> > I could see a case where it is possible to reduce a variance of a single
> > variable even in the 0-1 case. Let us say that black has about 5% chances of
> > winning. If we could (exactly) double the chances of black winning by
> > changing the nonuniform sampling somehow (say, enforce bad moves by white),
> > we could sample from that and divide the estimated black's winning chance in
> > the end by 2. This would of course be very difficult in practice. (A binary
> > random variable gives more information when the chances are closer to
> > 50-50.) This could be useful in practice in handicap games, by for instance
> > enforcing a black pass with 1% chance every move. Sampling would be
> > distorted towards white win, which is realistic since white is assumed to be
> > a stronger player, anyway.

> I don't understand this line of reasoning.

Let my try again using the handicap example. Let's say MC player is given 
a huge handicap. In the simulations, it is winning all of its games, so 
there is no information helping to select the next move. Using information 
theory, each play-out gives one bit of information if the chances are 
50:50, but if the chances are unbalanced, the information content is 
lower. (see http://en.wikipedia.org/wiki/Binary_entropy_function ) In the 
extreme case, there is no information at all.

Now, let us use distorted MC where we enforce black to pass with a few 
percent chance every move. White begins to win some of the simulations, so 
MC is useful again.

How this is related to reducing the variance?

Let us say that a black move leads to a white win with probability p very 
close to zero. Let us also assume that distorting the simulations doubles 
white's chances to 2p.
Using normal MC, the variance of our estimate of p using N samples is
p*(1-p)/N
and using distroted MC, the variance of 2p is
2p*(1-2p)/N
estimating p by using the estimate of 2p, the variance is divided by 4:
p*(1-2p)/2N which is less than p*(1-p)/N.

In practice, we cannot know that distorting would increase the chances 
exactly by doubling them, but if we use the same distortion to estimate 
all moves, we can still compare them.

Tapani
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