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

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A man has time to play roulette 5 times. He wins or loses a [#permalink]  13 Dec 2004, 18:10
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., $5). Find the number of ways the playing can occur. Can someone tell me how to solve this problem without using a tree structure. Manager Joined: 15 Jul 2004 Posts: 75 Location: London Followers: 1 Kudos [?]: 0 [0], given: 0 Roulette [#permalink] 14 Dec 2004, 05:34 It is 20. If all possibilities were valid: 2^5 = 32 different plays: WWWWW, WWWWL, WWWLW, WWWLL, â€¦., LLLLW, LLLLL. If the man wins in 3 consecutive plays: WWW??, gets 5$ 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 furtherâ€¦
Roulette   [#permalink] 14 Dec 2004, 05:34
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