I want to emphasize that the following dissertation is not intended to be mathematically rigorous. There are several assumptions made along the way which may or may not be valid, and some of the numbers used are little more than guesses based on long experience. What this analysis shows is a way to attack the very difficult problem of determining if a double is theoretically corrcect. When the derived equations are used for actual positions the results tend to be consistent with what our intuition tells us should be the answer, which indicates that the formulas may be a reasonable representation of reality.

Doubling Theory

The problem of determining whether or not it is theoretically correct to double in a position is very difficult. Ideally, what we would like to do is to look at all the resulting positions after each of the 1296 possible next two rolls. For each of them there would be three possibilities:

1) The resulting position got so strong that it is a pass. Then the cost of not doubling would be the equity difference between 1.0 and the equity of the resulting position (on a 2-cube).

2) The resulting position is now a double and a take. Then the cost of not doubling would be zero, since whether we had doubled last roll or this roll the result would be the same -- the opponent would have taken either time.

3) The resulting position is not strong enough to double. Then the cost of doubling is the equity difference between having a 1-cube in the center and having a 2-cube on the opponent's side.

If we then added all these together and divided by 1296, the total would give us a net cost (or gain) of doubling, and we could determine if it were correct to turn the cube.

The first question to ask is, what is the minimum take point. For this analysis we are not considering gammons, or to put another way we are assuming the equity from gammons have been normalized to simple win percentage. If there were no recube possible then the minimum take point would be .25 (this is win probability, of course). If the taker could be guaranteed that he would have perfect efficiency on his recube, then it has been shown that he could take with as low as .20 win percentage, since his recube vig will make up the extra 5%. In practice the actual figure is somewhere in between, since the taker does have recube vig but he will not necessarily get off a particularly efficient redouble -- he might lose his market or he might double well before his opponent has a pass. Naturally the type of game will determine how efficient the redouble might be. In a long race the redouble has a good chance to be very efficient, while if the taker's main equity is hitting a shot and after hitting the shot he is virtually gin then the redouble will not be efficient at all. The consensus opinion of experts appears to be that on balance the minimum take point is about 21.5%, so I will use this figure for the rest of the analysis.

If we translate this into equity, we see that the minimum take occurs when the doubler has an equity of .570. Thus, if after the next two rolls the equity is greater than .570, the doubler would have lost his market. To determine the cost of not doubling (or the gain from doubling) we would have to determine the equity of the position after the cube has been turned.

Suppose this equity is e (for simplicity we are always talking about the equity for a 1-game). Then, the cost of not doubling will be 2 * e - 1, since we would have won a sure point by not doubling and then cashing, but if we doubled the roll before our equity now is 2 * e. Note that e will not be the same as it would be cubeless, since now we have a new situation where the taker owns the cube, so he has access to it while the doubler does not. Therefore, the equity for the doubler is going to drop. Our task is to determine what the new equity is once the cube has been turned. If we have done so successfully, then if the old equity was .570 the new equity should be .500.

It is pretty much impossible to take into account all the recubes etc. which might take place. What we will do is as follows: Let's redefine the game so that the taker does not actually get to turn the cube back, but if his winning probabality in the game improves to a certain point he is automatically declared the winner -- he doesn't have to finish out the game. Thus, the doubler must continue playing until his winning probability is 1.00, while the taker need only continue playing until his winning probability is p, the point in question. If the taker were assured of a perfectly efficient recube it has been shown that p = .80 would be the proper value, but he is not assured of that efficient recube. Thus, in order to have our new game be equivalent in fairness to one where the taker actually owns the cube, it is clear that p will be somewhat higher than .80, and this will be a measure of the inefficiency of the taker's redoubles.

How do we determine the proper value for p? We can see that our new game is now defined as a race between the doubler and the taker -- the doubler wins the race if he reaches 1.00 winning probability first, while the taker wins if he reaches p first. Suppose the initial double was at the minimum take point, equity of .570 or 78.5% wins for doubler. For this to be the minimum take point, it is clear that from here the doubler will reach 1.00 winning probability first 75% of the time, while the taker will reach p winning probability first 25% of the time. Now, let's change the rules again, suppose the game has to be played to conclusion, and see how the doubler can win. There are two possibilities. First of all he might reach 1.00 winning probability before the taker reaches p. Secondly, the taker might reach p first, but the doubler could then pull it out. The sum of these two ways will be equal to .785, the doubler's overall probability of winning. So, assuming the doubler has x probability of winning when he reaches the point where he is forced to concede in our modified game, we have the following equation:

.75 + x * .25 = .785.

Solving for x, we get x = .14.

Thus, if our new game is defined such that the taker wins automatically if his probability of winning reaches .86, we have a game which is equivalent to the taker owning the cube normally.

Now, suppose the doubler has an expected winning probability w. We have to convert that to his real winning probability with the opponent owning the cube, which means that the taker will win automatically if he reaches .86 probability of winning. Using the same logic as before, if the game were played to conclusion the doubler could win either by reaching 1.00 probability of winning before the taker reached .86 probability of winning, or by having the taker reach .86 probability of winning first and the doubler then pulling it out. Therefore, if p were the probability that the doubler would win before the taker reached .86 probability of winning, we would have:

w = p + (1 - p) * (1 - .86).

Solving for p, we get: p = (w - .14) / .86

As a check, suppose w = .785 (the minimum take point). Then:

p = (.785 - .14) / .86 = .75

which is exactly what it should be.

