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As you know, the betting strategy with the highest probability of winning in the end is "double or nothing".

It is, on average, a losing strategy, but if you're going to gamble...

The brainless should not be in banking. — Willem Buitler

by Migeru (migeru at eurotrib dot com) on Sun Jun 7th, 2009 at 05:22:49 AM EST
[ Parent ]
the expectation of gain, is like other strategies, exactly zero. What you presumably mean is that you get a (n-1)/n chance of "winning" 1, by betting n, with n being an exponential of 2 (and a 1/n chance of losing n)

So the question, as expected, is whether you win before you run out of liquidity...

In the long run, we're all dead. John Maynard Keynes

by Jerome a Paris (etg@eurotrib.com) on Sun Jun 7th, 2009 at 08:24:40 AM EST
[ Parent ]
You're both right. It's true that the probability of winning 1 more than the initial amount in your pocket is 1 (ie certain), and in fact it is certain that this will occur within a finite number of bets. Moreover, by changing the monetary units, or by increasing the rate of growth of the bets, even bigger amounts can be won per bet, so the probability is maximal but the strategy is of course not unique.

However, the expected number of bets to achieve the outcome is infinite, since the strategy is nonintegrable, and while one game might(!) let you win 1 with certainty, in repeated games you will run out of money, or if you have infinite funds, you will fail to achieve any strictly positive rate of winnings in the long run (and depending on how you compute the rate of winnings, you could fail to achieve any finite negative rate of winnings either, ie the losses can't be usefully limited).

--
$E(X_t|F_s) = X_s,\quad t > s$

by martingale on Mon Jun 8th, 2009 at 05:12:13 AM EST
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