Re: Betting on {0,1} strings ... game theory ?

On Mar 2, 8:30 pm, olegtsodi...@xxxxxxxxx wrote:
On Mar 2, 7:43 pm, "Bart" <qjohnny2...@xxxxxxxxx> wrote:





On Mar 2, 4:27 pm, matt271829-n...@xxxxxxxxxxx wrote:

On Mar 2, 12:38 pm, matt271829-n...@xxxxxxxxxxx wrote:

On Mar 2, 4:22 am, olegtsodi...@xxxxxxxxx wrote:

On Mar 1, 10:24 pm, matt271829-n...@xxxxxxxxxxx wrote:

On Mar 2, 2:48 am, olegtsodi...@xxxxxxxxx wrote:

I may be wrong, but if you bet 1/2 of your money every time than on
average you are multiplying your money by p/2, where
1/p (>1/2) is your ratio. In the long run then, you should be
infinitely rich! This is by no means rigorous, but see if it is still
correct.

You mean bet half your money on zero every time? I don't think that
will work - not in general anyway. If the amount of money you have is
x then a loss takes you to x/2 and a win to 3*x/2. If you consider the
evolution of log(x) then a loss subtracts log(2) and a win adds
log(3/2). If I did my sums correctly then this will in the long run
cause x -> oo only if the probability of a zero is greater than log(2)/
log(3), or about 0.63. If 1/2 < prob. < log(2)/log(3) then x will tend
to zero.

Can you elaborate exactly on how you came up with the ln(2)/ln(3)
number?

Actually, let's do it more generally, and in the process partially
answer the original question (or answer a slightly different question
in fact). Suppose that at each turn the probability of getting a zero
is p, where p > 1/2, and at each turn you bet a fixed fraction of your
money on zero. Denote this fraction by a, where 0 < a < 1. In your
scenario a = 1/2.

If you have an amount x then a win takes you to x*(1 + a) and a loss
takes you to x*(1 - a). Considering the evolution of log(x), a win
increases log(x) by log(1 + a), and a loss "increases" log(x) by log(1
- a) (which is negative so this is actually a decrease). This is just
because log(x*(1 + a)) = log(x) + log(1 + a) and log(x*(1 - a)) =
log(x) + log(1 - a).

The expected change in log(x) is therefore given by p*log(1 + a) + (1
- p)*log(1 - a). If this quantity is positive then log(x) is expected
to increase without bound, so x is expected to increase without bound.
If this quantity is negative then log(x) is expected to decrease
without bound (to negative infinity), so x is expected to tend to
zero. So, the condition for winnings to increase without bound is

p*log(1 + a) + (1 - p)*log(1 - a) > 0 (1)

or, rearranging,

p > -log(1 - a) / (log(1 + a) - log(1 - a))

For a = 1/2 we have p > log(2)/log(3).

Looking at it round the other way, if we know p (and p > 1/2) then we
can find some number c, such that (1) is true for all a < c, and hence
guarantee (in probability) unbounded winnings. Some examples, rounded
to 6 d.p., are

p = 0.6, c = 0.389391
p = 0.51, c = 0.039989
p = 0.501, c = 0.004000
p = 0.5001, c = 0.000400

You can see that as p tends to 1/2 so c tends to zero. This means that
if you *don't* have an actual value for p, but merely know that
"(number of zeros / number of ones ) approaches a number greater than
1/2", then the best you can do with this strategy is choose a very
small value for a and hope. You can't guarantee unbounded winnings
because to cover every possible value of p you would effectively have
to choose a = 0, which would get you nowhere.

It did cross my mind whether somehow changing a adaptively (based on
observation so far) might be made to work, but I haven't thought about
it too much.

Ooops. I just realised that I misread the statement "(number of
zeros / number of ones ) approaches a number greater than 1/2" in the
original question. I read what I expected to read rather than what is
actually written. Hopefully it's intended to mean "(number of zeros /
number of ones ) approaches a number greater than ONE", which is how I
read it!

Also, although my explanation above was couched in terms of the
probability of getting a zero, rather than the limiting ratio of zeros/
ones, it works in pretty much exactly the same way if we are given the
limiting ratio. If this ratio is r, then to make infinite winnings we
need

-log(1 - a)/log(1 + a) < r

which can always be arranged by choosing a small enough, provided that
r > 1.

Now, I wonder if the following strategy might work in the absence of
foreknowledge of r. Keep track of the number of zeros and ones, and at
each turn calculate the ratio r. Then choose a so that a > 0 and the
above inequality is satisfied, and bet that proportion of what you
have on zero. It *seems* plausible, but I'm not exactly sure how to
prove it works.

Yes I put down 1/2 by mistake - I meant 1. I think there is a
strategy
that works.. I'm going to ask around some more.

Thanks.- Hide quoted text -

- Show quoted text -

Bart, Matt already wrote the solution for your ratio of 1- see above !

Matt, so what is the deal here- do you win the most if you bet
nothing ? :)
This has been my strategy all my life- I know it works.

Seriously, there must be a singularity there, at c=0, in some sense,
I'm guessing. What do you say?- Hide quoted text -

- Show quoted text -

Matt, you were operating on the assumption that you know the
distribution, in particular, that it's binomial.
However, you have only the infinite-time average (expectation value),
which is only one characteristic
of the distribution. One has to monitor such distribution based on the
past history and calculate p each time as
a function of past history and the long-term expectation value. I
haven't work on this kind of statistical problem
for some time, but I'm sure that there are standard techniques in the
field for such calculations.

oleg

.

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