Malmuth formula for variance/hand estimate.

[BillChen]

So I haven't posted for a long time, I apologize.

Mason's formula presented in Gambling Theory has become the standard for estimating variance/hr or variance/hand, and is used in numerous spreadsheets, including all of mine. In fact Mark Weitzmann includes a proof that the formula is unbiased. It's pretty cool and useful, and I hope this post helps you better understand it.

One thing that I noticed is that for some players, the variance estimator seemed to be higher than taking data per-hand, one explaination is a serial correlation between hands, but there lies another explaination.

So I realized that using the Malmuth-Weitzman formula
per session is an unbiased estimator of variance *only
if* the session lengths don't depend on results. I'm
actually gonna post this to 2+2 at some point.

Like if you just plan to play 200 hands and quit, then
V = (result-mean)^2/200 is an unbiased estimate of
per hand variance. However, this is not so if your
session is result dependent, that is for example,
playing until you win or lose 50 bets. I would hope
nobody is selecting quitting times based
on how much we are up or down, but our opponents may
be doing this. For example playing heads-up you often don't have a choice of when to stop.

Let us take this simple example. Suppose there is a
series of coin flips, equal probability heads and
tails and the result is either +1 or -1 (mean 0).
It's clear the variance per flip is one. If you agree
to play a fixed N-length session then

V_i = result^2/N will have mean=1.

Hence we can just average the V_i's, without regard to
length.

However let us say our session is to keep flipping
until the someone is up two units.

Then there is a 1/2 chance the session lasts 2 flips,
since you need running heads or tails. Similarly
there is a 1/4 chance of 4 flips , A 1./8 chance of 6
flips etc.

So the average estimate for

V_i = result^2/N = 1/2 * 4/2 + 1/4 *4/4 + 1/8*4/6 ...

The first term already sums to 1. The above expression
actually sums to 2*ln(2) = 1.4..

Anyway the point is the estimate is bigger than 1.
The problem gets slightly worse if we pick a stopping
time of +/- 3 bets, etc.

Bill

[AaronBrown]

This is true. If you follow another strategy and quit after the first flip different from the previous flip (that is, if you win the first you keep playing until you lose one; if you lose the first you keep playing until you win one), your expected variance per turn is 0.54.

Looking at things another way, you can imagine a graph of 10,000 hands with a certain variance per hand. If I subdivide this series and compute the average variance per hand over each interval, I can get a big variance by picking peaks and troughs of the series, and a small variance by picking middle points.

I doubt this matters much in practice, however. There are lots of factors that go into a decision to quit. Some tend to raise variance (playing until you get some result one way or the other) and some tend to lower variance (riding streaks). Moreover, all of them hit up against limits, you run out of time or money or other players.