Quiz : First Six Wins

[nabla]

In an hypothetical position where the game is (100%) won by the first player to roll a six, and no gammons are possible, what would be the correct cube action ?
I suspect that this theoretical quiz question must already be known, but even if it is the case I would then enjoy confronting my solution with the official one !

[Hardy_whv]

That's a nice quiz  

In any case, where the first player rolling a specific number, certainly wins, there is a 59% probaility for the player on roll. If he doesn't roll the required number, the other player then has a 59% probability of winning. And so on.

The cube handling depends on the match score (and also on the probability of gammons). So there is no general answer to this question. Let the gammon chances be very low and lets assume a money game or a "normal" match score (i.e. at the beginning of a longer match), it's a No Double -Take.

It's not that easy to generate positions, where such circumstances apply. One example is given in Walter Trice's book "Backgammon Boot Camp". It's a position that can't possibly happen, but it's a nice example:



Here the player rolling the first 6 wins nearly certainly (with only very few excemptions).

Knowing the number 59% for such cases is very helpful, as for many other positions with a similar situation (players needing a specific number) the probability of winning can be derived from that number.

Looking forward for more quizzes  

Hardy  

[Zorba]

If you mean two players with one checker on their sixpoint each and 14 men off, this is a well-known position for backgammon theory:

(Re)double, optional take.

In words:

Player needs to roll 6 pips or more to get off. That is any roll except 1-1, 1-2, 1-3, 1-4 and 2-3, so 27 out of 36, 3/4 or 75% to win the game right away.

If Player misses, then Opponent needs to roll 6 pips or more, again a 75% chance. If Opponent misses too, then Player will always win: at worst he rolled a 2-1 and moved 6/3, so he will always bear off his checker on the second turn.

What does this mean for the cube?

If Player on roll doesn't roll six pips or more, Opponent has a perfectly efficient redouble (Player has an optional take). The original Player's takepoint (TP) on that recube will then be 25% since he won't have  a chance to make an effective re-redouble himself anymore. So when Oppoent is deciding on taking the Player's original cube, he should realize that his recube vigorish turns his 75% winning chances after a miss into a straight win.

Now, without any recube vigorish, your TP is 25% in money games. Here, the recube vigorish is such that it turns 75% winning chances into 100% for Opponent, so with use of the cube, he wins 4/3 as much. That means he only needs 3/4 of 25% to take: 18.75% The recube vigorish turns that into 25% effectively.

Since Opponent has exactly 18.75% winning chances in the position without considering the cube, it is an optional take and a mandatory double for Player.

In numbers:

Player wins on first roll: 3/4 or 75%
Player does not win on first roll: 1/4 or 25%
Opponent gets to roll and wins: 1/4 * 3/4 = 3/16 or 18.75%
Opponent gets to roll and does not win, so player wins on second roll: 1/4 * 1/4 = 1/16 or 6.25%

So, after a double/take (cube on 2), this will happen:

Player wins: 3/4 * +2 points = 1.5 point
Player does not win on first, opponent redoubles:
Either Player passes for an immediate 1/4 * -2 points = -0.5 point, or:
Player takes, opponent wins: 1/4 * 3/4 * -4pts and opponent does not win: 1/4 * 1/4 * +4pts, combined: -12/16 + 4/16 = -0.5 point also: optional take.

So overall, Player wins 1.5 - 0.5 = 1 point after double/take.
If opponent passes, Player also wins 1 point.

Therfore, it is an optional take for opponent, even with only 18.75% cubeless winning chances.

Of course the double is hugely correct, without it Player would win:

3/4 * +1 point = 0.75 point
1/4 * -1 point = -0.25 point (optional pass, taking amounts to the same equity)

0.75 - 0.25 = 0.5 point

So after no double, Player only wins +0.5 instead of the +1 after a double; it is a very strong and perfectly efficient double!

[nabla]

Whooops sorry, I forgot to say "Money game" in the quiz question !!

Zorba, your position is very interesting too, but the illegal position that Hardy set up was the one I was describing - with the exception that I would have added two men off to discard any gammon.
But actually, I was rather thinking of a game with dices only, of "who-will-roll-the-first-six".

I agree with the 59%. But I think that it is a redouble-take and that I can prove it. Did Trice say it was a no-double ?

[Hardy_whv]

I agree with the 59%. But I think that it is a redouble-take and that I can prove it. Did Trice say it was a no-double ?

Well, Trice said:

"The concepts involved in deciding what to do with the cube here have been driving gamblers to bankrupcy and madness for centuries. Let it suffice to say that unless your pockets are extremely deep, you just don't want to play a game in which this kind of decision comes up."

