Problem of the Week #91: January 9


Tournament match to 7 points. Black leads 4-0 and owns a 2-cube.




(a) Assume you are Black and playing a very strong player. Should Black double? If Black doubles, should White take or drop?

(b) Same question, but now you are Black playing a weak player.

Hi Bill,

a nice question.

Theoretically this position is exactly in the narrow doubling window, with the chance to win for white on about 9,5% (21/216 to be exact).

So when playing against you I would double and hope to win a match against a former world champion, you would with no doubt survive the crawford game and double quickly in the next game and win a backgammon by some luck, but against weak players I would take the almost security of the crawford game rather than giving him the chance to catch up,

greetings k.

I'm a backgammon newbie, and I love these problems! Thanks, Bill. I've read "Backgammon for Winners," have a few others on order, and have been playing a lot. This is a round about way of saying I'm out of my league on this problem, but I'm going to take a crack at it anyway...

This situation reminds me a lot of poker situations where one player has a skill disadvantage. The player with the disadvantage wants to threaten the better player with variance by raising bigger, or by betting a lot relative to the pot size with even their barely good hands (like TPTK on wet board, for instance). When the weaker player feels he has even a slight advantage on a flop or pre-flop, he should create a big pot so he is not out-skilled by the better player when the next board card comes. He wants to get his money in as fast as possible when he's ahead. The same sensibility applies to short-stacking -- it takes a lot more skill, post-flop especially, to negotiate 100 big blinds than 20-50, so if you are outclassed in a game, buy in for less and see if you can jam all your money in on the flop or turn with your vulnerable TPTK, two pair, or whatever. It's as if every hand played by the weaker player in poker suffers from reverse implied odds!

It is in this spirit that I can see why a weaker black player would double here and why a stronger player wouldn't. The weaker player wants to threaten the stronger one with unacceptable variance, especially when we're talking about a match to 7. The stronger player might comfortably concede the 2 points instead of risking giving up 4, with the confidence he will most likely grind out a win in the match against the weaker player. When I'm in a hand with a weak player in poker, or someone I perceive to be less skilled than I am, I keep pots small and chip away, instead of blowing up the game and giving him a chance by building pots for him.

Of course, in this problem the photo negative would apply to when the stronger player is black -- he'll want to see if he can book the 2 points instead of bloating the stakes for a weaker player who will surely take, and have a much better chance of shipping the match should he get off to a 4-0 lead.

Black's chance of winning this game is 84.72%
I figured this out as follows:

Black wins immediately with 4-4, 5-5, or 6-6 (3/36 chance).
If black doesn't roll one of those (33/36 chance) AND white rolls any double (1/6) white will win

33/36 * 1/6 = 15.28% white will win. So black wins 84.72% of the time.

The thing I am not sure of is how to evaluate black's chance of winning the match if he doubles/white drops and he takes a 6-0 lead.
The problem doesn't state if the crawford rule is in effect but I assume that it is.

a) Against a strong player, black should double and the strong player should drop. The reason is that if the white takes, black will win the game and the match 84.72%. Of the remaining 15.28%, black is 50-50 to win the match which is another 7.64%
So with a double/take, blacks wins the match 92.36%
I expect that a strong player can come back from down 6-0 more than 7.64% of the time assuming the crawford rule is in effect.

b) Against a weak player, I don't think black should double. The weak player could very well take the double thinking that if he is down 6-0 he has no shot of winning anyway. The extra equity black gets from the double/take is not enough to offset the times he loses 4 points against a weak player that probably has almost no chance of winning once the match is 6-0. Against a weak player, black wants less volatility and more games to exploit his advantage.

I evaluated the position wrong. This game is even worse for white because black also wins immediately with a 2-2 and 3-3. That changes the numbers as follows:

White's chances of winnings the game: 31/36 * 1/6 = 14.35%
Black wins the game 85.65%

Against a strong player, if there is a double/take, black wins the match 92.825% (85.65 + 14.35/2) of the time.

This is incorrect (assuming white is not a complete donkey ).

Why? If the match is tied 4-4 and it is strong player vs player, why should the odds be different from 50-50?

If you were in white shoes and was behind 4-0 to 7 holding a 4-cube before your last roll what would you do?

Whoops! Forgot about that!

Ok, first, white wins IFF black rolls anything but 22+ (31/36) and he rolls any doubles (1/6), so he wins 14.35%. If white gets to roll, he wins 16.67%. Winning from 6-0 is about 91%. Winning from 4-2 is about 65%.

If black doesn't cube, he gets to 6-0 (91%) 85.6% of the time and 4-2 (65%) 14.4% of the time for an expected MWC of 87.2%.

Double-drop is 6-0 (91%) 100% of the time, so expected MWC 91%.

