Cube Efficiency, Part I
by Rob Adams – July, 2005
Let’s consider a simple money game position with both players having two checkers each on their 3-points. So each side has a 6-pip position. The side on-roll owning a 2 cube (we’ll call him X) has a redouble, and O has a proper drop as she doesn’t quite win 25% of the time. To win, O needs X to miss (19/36) and then O not to miss (17/36), which gives a product of 323/1296. All the games will end with the cube on 4 after a take since O would be a slight underdog on-roll and so should not redouble to 8. This 323/1296 is 1 number short of the 324/1296 that would be 1/4. So O drops and loses two points.
OK, simple enough so far, but now let’s consider a similar position with each player having two checkers left, one on the 4-point and one on the 2-point. So again, each side has a 6-pip position. Again, with X on-roll owning a 2 cube: for O to win, X must miss (13/36) and O must not miss (23/36), which gives a product of 299/1296. This is well short of 324 so a drop, right?! Not so fast! Dropping here would be a mistake.
The reason is that O will be redoubling at every chance. So all of her wins will be on an 8 cube while most of her losses come on a 4 cube. To illustrate this precisely, let’s look at a sample of 36 × 36, or 1296 games:
First of all, if O drops, she loses two points every time, so she loses 2592 total in the 1296 games.
Or, after taking the 4 cube, O loses 4 points 23/36 right off. These total 828 of our 1296 games. And even if X misses he will be taking the coming 8 cube as he still wins 13/36, so 13/36 × 13/36 = 169 games where O loses 8 points. However, in 13/36 × 23/36 = 299 games O wins 8 points. So O wins 299 × 8 points or +2392, but she loses 828 × 4 + 169 × 8 = −4664 points. So in all, over the 1296 games, she loses 2272 points.
But, 2272 is less than the 2592 O loses if she drops every game so O should take. Indeed, by taking, O loses only about 1.75 points per game rather than 2 points. That’s quite an improvement, especially for a position that wins less often than our first position, which was a proper drop. Why is this?
The answer is Cube Efficiency. This can be a difficult concept. I’ll try to illuminate it a little bit in what follows. Let’s consider one more 6-pip position:
Here, each side has one checker left on the 6-point. Now for O to win X has to miss (9/36), and O needs not to miss (27/36), which gives a product of 243/1296. This is more simply calculated as 1/4 × 3/4 = 3/16. So O wins only 3/16 = 18.75% of the games and yet has a borderline take/drop here. This is a rare position of perfect Cube Efficiency. Every time O gets a turn, she will be able to offer a perfectly efficient redouble. By perfectly efficient I just mean one where X can either drop or take for the same equity. More generally, an efficient double is one where your opponent has a close decision on whether to drop or take (whether it is technically either a drop or a take). This is a difficult concept to quantify, but that doesn’t stop the mathematicians. And it’s a good thing too, or they wouldn’t have found the following very useful formula:
Take point = Naïve Take point − [Naïve Take point] × [Opponent’s Take point on the coming redouble] × [Recube Efficiency]
This formula really shows its strength in match play, but it works for money too. The Naïve Take point in money play in a simple race without gammons is just 25%. This is found by comparing risk versus gain. Naïve Take point = risk ÷ (risk + gain). For money, taking a 4 cube risks two points (4−2) to try to gain 6 points (4 − (−2)). This gives 2 ÷ 8 or 25%.
Using our formula and looking back at the first position where O got no use out of the cube: this is a Recube Efficiency of 0. So the take point is just the Naïve take point or 25%. Since O didn’t win quite that much, she had to drop.
Looking at our last position (with the perfect Recube Efficiency): this is a Recube Efficiency of 1. So the actual take point using our formula is 1/4 − [1/4]×[1/4]×[1] = 1/4 − 1/16 = 3/16. And since O won exactly 3/16, she had a borderline take/drop.
What about our second position? Well, O gets to offer a fairly efficient redouble every time. I’m not sure how to calculate the exact figure, but if O won 27/36 she could offer a perfectly efficient double for a Recube Efficiency of 1. And if O won 18/36 she would get no use for a Recube Efficiency of 0. O does win 23/36 so perhaps a guess of 5/9 would be a good approximation (23 is 5/9 of the way from 18 to 27). So using our formula, O’s actual take point would be 1/4 − [1/4]×[1/4]×[5/9] = 31/144 or about 21.5%. Since O does win about 23%, it is a solid take… the formula works!