I think, 59% is far from doubling. Lets assume, the player doubles, the other player takes. No 6 is rolled (69%). Now you have the same situation for the other player. He doubles, the other player takes, no 6 rolled (69%). And so on. This happens 6 times. The probability that there was no 6 for six consecutive rolls, is about 11%. The Cube now is on 64! And that can continue to bounce back and forth for much longer. Well, thats not Backgammon, thats pure gambling.

In money game, if there is no additional gammon threat, I wouldn't double with 59%.

Hardy  

[Zorba]

Sorry, I see I have misinterpreted the original post! I remember seeing something similar to Hardy's position and what you describe in rec.games.backgammon, many years ago. The discussion is archived here: Undefined equity

The answer there, with lots of gammons in the air, seems to be that both players should double and take until the six is rolled: they are a favourite to win when on roll, and 11/36 rolls lose the market by a mile.

This situation opens up unlimited cube values, and if you try to calculate the cubeful equity, you will get a series with a sum that won't converge. So, the cubeful equity will be undefined!

In your theoretical situation with no gammons, a quick try gets me this. I could well be wrong though!

396/1296 win the game right away (11/36)
275/1296 lose the game on opponent's next roll (25*11/36^2)

for a net + 121/1296 points in these 671/1296 situations

and

625/1296 repeat situation (25*25/36^2)...

If you (re)double, opponent redoubles after your miss, so your immediate wins are doubled but your losses quadrupled, then the situation repeats and it's consistent to cube to 8 then.

Now you have:
396/1296 wins for +2 points
275/1296 losses for -4 points

for a net -308/1296 points in these 671/1296 situations

and 625/1296 repeat situation with cube value four time higher

Maybe I'm missing something here but it sure looks like (re)doubling is very wrong, going from +121 to -308 for 2-roll sequences and essentially repeating this every two rolls when the game is not over yet, with an ever quadrupled cube?!

[socksey]

My brain went on "tilt" about 3 posts ago.  No wonder my rating doesn't seem to move.  

socksey



"Morcelli has four fastest 1500-metre times ever. And all those times are at 1500 metres." - David Coleman, sportscaster (this, I can understand)    

[nabla]

Wow, I did not expect so much input on this subject, all very interesting !

Hardy, I have to agree with Trice's recommendation not to double in a real-world situation. An economist would say that we have to take into account a non-linear utility function, i.e. winning $2'000'000 is not twice as good as winning $1'000'000 (at least for me - everyone has a different utility function).

But here is why it is a double when you only seek to maximize the mathematical equity :

Suppose that this position is NOT a redouble. Suppose now that I am dumb and that I still redouble. By hypothesis, my opponent, being in the same situation as I was, does not have a redouble, and will never have one, which means that he owns a dead cube ! (Recall that if he rolls a 6, he is a 100% winner and the cube is useless). So I have sent a dead (=useless to my opponent) cube while having a positive equity, which mean that I have just doubled my equity, and have been correct to redouble. This contradicts the hypothesis that the position was not a redouble.
What I find funny is that this proves that redouble is correct without comparing the actual equities of doubling or not doubling.

Zorba, thanks for the link ! But now I have to look into your calculations to find where the mistake is  
At first sight it seems to me that you wrongly discard the gain made in repeating the situation with the cube four times higher - which is a gain as the repeated position is a 59% win !

[nabla]

Zorba, actually bobk's post in the link you gave casts some doubt on my nice solution !
Everything is all right given that the equity of the position exists, but it looks like it doesn't. I must think about this again, didn't expect my own quiz to be a brainteaser for myself    

[dorbel]

Just curious. Why do people spend hours pondering this sort of thing, when there is so much to be learned about backgammon?  

[webrunner]

Post a better quizz then Dorbel

[lewscannon]

Interesting thread here (sorry, Dorbel). But it seems you guys are way complicating this. If the initial poster was describing the impossible board described (though I could swear this game happened to me once with a bot, and I lost), it's really very simple. Hardy, I don't know where your 59% comes from. It's a question of rolling a 6 with one die or the other, and that's a 33% shot. What makes it a potential cube is the possibility of then covering the 6 point. Doesn't look like a cube to me, though.

[nabla]

59% is the probability to win the game (cubeless).

It comes from the formula 1-p=25/36*p (the winning probability for the second player is the probability that he survives the first roll, times the winning probability for the first player), giving p=36/61.

For me, this question was first meant to be an idealisation of the common situation where you want to be the first to hop your opponent's prime (but are not sure to win then). I don't think that it is a useless one. The problem is that it doesn't seem to have a solution !

[Zorba]

But here is why it is a double when you only seek to maximize the mathematical equity :

Suppose that this position is NOT a redouble. Suppose now that I am dumb and that I still redouble. By hypothesis, my opponent, being in the same situation as I was, does not have a redouble, and will never have one, which means that he owns a dead cube ! (Recall that if he rolls a 6, he is a 100% winner and the cube is useless). So I have sent a dead (=useless to my opponent) cube while having a positive equity, which mean that I have just doubled my equity, and have been correct to redouble. This contradicts the hypothesis that the position was not a redouble.

