Backgammon Theory

Theory

Drop rate on initial doubles

From:   Gary Wong
Address:   gary@cs.arizona.edu
Date:   29 July 1998
Subject:   Re: Cube theory question
Forum:   rec.games.backgammon
Google:   wtiukgqzdc.fsf@brigantine.CS.Arizona.EDU

Phill Skelton writes: > Does anyone have any idea roughly what fraction of doubles should be > taken or dropped? I know this is a pretty vague question, so I'll > try to explain what I mean. Actually I think it's a nice and precise question -- especially compared to a few of the others that have been asked here recently :-) (WARNING: Boring mathematical content included. Persons of a boredom-sensitive or non-mathematically inclined nature should skip to the last paragraph. Discontinue reading my articles if pain persists.) > Given that large swings in equity can occur in a few rolls, it is > (as we all know) possible to go from not being good enough to double, > through to a double/drop position in just one roll per player. In > what fraction of games does this occur? I am asking because I have > been analysing my results recently,and I am looking for evidence that > I am on average doubling too early (more doubles taken by my opponents, > less dropped) or too late (more doubles dropped) on average. Do any > of our resident experts have any feeling for what these numbers ought > to be? I'm more of a resident than an expert, but I'm going to inflict my feeling on the newsgroup anyway :-) I was thinking about this problem earlier, in trying to present a reasonable model of the distribution of points won per game. In the end I decided it was too hard to reason about exactly, gave up, and was going to assume an empirically derived take/drop ratio and see whether that assumption led to a reasonable agreement with observations of points won in real games. I don't have a complete answer, but here's some thoughts that might help you. Assume money games, no Jacoby rule, and ignore last-roll situations (they're `easy' to decide anyway, simply by enumerating all 36 or 1296 dice rolls over the next exchange). Ignore gammons, to keep things simple. Assume both players are `perfect'. Assume you're making an initial double, not a redouble (you're more reluctant to double if you already own the cube, so P would be higher in that case). The question we now ask is: Assume that it is correct to double this turn. What is the probability P that it is also correct to drop? Since you have the opportunity to double, we know you didn't double last turn. And since we're also assuming perfect play, we know it would have been _incorrect_ to double then. Since we're not considering gammons, the only reason it wouldn't have been correct to double was that your position was not good enough. Therefore, the position last turn was no double/take, and this turn is double/(we don't know). To be a take last turn, we know your equity E then must have been less than your opponent's drop point D. In practice D will vary slightly from roll to roll (since it depends on the recube vig, which in turn depends on the cube efficiency, which varies...) but let's ignore that for now. To be a double this turn, we know your cubeful equity after doubling immediately must be higher that your cubeful equity from holding the cube for one more turn and then deciding. This is loosely equivalent to claiming that the cost of doubling now (and then rolling junk and not losing your market after all) is less than the cost of _not_ doubling (and then rolling well and losing your market). Calculating these costs is a complicated question in itself, but let's go with the conventional rule-of-thumb for now and assume (given volatility V) you double if E + V > D, and your opponent drops if E > D. The question we previously asked (given that it wasn't a double last turn, and it is a double this turn, what is the probability it is now a drop?) becomes: given E + V <= D, and E' + V' > D, what is the probability that E' > D? (E and V are last turn's equity and volatility; E' and V' are this turn's). E and E' are obviously related: E' = E + delta, where delta is assumed to be normally distributed with mean 0 and standard deviation V. There are too many inequalities in there for me, so let's simplify things by making yet another assumption. Assume that last roll, it was _almost_ a double (E + V = D). It's obviously easier to lose your market from that position than from being nowhere near a double, so any conclusion we make will be an upper bound on P. From the definition of E', P( E' > D ) is P( E + delta > D ), and since we assume E + V = D, that's the same as P( delta > V ). That's just the plain old normal distribution P( z > 1 ), which is about 0.16. However, we haven't yet taken into account that we also assumed that E' + V' > D. Since E + V = D, for E' + V' to exceed D then either E' must be greater than E; V' must be greater than V; or both. Measuring the change in volatility gets pretty messy, so let's just assume that E' must exceed E. Under a normal distribution this will happen exactly half the time, so although the probability of losing your market is only 0.16, the probability of losing your market _given that your position improved_ is twice that, or 0.32. Remember the assumption we made in the middle of this paragraph will tend to overestimate P, so we conclude that the probability that any particular double should be dropped is no more than 32%. The other assumption we made was to ignore the change in volatility -- unfortunately there's little you can say about this in general (it's too hard for me to think about, anyway! We measure the equity, and call the variance in equity the volatility. What do we call the variance in volatility?) It depends on the position -- for instance, in prime-vs-prime games the volatility is very high, and it is often correct to lose your market without doubling (ie. P will be higher). In long races, the volatility is very low, and with correct play you should get the opportunity to make an efficient double (ie. P will be lower). The overall P depends on the relative probability of these situations occurring. Our conclusion must be that the _average_ P is under 32%, but P will vary from position to position. Well, that's quite a long-winded non-answer :-) But I hope the reasoning behind it helps in any case. I feel kind of cruel making you read all that without giving you some sort of independent verification that I'm vaguely on the right track, so here goes: a few years ago, Stig Eide posted an article with statistics from 1000 games of Jellyfish playing itself. Among the results was that 35% of games resulted in a single point win -- assuming the only way you can win a single point is for your opponent to drop an initial double, that would indicate that 35% of doubles ought to be dropped. This is a little higher than our theoretical P of 32%, so it looks as if some of our assumptions led us to underestimate P slightly, or Jellyfish drops slightly more doubles than it should, or it waits slightly too long before doubling, or that sometimes it is correct to win a single point without ever turning the cube, or the difference can be attributed to sampling error in the experiment. (NB: Boring mathematical content ends.) Whether you believe the wobbly acalephoid computerised fish or me, it looks as if P ~= 1/3 (2 takes to every drop) is in the right ballpark. Cheers, Gary. -- Gary Wong, Department of Computer Science, University of Arizona gary@cs.arizona.edu http://www.cs.arizona.edu/~gary/

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