The Recube Efficiency is always a number between 0 and 1. And for money games in a long race, I’ve found a guess of 1/2 is reasonable, giving a racing take point for money of 7/32 or just under 22%.
OK, but what about match play? Well, an efficient double is still one where your opponent has a close decision on whether to take or drop, but his take point varies depending on the match score and cube location.
Let’s look at an example. Assume you are X on-roll in a race ahead 1-0 in a 5 point match with the cube still in the center. At what point can your opponent take, and when will he do better to drop? First, let’s look at some match equities to calculate the Naïve Take point.
If you double and O drops, you will be ahead 2-0 or −3,−5. I’ll use Snowie’s match equity table, which is very good for strong, equally skilled players. Other assumptions and/or tables may give different results, but they should be close. Snowie gives −3,−5 as 64.5% MWC.
If you double and he takes and you win, you will be ahead 3-0 or −2,−5 for 73.5%. And if you double, he takes, and you lose, you will be behind 1-2 or −4,−3 for 42.8%.
So by taking, your opponent is risking 73.5 − 64.5 = 9% to try to gain 64.5 − 42.8 = 21.7% for a Naïve Take point of 9/30.7 = 29.3%. That may seem high, but if O doesn’t get any use out of the cube, it is probably fairly accurate.
To use our formula, we need X’s take point on the coming redouble to 4. In that case, dropping again leaves −4,−3 for 42.8%. Taking and winning the 4 cube would win the match for 100%, and taking and losing would leave −4,−1 and the Crawford Game for 18.1% MWC according to Snowie. So taking the 4 cube risks 42.8 − 18.1 = 24.7% to try to gain 100 − 42.8 = 57.2% for a take point of 24.7/81.9 = 30.2%.
OK, so let’s reconsider O’s take point assuming a Recube Efficiency of 1/2: using our formula we have 29.3% − [29.3%]×[30.2%]×[1/2] = 24.9%. That seems a little more reasonable, but remember that is higher than for money, which in a long race had a take point of about 3% less.
Also remember that specific positions need to be considered as the match gets closer to the end. A perfectly efficient double is different than the 25% for money.
Consider first a position with X on-roll, still with the cube centered, ahead 1-0 in the match to 5 with two checkers left on the 2-point; O has just one left on the ace point. So X will win 26/36 or 72.2%. O will get no use out of the cube and wins 10/36 or 27.8%, which is less than 29.3%, so he has a drop.
But, then consider the position with both X and O having two checkers left with one each on their 2- and 3-points (a 5-pip position each). For O to win, X must miss (11/36) and O must not miss (25/36) for a product of 275/1296 or 21.2%. This is much less than the previous position, which was a drop: could this possibly be a take? YES! The reason is that O is able to give an almost perfectly efficient redouble every time. So O’s take point is not 29.3% or 24.9% but rather 29.3% − [29.3%]×[30.2%]×[1] = 20.5%. Even if you adjust a little, since the recube isn’t perfectly efficient, and guess at a take point of 21%, it still looks like a proper take as O does slightly better than that.
So in summary, a simple answer to your take point just doesn’t exist. Depending on the future Recube Efficiency, you need to give a range. For money, without gammons, your take point is somewhere between 3/16 and 1/4 with a good guess at a racing take point of 7/32. At −5,−4 in a match with the cube centered, your take point is somewhere between 20.5% and 29.3% with a good guess of 24.9% in a long race without gammons. Other match scores can be worked out similarly. And when the game reaches the last few rolls, you can make a very good guess at your exact take point by using the formula: Take Point = Naïve Take Point − [Naïve Take Point] × [Opponent’s Take Point on the future redouble] × [Recube Efficiency]. And an efficient double is one where your opponent has a close decision on whether to take or drop while an inefficient one is a double where his decision is an easy one.
Proceed to Cube Efficiency, Part II to see how the formula affects match play and more.