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A man has time to play roulette 5 times. He wins or loses a

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Intern
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A man has time to play roulette 5 times. He wins or loses a [#permalink]

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New post 13 Dec 2004, 19:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A man has time to play roulette 5 times. He wins or loses a dollar at eacy play. The man begins with $2 and will stop playing before the 5 times if he loses all his money or wins $3 (i.e., $5). Find the number of ways the playing can occur.

Can someone tell me how to solve this problem without using a tree structure.
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Roulette [#permalink]

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New post 14 Dec 2004, 06:34
It is 20.

If all possibilities were valid: 2^5 = 32 different plays: WWWWW, WWWWL, WWWLW, WWWLL, …., LLLLW, LLLLL.

If the man wins in 3 consecutive plays: WWW??, gets 5$ and stops, we loose 2^2=4 possibilities, so we count 1 instead of 4.

If the man looses in 2 consecutive plays: LL???, we loose 2^3=8, so we count 1 way for LLL, but loose 8 .

So far: 32-3-7= 22

Now there are other plays where we get to 0$ before playing the 5th time: LWLL (we loose 2 possibilities, nett:1way), and WLLL (nett: 1 way). 22-2 = 20 ways.

My problem is that I got to 22 quite easily, but to get to the other dead ends I had to draw the tree. :?

I hope somebody can help further…
Roulette   [#permalink] 14 Dec 2004, 06:34
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A man has time to play roulette 5 times. He wins or loses a

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