Double-take, which will always result in playing for the match one way or the other, you win 85.65%.

So ND-reship is the optimal line for equal skill, but if you think the skill difference is enough to change these EMWC by 2%, double against the guy much better than you, and definitely keep the cube against the weaker player (unless he's so awful you think he drops more than a third of the time, or will forget to reship a lot). It's hard to believe you'd be playing somebody so much better than you where he'd want to drop this and play the 9% instead of the 14.35%, but maybe that exists.

Let's see...1c-7a is 91%, 3a-5a is 65%. So an undoubled win gets to 91% and a loss drops to 65%. After doubling, wins and losses are 100% and 0%, respectively. So potential gain is 9% and loss is 65%.

Game equity shouldn't be too hard...white needs to avoid 22+ and then roll doubles, so he gets 31/36*1/6 equity=31/216 which is about 1/7 or 14%. White can take and reship with >9% so if doubled we know he can take and reship. The question is, can we double...We're risking 65% to gain 9% which is 7.x/1 and we're not quite that big of a favorite, so optimal line is to keep the cube. Vs a world class player i WOULD double this and vs anyone else I wouldn't.

Grunch.

A redouble triggers, if White get the chance, an 8 cube.

Underlying the Heinrich/Woolsey MET, we get:

White passes, 6away 1away for 10%
White takes and win for 100%
White takes and loses for 0%

Therefore he gains 90% when he takes and win, and losses 10% when he lose the game. Therefore he is getting 9 to 1 odds on the take. So he must win more then 11.1% to justify a take.

No double, Black wins, 1away 6away for 90%
Double, Black wins Match for 100%
No double, Black loses 3away 5away for 66%
Double, Black loses Match for 0%

He is giving 66 to 10 odds on the redouble, so he must be at least a 66 to 10 favorite in order to consider redoubling. So he must have at minimum 84.85% to consider doubling. Therefore doubling window goes from 84.85% to 88.9%

Let us now consider the equity of the position.

Black wins with 5 doubles out of 36 instantly.

The next move White wins only with doubles, 6 out of 36.

So from 1296 games White wins 6 x 31 games = 186 games.

1860000 : 1296 = 1435 so we have 14.35% winning chances.

Black has 85.65% for the win and needs 84.85%, so he gains barely at the correct double. White has 14.35% and needs barely 11.1% so he gains more, his take is definitely correct.

This is overall correct, if we assume that the opponents are equal. For a match with unequal opponents we have to use a MET which would take the rating difference into account, to get “correct” numbers. The only approach i know, is from Jakobs and Trice. We then have to know how much the rating difference is, because a 100 point difference is completely different to a 300 point difference.
Nevertheless, utilizing the Jacobs/Trice approach we get, without computing it exactly:

a) As more stronger our opponent is , the barely correct double turns more and more into a strong double, and the correct take can turn into a pass, depending on the rating difference.
b) The barely correct double turns instantly into a blunder, so I would reject it, on the other side, the weaker player should grab the cube, which is not a weapon, as a gift.

Cubeless Equity in a Cash Game
In a cash game, this one is easy. Black cashes, regardless of the relative strengths of the players. Look at the odds:

• Black wins on his first roll with any double higher than 2-2. Otherwise, Black wins on his second roll unless he rolls 2-1 back-to-back.
• If the game goes that far, White wins with any double on his first turn. Otherwise, White wins when Black rolls 2-1 back-to-back.
• White’s cubeless chance of victory is (31*6*36 + 2*30*2) / (36*36*36), or about 14.6%.
• Black’s cubeless chances are (100% - 14.6%), or 85.4%.

If Black cashes on his second turn, he avoids the disaster of rolling back-to-back 2-1s. His chance of winning then improves to 85.6%. Of course, if Black cashes now, his chance of winning is 100%.

Equally Skilled Players in a Match
In a match between equally skilled players, Problem 91 is a no-double/take. Let’s work it out:

If Black doubles, and White takes, then Black will be playing for the match. If he gets a turn, White has a mandatory redouble. Thus, both sides will be playing for the match. Black will win 85.4% of the time.

If Black doubles, and White passes, then the score will be 6-0, with the Crawford game upcoming. Assuming there are no gammons or backgammons, White will have to win the next four games in order to win the match. His chances are 50% raised to the fourth power, or 6.25%. Even if you give him one gammon victory, White still must win three games in a row. In that case, his match winning odds are 12.5%. To be more precise, let us assume that the chance that White will win a gammon or backgammon in any single game is one in eight. Then we have:

• The probability that White wins any single game is 50%. Of these, 37.5% are single games, and 12.5% are gammons or backgammons.
• The probability that White wins four games in a row without a gammon or backgammon is 37.5% raised to the fourth power, or almost 2%. (1.98%)
• The probability that White wins a gammon or better in the first of three games, and wins single games in the following two is (12.5%)*(37.5%)*(37.5%), or 1.76%.
• The probability that White wins three games in a row, with one gammon or backgammon somewhere in the mix, is three times 1.76%, or 5.27%.
• The probability that White wins two gammons in a row or better is 12.5% squared, or 1.56%.
• The probability that White wins the match is the sum 1.98% + 5.27% + 1.56%, or 8.81%. So, lets give him 9%, and estimate Black's match winning chances (MWC) at 91%.

