost, here's the scenario...

i have bet on 3 contestants in a gameshow, there are now 14 left total, let's say all are the same skill level so all have the same odds, what are my chances of getting a least 1 person into the final of 3 contestants.

send help

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10-08-2016 , 11:18 AM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

I think this is handled by the Hypergeometric distribution.

N total possibilities
K are considered successes
N-K are considered failures
n are chosen without replacement

Prob that you get exactly k successes in your n draws is P(k):

P(k) = C(K,k) * C(N-K,n-k) / C(N,n).

So you want to P(1)+P(2)+P(3).

P.S. There is a simple way of getting this too.

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10-11-2016 , 11:18 AM

timbuk40

stranger

Join Date: Oct 2016 Posts: 7

You have to miss 3 draws to fail. 11/14, 10/13, 9/12

Multiply them to get 45.33% fail rate. You will one in the finals 54.67% of the time.

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10-11-2016 , 12:56 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

Quote:

Originally Posted by timbuk40

You have to miss 3 draws to fail. 11/14, 10/13, 9/12

Multiply them to get 45.33% fail rate. You will one in the finals 54.67% of the time.

Nice. Welcome to the 2+2 Probability Forum!

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10-11-2016 , 05:35 PM

Buzz

gramps

Join Date: Sep 2002 Posts: 18,254

Quote:

Originally Posted by RJ91

14 choose 3 = 364 ways to choose three of fourteen contestants.
1 way for all three of our favorites.
3*11=33 ways for two of the three of our favorites
3*11*10/2=165 ways for just one of our three favorites.
1+33+165=199 ways for at least one of our favorites to make the final three.

199/364=54.67% chance for at least one of our favorites to make the final three.

(I haven't looked at the other replies yet).

Buzz

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10-11-2016 , 06:20 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

Since the thread keeps getting bumped ....

Pure logic (or thinking through the logic behind the hypergeometric distribution given above) tells you that at least one of your favorites making the final three is the complement of none of your favorites making the final three. None of your favorites making the final three means that all of the finalists are from your non-favorites.

So, this prob is given by:

= 1 - C(11,3)/C(14,3)

= 1 - 165/364

= 199/364.

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10-14-2016 , 05:59 AM

RJ91

enthusiast

Join Date: Jan 2012 Posts: 83

Quote:

Originally Posted by timbuk40

You have to miss 3 draws to fail. 11/14, 10/13, 9/12

Multiply them to get 45.33% fail rate. You will one in the finals 54.67% of the time.

thanks, this actually helped a lot, ty for help all

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