I understand that the possibility of ‘market loss’ is necessary for a player to have an optimal double. In other words, if you can’t lose your market, you shouldn’t double. To see this, suppose that that you can't lose your market, and consider refraining from doubling this time. If your position becomes worse over the next sequence, you will (presumably) be happy that you avoided doubling. However, if your position becomes better, you can still double now -- so haven't lost anything. Therefore, doubling earlier would have been a mistake.

While I have some idea of why market loss is necessary for doubling, I don't understand why some market loss -- or even a large possibility of market loss -- is sufficient for doubling. Can someone explain this to me?

To explain my confusion, consider a symmetrically volatile position with lots of market loss. (So my equity could go up a lot this sequence, but it could also fall dramatically.) Obviously, I gain a lot from doubling in sequences when my equity rose and I lost my market. However, I would appear to lose a lot from doubling when my equity falls. So is doubling such a good idea?

(To be clear, I do see that doubling is a good idea in situations where my equity will either improve or stay the same — could this be the key to understanding the problem?)

'Some market loss' isn't necessarily sufficient for doubling. The amount of market loss that's sufficient for doubling depends on the position. Show us a position and you'll get a clearer answer.

Yes, I understand that; however, it is commonly held that one should double if the chance of market loss is "very substantial" (say, above 75% for concreteness). Can anyone provide me with a general explanation as to why this should be the case?

I understand that, if volatility is very high, then there's a good chance that you will lose your market -- and you will then regret not having doubled. However, in such situations, your position may also become a lot worse -- and after such sequences, you will regret having doubled. As a result, I am not quite seeing why volatility tends to generate optimal doubles (if this is indeed the case).

I apologise if this question is very basic. I think that I can get a handle on doubling theory in a setting where probabilities change continuously (as in Keeler and Spencer). However, I remain a bit confused about what happens when we introduce volatility (as in real backgammon!)

O'Hagan's Law says if (1) you have about 25% market losing sequences, and (2) you are doing at least OK on the other 75%, you probably have a cube.

I've also heard it stated that you should double if you have 25% net market losers, i.e., if you think you have 50% market-losing sequences and 25% terrible sequences, your "net market loser" percentage is 50%-25%, or 25%, so you probably have a cube.

So you're right that you have to consider both the good and bad sequences. The keyword "O'Hagan's Law" should help you find some of the stuff that's been written about this. Unfortunately I don't think you'll find a very scientific explanation for where the 25% number comes from, other than it explains lots of XG results and seems to make accurate predictions.

IÂ’m not sure that the premise of your question is correct. Having significant market losers is not always sufficient to have a correct double. You must also have greater equity after doubling than you would if you had not doubled.

For example I can come up with a position where you have significant market losers (defined as >25%) but are an underdog. Suppose you have a single checker on your opponentÂ’s ace point and a closed home board. Your opponent has all his home board points made with spares on the 3,4,5 and 6 points plus a blot on his bar point.

Now, if you roll a 6, you hit his blot and have obviously lost your market. This happens with probability of 11/36, about 31%. Simplifying, assume you lose when you don’t hit and win when you do. (I am aware that there is a small likelihood of losing after a hit or winning after a miss, but those likely will roughly cancel each other). This means you win with probability 0.31 and lose with probability 0.69 for an equity of -0.38. (Again I am simplifying by ignoring gammons). If you double, assuming incorrectly that cube ownership has no value, you stand to lose -0.76 points per game. In reality your situation is worse than that after accounting for cube ownership. (Which I cannot even pretend to be able to do in a “back of the envelope” type analysis like this).

Obviously you should not double in this position. While contrived, I am sure there are plenty of positions with market losers where you either are not favored or only very slightly favored that would be correct no doubles. Blindly doubling simply because you have market losers would seem to be a mistake IMO.

I’ll point out that the “net market losers” approach to O’Hagan’s law handles these cases OK.

Hello!

Can someone explain it to me, step by step, where the 77,6% came from?
What is the calculation method?

This is doubling theory and market looser, by Hank Youngerman (https://bkgm.com)

Position: money game

Blue: 1 checker on 2 point
1 checker on 3 point
White: 1 checker on 24 point
1 checker on 21 point

Should Blue double? Should White take?
"Blue has 25 rolls that win the game immediately (any roll that doesn't contain an ace). If he fails to do so, then White has 29 rolls that will win him the game (everything but 1-1, 2-1, 3-1, or 3-2). Blue's chances of winning are 25/36 plus 11/36 times 7/36, for a total of 77.6%. Clearly he should double, and White should pass, right?"

25/36 + 11/36 x 7/36 = 977/1296 = 75.4%. I’m not sure where 77.6% comes from, but it should not affect the result. It’s still double/take based on the fact that white will have a redouble/pass when blue fails to win immediately. Blue only has 75.4 winning chances while owning the cube. His winning chance after doubling is 25/36 = 69.4%, so white has 30.6% winning chance after taking, more than enough to make it a take.

Thank you for your help