If Black does not double, the computation is more complex! Black will arrive at the Crawford game, leading 6-0 on the scorecard, 85.4% of the time. As we just determined, he will go on to win about 91% of the matches from there. When Black loses this game, however, the match score will be 4-2 with Black 3 games away from victory, while White is 5 games away. By consulting a match-equity table, such as the one prepared by Kit Woolsey, we learn that Black’s match winning chances will then be roughly 66%. By combining all these numbers, we find that Black’s probability of winning the match when he does not double is (85.4%)*(91%) + (14.6%)*(66%), which is about 87.4%.

• If Black doubles, his MWC is 85.4% when White accepts and 91% when White passes. Clearly, White should take if doubled.
• If Black does not double, his MWC is 87.4%. Because White's take will reduce it down to 85.4%, Black should not double.

A Match between Unequal Players
The situation changes when the players are of greatly differing skill levels. If the weaker player has the Black checkers in Problem 91, he should double in a shot. Playing White, however, the stronger player prefers not to gamble the outcome of the entire match on a few shakes of the dice. He may reasonably choose to pass the double he would take against another expert. Conversely, the stronger player as Black must not double. If he does, we will strip him of his expert rating. As White, of course, the weaker player should gobble up the cube, and throw it back at the first opportunity.

My solutions:
Part (a): Double, pass.
Part (b): No double, take.

For the Record
I am so often wrong that I like to post my record in these messages. It's kind of a truth-in-advertising thing. Grunch: I have been answering these problems without the use of a bot, and before checking the excellent solutions of others, since Problem 28. My record at this writing is 51%.

This I only correct, i think, if we see this position as one roll situation. Considering not to double in a true two roll position, we get 84.85 x 90% and 14.35 x 66% for overall 86.55%. Taking this “rerecube vigorish” into account, i would basically not double and take. For b) it is instantly not a double.

I really want to congratulate you. Without messing with all the math, you've gotten right to the heart backgammon, and gotten it right.

And you two guys have my total respect. With about one-fourth the verbage, you say everything I had to say, and then some... Very nice.

Once again, here is a precision I aspire to. Well done.

All of you omitted the possibility that Black could roll 2-1 twice, and allow White to win on his second roll, and I think that, and the fact that the rest of us are using 91% MWC for 1c/5a, is why higonefive comes up with D/T between equal players. But this is just a small detail.

Thanks, again, for your contributions to this board.

2+1+2+1=6 which is just enough

9:1 odds = 1/10 not 1/9.

You probably already noticed that it's 7-away not 6-away.

My bad. Thanks for the correction. Apologies to other posters. For some reason, my brain locked, and saw Black playing his first roll of 2-1 as a forced move 4/1!

Here are the corrected numbers:

• White's cubeless probability of winning is (31*6)/(36*36), or 14.35%.
• Black's cubeless probability of winning is 100% - 14.35%, or 85.65%.
• If Black does not double, his MWC is (85.65%)*(91%) + (14.35%)*(66%), or 87.41%.

The conclusions for a match between equals are the same even with the new numbers:

• If Black doubles, his MWC is 85.65% when White accepts and 91% when White passes. Clearly, White should take if doubled.
• If Black does not double, his MWC is 87.41%. Because White's take will reduce it down to 85.65%, Black should not double.

Should this be: 66 / (66 + 10) = 66 / 76 = 86.84% is the minimum needed to double?

Alternatively, if you use 91% as the MWC for 1away 7away, would the same calculation be:

66 / (66 + 9) = 66 / 75 = 88% as the minimum needed to double?

If so, that would push you into the ND/T camp for a match between equal players.

It is not just that I made many small errors in my first post. Somehow, I got sidetracked into developing my own match-equity table from scratch! So, here is the redo, with conventional references to METs and an expanded conclusion that you may want to skip ahead to.

CUBELESS EQUITY IN A CASH GAME
In a cash game, this one is easy. Black cashes, regardless of the relative strengths of the players. Look at the odds:

• Black wins on his first roll when he rolls doubles two or better.
• If Black rolls 1-1 or a non-double, White can win by rolling any double on his turn.
• Otherwise, Black wins.
• White’s cubeless chance of victory is the compound probability that Black does not roll double twos or better (31/36) and that White rolls any double (1/6). The product is 31/216, or about 14.35%.
• Black’s cubeless chances are (100% - 14.35%), or 85.65%.

