an urn contains 17 balls marked lose and 3 balls marked win. you and an opponent take turns selecting a single ball at random from the urn without replacement. The person who welects the third win ball wins the game. It does not matter who selected the first two win balls.
I got part a and b but part c is:
If you draw first, what is the probability that you win? hint: you could win on your second, third, fourth,.., or tenth draw, but not on your first.
Hmm labeled according to the turn in which you would win, where the part in brackets can be permutedOriginally Posted by calculuskid1
2) [WW]W
3) [WWLL]W
4) [WWLLLL]W
...
Probability for (2) should be 1/C(20,3).
Probability for (3) should be C(3,2)*C(17,2)*4!/P(20,4)*1/16. (According to standard order of operations; parentheses are not needed.)
where C(n,k) is number of k-subsets of {1,2,...,n} and P(n,k) is number of k-permutations of {1,2,...,n}.
There may be an easier way.. but following the same pattern it's easy to write terms (4)-(10).
Well *slightly* easier is to look at the sequence backwards, then for (3) we can write
[WWLL]W[LLLLLLLLLLLLLLL]
But we don't care about the [WWLL], and it comes out to C(17,15)/C(20,15)*3/5, which is equal to the expression I gave in post #2.
Edit: Additionally I just realized that C(3,2)*C(17,2)*4!/P(20,4)*1/16 can be simplified to C(3,2)*C(17,2)/C(20,4)*1/16.
I think you have two ways now that are relatively simple. But if someone sees a simpler way I'll be happy to see it.
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