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The million dollar question is simple enough: out of all the games that could result from playing this position, how many do we win (and how many of our wins and losses are gammons, and how many are backgammons)? The model is exactly the same as if we had an urn with a googol balls in it (it's a big urn), and many of the balls have win written on them, and some say gammon loss, and if we look hard enough there are a few that read backgammon win, and so on. ((Balls and urns are to probability theorists what teapots and checkerboards are to computer graphics researchers, or squeamish ossifrage is to cryptographers – they seem to come with the territory.)) Instead of having the patience to count the googol balls, we just give the urn a really good shake and then pull 100 balls out without looking, and say for instance "Well, I got 53 wins, 31 losses, 9 gammon wins, 6 gammon losses, and a backgammon win – looks like my equity's roughly +0.26." and go home. If we were a bit more thorough, we could go a bit further and figure out that by cheating and measuring the sample proportions instead of the population proportions, we introduced a standard error of 0.06 into our result. (Of course, the trick is to select a sample size that's big enough that you reduce the standard error to a tolerable level, but small enough that the answer arrives before you get bored.)
It will come as no surprise that a rollout with a limited number of trials follows exactly the same procedure. It's sufficient to say that the proportion of wins/gammons etc. that come up when GNU Backgammon plays against itself (say) 1296 times, aren't likely to vary all that much from the proportion we would get if we measured the proportion of results in every game we could possibly get of GNU Backgammon playing against itself. (Of course, there may still be some doubt whether the results of GNU Backgammon vs. GNU Backgammon are representative of the results of a perfect player vs. a perfect player, or of you vs. Joe Average, but that's another story.)