Of course, if Black cashes now, his chance of winning is 100%.

EQUAL PLAYERS IN A MATCH
In a match between equally skilled players, Problem 91 is a no-double/take. Here is the breakdown:

If Black doubles, and White takes, then Black will win the match with a victory in this game. If he gets a turn, White has a mandatory redouble. Thus, both sides will be playing for the match. Black will win 85.65% of the time.

If Black doubles, and White passes, then the score will be 6-0, with the Crawford game upcoming. Assuming there are no gammons or backgammons, White will have to win the next four games in order to win the match. His chance of doing that is 50% raised to the fourth power, or 6.25%. Even if you give him one gammon victory, White still must win three games in a row. His match winning odds are then 12.5%. And with a lucky backgammon, White might win the match in just two games. A match-equity table (MET), such as the one found at http://www.bkgm.com/articles/GOL/Oct03/met.htm, provides a statistical average of all these possibilities, giving White an overall 9.1% chance to win the match. To find Black’s match winning chance (MWC), we can look in the match-equity table, or we can just subtract. The difference between 100% and White’s MWC is 90.9%.

If Black does not double, the computation is more complex. The game will be played to its conclusion, with Black winning some 85.65% of the time. A win will take Black to the Crawford game, leading 6-0 on the score sheet. From there, as we determined above, Black will go on to win the match about 90.9% of the time. On the other hand, Black will lose this game 14.35% of the time, taking him to a match score of 4-2. In the parlance of match-equity tables, Black will then be 3 games away from victory, while White is 5 games away. At the so-called 3-away/5-away score, the match equity table gives Black a 64.6% chance of victory. By combining all these numbers, we find that Black’s probability of winning the match when he does not double is 85.65% * 90.9% + 14.35% * 64.6%, which is about 87.13%.

So, here is what we now know:

• If Black doubles, his MWC is 85.65% when White accepts and 90.9% when White passes. Clearly, White should take if doubled.
• If Black does not double, his MWC is 87.13%. Because White's take will reduce it to 85.65%, Black should not double.

A MATCH BETWEEN UNEQUAL PLAYERS
The situation changes when the players are of greatly differing skill levels. If the weaker player has the Black checkers in Problem 91, he should double in a shot. Even when his opponent takes, he loses only about 1.48% of his match equity. But what a deal! For this price, he buys an 85.65% chance to win the match based on dice alone against his world-class opponent. Playing White, the stronger player may not wish to risk the entire match on a few shakes of the dice. He may reasonably choose to pass the double he would take when playing against another expert.

Conversely, the stronger player as Black should not double. Just look at the numbers. He is not supposed to double even against a normal opponent. Why would an expert want to amp up the luck factor against a novice or intermediate? If he does, he had better have a rock-solid read on his opponent, and know that he will pass. If the expert’s double is accepted, then we will strip him of his expert rating. As White, of course, the weaker player should gobble up the cube, and throw it back at the first opportunity.

MY SOLUTIONS
Part (a): Double, pass
Part (b): No double, take

Grunch.

This is similar to a spot I posted where if you do redouble you instantly play for the whole match that I was questioning. Without delving into all the match equity numbers that were brought up in that thread, I'm gonna say that against a tough player we should probably push our immediate edge, double and hope to kill the game off as we are going to most likely be behind in future games and have little interest in reducing variance as that favours the better player susbtantially.

Against the weaker player, we should probably hold onto the cube, as our edge will exist over numerous future games, and we don't want to give them a cheap shot at playing for the whole match when they send the cube straight back on 8 next turn. Just like poker and pool, when playing strong players we have to try and take any edge we can see as future edges will probably be very hard to come by; but against weaker players we should err towards lowering variance, as our greater skill edge will reveal itself quite quickly over any kind of sample size.

a) double, pass (although this is v relative to how large skill divide is imo - prob a take between equals).
b) no double, take.

Thanks for the words of encouragement, Taper, and wtg on your in-depth posts itt!

I don't read any answers till i post, so sorry if this has been asked.
Is the Crawford rule in play at this tourney ? (my answer will presume yes)
...(a) Assume you are Black and playing a very strong player. Should Black double? ... NO!
If Black doubles, should White take or drop? ...
YES! BEAVER!
...(b) Same question, but now you are Black playing a weak player. ...
If weak means white won't beaver then yes double and since white is weak they would (take).
At best white is 1 out of 6 to tie and black has 5 rolls to win it all now.

I'll just point out that there are no beavers in tournament play, and the Crawford rule is standard in all matches.