Re: martingale betting system question
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Wed, 15 Jun 2005 09:36:31 -0500
On Wed, 15 Jun 2005 15:27:49 +0100, John <spam@xxxxxx> wrote:
>Hi al,
>
>I recently found out about the martingale betting system (start with a
>1 unit bet, and then bet 1 unit when you win or double your previous
>bet when you lose). And when i decided to simulate it by using a
>computer program much to my surprise it seemed to be a winning
>strategy given a sequence of random 0 or 1 inputs. This struck me as
>seriously weird. I always figured there is no way to win this type of
>coin toss game in the long run.
>
>Obviously this system doesn't work in a casino scenario because of the
>exponential growth of the bet and the betting limit. But beating a
>random number generator in a betting game seems odd anyhow.
>
>I mean this almost seems to imply that you can predict the next random
>0 or 1 with a more than 50% accuracy. Which obviously is impossible
>and a sure sign i'm making some kind of thinking error here. But so
>far i haven't been able to find it.
>
>If anybody could help me make some sense of this, i would really
>appreciate it.
If there's no limit on the size of the bet and you have infinitely
much money then the martingale _is_ a winning system, in the
sense that (with probability 1) you will eventually win a bet,
and at that point you will have made a profit of one unit.
(Let's ignore the fact that if you have infinitely much
money you haven't actually gained anything by winning one
more unit.)
That's meaningless in any real situation because you _don't_
have infinitely much money to bet - if you have only
finitely much money it's true with probability 1 that
you will eventually lose it all. Now, how long this takes
depends on how much money you start with. If you adjust your
simulation so you start with a limited amount of money then
the simulation _will_ eventually go broke, but it could
take a very long time, depending on the amount you start with.
(If you start with a billion dollars you will have a _very_
good chance of winning one additional dollar before going
broke. But just as winning one more dollar is meaningless
if you start with infinitely much money, winning one more
dollar is probably not all that exciting to the typical
billionaire.)
Specifically, say you start with 2^n - 1 dollars and you
play until you either go broke or win 1 dollar. You
win 1 dollar unless you lose n tosses in a row (for
example if n = 3: You're starting with 7 dollars, so
you win 1 dollar unless you lose 3 tosses in a row,
which wipes you out). So the probability that you
lose everything is p = 1/2^n, and the probability
that you win 1 dollar is 1 - p. So your expected
gain is
E = (1 - p)*1 - p*(2^n - 1) = 0.
In the long run you break exactly even.
>Cheers,
>John
************************
David C. Ullrich
.
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