Re: Programming to Beat the Odds in Gaming

Richard Heathfield <rjh@xxxxxxxxxxxxxxx> writes:

I asked A to always bet on Red. I asked B to always bet on Black. I asked C
to start on Red and alternate strictly between Red and Black. I asked D to
start on Black and alternate strictly between Black and Red.

We then played through 20368 spins of the wheel, at a dollar a spin (each).
(This was not an Atlantic City casino - just a cheapjack back-of-beyond
place that you've probably never heard of, and cheapjack enough not to
turn up its nose at a single dollar.)

How did we do? Well...

A won 9902 times, losing 564 dollars altogether.
B won 9885 times, losing 598 dollars altogether.
C won 4985 times, losing 10398 dollars altogether.
D won 4903 times, losing 10562 dollars altogether.

The house won (rolled 0) 581 times.

(These compare very well with the theoretical distribution, which I make
out to be 9900/9900/4950/4950/550.)

As you can see, the alternation strategy is disastrous! It is approximately
half as successful as sticking to one colour. (I must admit that this
result surprised me - I thought it would be /as/ successful, but no more
so.)


I think you should have gone with your first instinct, and
stayed surprised. This can't be right.

Given your setup, if C is betting red, then D is betting black,
and vice versa. This means that unless the house wins (rolled 0),
exactly one of C and D will win for each spin.

Therefore, we must have C wins + D wins = total spins - house wins.

(C and D are must not really be your friends, since they've clearly
colluded to skim off some of your winnings.)


The expected value for alternation in the long run is the same
as any other strategy because, like you say, the roulette wheel
can't remember patterns.


(I love this thread.)

--
Mark Jeffcoat
Austin, TX
.