So, suppose we have an equity e for a player with no cube, and want to turn it into the proper equity with the opponent owning the cube. The procedure is as follows:

1) Turn this equity into win percentage, using the formula w = (e+1) / 2
2) Turn this win percentage into a modified win percentage, using the formula: p = (w - .14) / .86
3) Turn this modified win percentage back into equity, using the formula: E = 2 * p - 1.

As a check, it is easy to see that a cubeless equity of .570 turns into a modified equity of .500, which is what it should be.

We can now determine the cost of not doubling when we lose our market. Since losing our market involves arriving at an equity greater than .570 (thus a modified equity greater than .500), and we are talking about the equity on a 2-cube vs. claiming one point, the cost is 2 * E - 1 where E is the modified equity.

What about the value of owning the cube vs. having the opponent own the cube. This will be important in order to determine whether or not we should be redoubling. As before, turn the equity into win percentage w. If the person with w win percentage owns the cube, then his opponent's probability of winning cubeless is 1-w. In the modified game (where the player owning the cube wins automatically if he reaches win percentage of .86) we can see that the player not owning the cube can win the cubeless game either by reaching win percentage of 1 first, or by having his opponent reach his .86 win percentage and then pulling it out. So, if p is the modified win percentage for the player not holding the cube, then we have:

1 - w = (1 - p) + .14 * p.

Solving for p gives us p = w / .86.

Once again, if we start with a given equity for the player holding the cube we can turn this into a modified equity which takes his cube ownership into account by changing the original equity to win percentage, using the above formula to get a modified win percentage, and changing this modified win percentage back to equity.

Finally, suppose the cube is in the center, and suppose a player's win percentage is w. How do we determine his modified win percentage, taking into account that both players have access to the cube. It is intuitively clear that the give up point (i.e. the point where we decide an opponent concedes in order to take into account our access to the cube) should be lower than the .86 figure we have been using after the cube has been turned for two reasons:

1) Players will be doubling earlier with the cube in the center than if they own the cube.
2) Initial doubles tend to be more efficient than redoubles. This is because an initial double generally occurs in the developmental stages where a player builds up a slow advantage, while a redouble often occurs when the underdog turns the game around suddenly by hitting a shot or rolling one lucky number. Thus, you are more likely to lose your market on your redoubles than on your initial doubles.

After examining the relevant numbers, I estimated that .835 is the give up point with the cube in the center. Thus, with both players having access to the cube the game is equivalent to a cubeless game where the player who first reaches a win percentage of .835 automatically wins.

Assuming this to be correct, suppose a player has a win probability of w. What are his chances p of winning the modified game? Clearly he can win the cubeless game in one of two ways:

a) He reaches win probability of .835 first (probability p), and then goes on to win from there.
b) His opponent reaches win probability of .835 first (probability 1-p), but he manages to pull it out.

Putting this together, we have:

w = p * .835 + (1 - p) * .165

Solving for p gives us: p = (w - .165) / .670

As before, we get a modified equity by turning an equity into win percentage, using the above formula, and turning it back into equity.

The formulas we have just derived will be key to determining when we should be doubling or redoubling, and the cost of doubling when we should have waited. To get a feel for this, let's look at some examples. In each case, let e0 be the modified equity with the cube in the center, e1 the modified equity with the opponent owning the cube, and e2 the modified equity with the player owning the cube (all equities are on one cubes).

```Equity       e0       e1        e2

.40         .597     .302      .628
.39         .582     .291      .616
.42         .627     .326      .651
```
It is clear that if e1 > e0 / 2 it is correct to double from the center, and if e1 > e2 / 2 it is correct to redouble. This is because we are doubling the stakes, so if the equity (on a one cube) with the opponent owning the cube is greater than half of what it would have been had we not doubled it is correct to double. Thus, on average it is correct to make an initial double with an equity of .390, and to redouble with an equity of .420. This is pretty consistent with what our intuition would tell us. Of course it does not tell us automatically that we should double a given position if our equity is above .390 or .420; that decision is dependent on the volatility of the position. The give up points of .860 and .835 are really just our best guesses of how to represent average volatility properly.

Now, to answer the question of what the cost of doubling and not losing your market is. Let's suppose the cube is in the center. After the next exchange of rolls, there are three possibilities:

1) The equity is above .570. Then we have lost our market, and not doubling has been costly. We have already seen how to compute this cost.
2) The equity is between .390 and .570. If this happens, we can assume that we would turn the cube now, and our opponent will have a take. This won't always be accurate due to differing volatility in the position, but it is as good an assumption as we can make. Therefore it didn't matter whether or not we doubled last roll -- the result would have been the same.
3) The equity is below .390. Now we can assume that we would not be doubling, therefore it was costly to have doubled the roll before. Again this assumption may not be 100% accurate, but it is the best we can do. In this case, the cost of doubling is e0 - 2 * e1, since e0 is the equity we would have with the cube in the center, and 2 * e1 is the equity with our opponent owning a 2-cube. If we are talking about a redouble, then we will have cost by doubling if the equity is below .420, and the cost will be e2 - 2 * e1 by the same reasoning.

As an illustration, let's suppose that after the next exchange of rolls our equity is .200. How much have we cost by doubling or redoubling. The modified equities turn out to be:

e0 = .299
e1 = .070
e2 = .395

Therefore, the cost of doubling is .299 - 2 * .070 = .159. The cost of redoubling is .395 - 2 * .070 = .255

We are finally in position to answer the big question -- should we turn the cube. The procedure is as follows: Look at the 1296 possible pairs of initial rolls and responses, and for each of them determine the expected equity. For each of these equities, determine the gain or loss from doubling. Add them up, divide by 1296, and the final result will be the overall gain or loss from doubling.