A hypothethical staking arrangement

[pzhon]

To analyze this problem exactly, it helps to know one standard result from combinatorics.

Reflection Lemma: The number of ways of arranging W wins and L losses so that the difference losses-win is never B is (W+L choose W) - (W+L choose W+B) when L < W+B, and 0 when L >= W+B.

I found proofs of more complicated versions on the web, but not this one, so here is a proof.

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When L &gt;= W+B, it is trivial that the count is 0, since the ending point is not allowed.

When L &lt; W+B, there is a bijection between the arrangements of W+B wins and L-B losses (set S) and the arranglements of W wins and L losses where there are B more losses than wins at some point.

For any arrangement in T, let the first time there are B more losses than wins occur at A+B losses and A wins. Reverse these 2A+B wins and losses to produce an arrangement in S. For any arrangement in S, let the first time there are B more wins than losses be at A+B wins and A losses. This exists by the assumption that L&lt;W+B. Reversing these 2A+B produces an arrangement in T.

There are W+L choose W+B elements of S. The number of arrangements of W wins and L losses not contained in T is (W+L choose W) - (W+L choose W+B). [img]/images/graemlins/diamond.gif[/img][/list]
There are two types of results: Bust out at precisely B+2K steps, or finish without busting out. While the reflection lemma lets us compute the probability of each type, we can ignore the probabilities of busting out by computing the expected size of the ending bankroll, so that busting out has a value of 0. This means we have to compare the starting bankroll B with the expected ending bankroll.

B = $200
1: $180
2: $192
3: $194.4
4: $204.48
5: $209.95

In words, if you start with only enough to lose once and keep half of any profits, you have to make 4 bets or more to expect to profit as a 3:2 favorite on each bet. As a 51:49 favorite with a bankroll of 1 bet, you need to be able to place 538 bets to come out ahead on average.

The expected value grows unevenly since you always gain from the ability to make one more wager unless you are exactly even, and you can't be even after an odd number of bets.

B = $400
1: $380
2: $408
3: $405.6
4: $426.24
5: $429.98

As wtfsvi stated out, with a bankroll of $400 (or greater), you only need 2 bets as a 3:2 favorite. With $400, you need 522 bets as a 51:49 favorite. With $600, you need 505 bets as a 51:49 favorite. With an unlimited bankroll, you need at least 190 bets as a 51:49 favorite.

With an unlimited bankroll, the number of bets needed for the wager to be +EV grows quadratically with the reciprocal of the advantage. You want to play if breaking even is about 0.2760 or more standard deviations below average, at which point you would be at least a 60.87:39.13 favorite to profit, and your average wins conditional on winning would be at least 1.285 times your average losses conditional on losing.

If your opponent lets you make up to n bets, but does not force you to make all n if you don't want to, then you should quit early sometimes when you are even. By exercising this option, it takes fewer rounds to be profitable.

[wtfsvi]

Thanks pzhon. This is very good.

[ QUOTE ]
If your opponent lets you make up to n bets, but does not force you to make all n if you don't want to, then you should quit early sometimes when you are even. By exercising this option, it takes fewer rounds to be profitable.

[/ QUOTE ] Can you explain why?

edit: never mind. I think I get it. You should quit if you have 1 bet left and are even, and if you are a smaller favourite than in the OP you should quit with more bets left.

edit again: no, I don't get it. And I don't get how you should only quit when you are even, and not if you are down. Feel free to enlighten me.

[pzhon]

[ QUOTE ]

I don't get how you should only quit when you are even, and not if you are down. Feel free to enlighten me.

[/ QUOTE ]
When you are behind, playing one more round risks $200 to win back $200 of your losses. As a 60:40 favorite, you are happy to play.

When you are ahead, playing one more round risks $100 (50% of $200) to win $100. As a 60:40 favorite, you are happy to play.

Only when you are even might it be bad to play one more round. In that case, you are risking $200 to win $100 as only a 60:40 favorite. It might be worth playing more if you expect that taking a loss here will set up enough good opportunities in the future, and this happens if you get to play at least 4 rounds with a bankroll of 1 bet, or at least 2 rounds with a bankroll of at least 2 bets.

For example, suppose you get to play up to 3 rounds with a bankroll of 1 bet as a 60:40 favorite. If you always play as many rounds as possible, you expect to end up with $194.40 on average, for a $5.60 loss. However, the ability to quit after 2 rounds if you win and then lose is worth 24% * $20 = $4.80. If you quit if you are even after two rounds, then you only only expect to lose $0.80. With a slightly greater probability of winning, this would be the difference between whether it is profitable to play or not.

If you have 2 more rounds to play, the ability to quit is worth only $8, not $20. If you can quit in the 24% of the cases where you win and then lose, then playing up to 4 rounds is worth an extra 24% of $8 = $1.92, so your profit would be $6.40, not $4.48.

Of course, in the OP's scenario, you are staking someone who is not going to quit when it is in your interest to quit.

[wtfsvi]

Thank you.