Given: A man has time to play roulette at most five times. At each play he wins or loses one dollar. The man begins with two dollars and will stop playing before the five times, if he loses all his money or if he is up three dollars (i.e. if he has five dollars).
Asked: What is the total number of ways that the betting can occur?

Let W denote a win and L denote a loss.

Number of ways he can lose all his money (2$) = {"LL","WLLL","LWLL","LLWL","LLLW"}: 5 ways
Number of ways he is up 3$ (he has 5$) = {"WWW","LWWWW","WLWWW","WWLWW","WWWLW","WWWWL"} : 6 ways
Total number of ways = 5 + 6 = 11 ways

IMO B

Bunuel
It is not clear whether he has to stop playing if either of the 2 conditions is met or he can continue playing.
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If we don't permit LLLW then IMO A
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Kinshook, even though you have the same answer, I don't understand how you obtained 11.

Number of ways he can lose all his money (2$) = {"LL","WLLL","LWLL","LLWL","LLLW"}: 5 ways. The last 2 in this sequence dont make sense, because 2xLL means loss of money, so he shouldnt keep playing

The same goes for Number of ways he is up 3$ (he has 5$) = {"WWW","LWWWW","WLWWW","WWLWW","WWWLW","WWWWL"} : 6 ways. The last 2 show a gain of 3 and yet he keeps playing, even though it says that he stops playing.

The answer is 11 ways, but something is off

Thanks Kinshook

According to me, none of the answer choices is correct, if he has to stop playing after any of the 2 conditions is met.

Please provide your reasoning and the answer.



Posted from my mobile device
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I don't know how to take a shortcut with this question, so I tried to draw all possible plays. To make this easier, i'm going to display overall money per play. I will omit the fact he starts with $2, so first number in the sequence is either 1(L) or 3 (W)

1: 1,0
2: 1,2,3,4,3
3: 1,2,3,4,5
4: 1,2,3,2,1
5: 1,2,3,2,3
6: 1,2,1,2,1
7: 1,2,1,2,3
8: 1,2,1,0
9: 3,2,1,0
10:3,4,5
11:3,2,1,2,3
12:3,2,1,2,1
13:3,2,3,2,1
14:3,2,3,2,3
15:3,2,3,4,3
16:3,2,3,4,5
17:3,4,3,4,3
18:3,4,3,4,5
19:3,4,3,2,1
20:3,4,3,2,3

IMO 20: D, however OA is 11: B, so i'm doing something wrong

We can look at this from the perspective of his money at the end.

End with $0:

Force stop by losing twice: -1, -1

From here we can expand into -1, +1, -1, -1. Or +1, -1, -1, -1. 3 cases total.

End with $5:

Force stop by winning three times: +1, +1, +1.

From here we can expand into -1, +1, +1, +1, +1 or +1, -1, +1, +1, +1 or +1 +1, -1, +1, +1. 4 cases total.

Now we count the cases where he doesn't land on $0 or $5. If we don't land on $0 or $5, this means we go all 5 rounds because there is no condition where we are forced to stop on other numbers. Note landing on $2 or $4 is not possible because we need an even number of games to do so.

End on $1 or $3
To end on $2 on after round 4, we need to shuffle -1, +1, -1, +1. There are 4C2 = 6 ways to do so however we need to eliminate -1, -1, +1, + 1 as that is the only way to terminate the game early. Then either +1 or -1 at the end would let us land on $3 or $1. That is 10 cases in total.

We also have the cases where we land on $4 after 4 rounds, we can take the last 3 cases in "end with $5" and replace the last +1's with -1, that is 3 more cases.

Finally, we have 3+4+10+3 = 20 cases in total.

Ans: D
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Replying to a DM from petroshatz , (Thanks for the question!)

When I mentioned "shuffe -1, +1, -1, +1", the 5 ways of doing so were (1) +1 +1 -1 -1, (2) -1. +1. -1. +1. (3) +1. -1. +1. -1 (4) +1. -1. -1. +1. (5) -1. +1. +1. -1. We land back on $2 after round 4 and round 5 either +1 or -1 gives us a viable result.
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