I think there's a logic error in your argument: you use as a premisse that the position is not a redouble (not for you and thus, not for your opponent). But then you also assume positive equity. The two are contradictory for this position, so you can't make both of these two assumptions to construct an argument, trying to prove that it is a redouble after all!

[Zorba]

At first sight it seems to me that you wrongly discard the gain made in repeating the situation with the cube four times higher - which is a gain as the repeated position is a 59% win !

What I tried to show is that if you (re)double, then opponent should also redouble when he has the opportunity of course (symmetrical situation) and then the cubeful equity is negative (it is not uncommon to have negative equity if you wrongly double a 59% wins position!)

Here, the result of doubling is that that your immediate wins will be doubled, but your immediate losses quadrupled, with a negative (cubeful) equity as a result.

The situation repeats if neither player rolled a six, and since the situation is still the same, both players should keep redoubling. But if redoubling only leads to negative equity on the first two plies if the game finishes, then this should also be true for any future two ply sequence.

So basically, the 59% wins seem to lead to negative equity if you cube; there are not enough marketlosers to make redoubling corrrect.

The position in the forum archive I referred to has a lot of gammons floating around (I tihnk more than half of the wins after a six will be gammons there), which makes the market-losers much bigger, turning that position into a correct (re)double/take.

[nabla]

I think there's a logic error in your argument: you use as a premisse that the position is not a redouble (not for you and thus, not for your opponent). But then you also assume positive equity. The two are contradictory for this position, so you can't make both of these two assumptions to construct an argument, trying to prove that it is a redouble after all!

You are correct that I made a logic error, but it was a bit more subtle. The error was to assume that the position had an existing equity. If yes, this equity is clearly positive, because in the worst of case you can choose not to double, cash 11/36 of the games and leaving your opponent in the same situation as you were (I am speaking of the cube-in-center case here, but cube in hand is better or equal to cube in center).
In other words, E is greater or equal to 11/36-E, so E is greater or equal to 11/72.
The only problem is that in fact E does not exist. I wonder whether an argument for doubling could still be found, but it must not state the problem in terms of equity - in which terms then ?

[nabla]

Here, the result of doubling is that that your immediate wins will be doubled, but your immediate losses quadrupled, with a negative (cubeful) equity as a result.

No, doubling has the two effects you mention, but it also has a third one. If no one wins for the next two plies, the first player will be very happy that the cube shows 4 instead of 1 in a position that he is 59% favorite to win. He will be happy whether he now chooses to double or not.
The equity when both opponents double at each turn is not negative, but it is not positive either, nor null. It doesn't exist, because it is defined as the sum of a non-convergent series :
11/36*(1-2*25/36+4*(25/36)^2-8*(25/36)^3+-.......)
where unfortunately each term is greater than the previous one.

[Hardy_whv]

Interesting thread here (sorry, Dorbel).

I agree  

But it seems you guys are way complicating this. If the initial poster was describing the impossible board described (though I could swear this game happened to me once with a bot, and I lost), it's really very simple. Hardy, I don't know where your 59% comes from. It's a question of rolling a 6 with one die or the other, and that's a 33% shot.

I don't agree here. It's not a question of rolling a 6 with your next roll (and that wouldn't be 33%, but 30,5555%, btw). It's the overall probability that you roll a 6 before your opponent, and that's:

- Rolling a 6 with your 1st roll

plus

- not rolling a 6 on your 1st roll, and your opponent doesn't either, but you roll a 6 on your 2nd roll

plus

- not rolling a 6 on your 1st roll, your opponent not rolling a 6 on his 1st roll, you not rolling a 6 on your 2nd roll, your opponent not rolling a 6 on his 2nd roll and you rolling a 6 on your 3rd roll

plus

- and so on, and so on.

This sums up to nearly exactly 59%.


Hardy    

[nabla]

Yes, hardy, this sums up to 59%=36/61 for the reason I gave (a bit too shortly probably) 6 posts ago.

Zorba, maybe I managed to use your calculus to prove that the position is a cube after all ! For this, I will carefully avoid to compute any equity where the players may double forever. At first I will compute the cubeless equity.
It is easy to calculate or to derive from the 59% probability. It must satisfy E=11/36*-25/36*E, which solves to E=11/61=0.18 (about)
Now I want to know the equity in the situation you mentioned, that is if it is decided that I will double, my opponent will double too if given the chance, and then I will not double any more and keep the cube forever.
There are three possibilities :
1) I win immediately with probability 11/36, cashing 11/36 equity.
2) I don't roll a six, my opponent does, here the probability is 25/36*11/36 and the cube is on 2, that gives me an equity of -550/1296
3) Nobody rolls a six (probability (25/36)^2, the cube is on 4, I win four times my cubeless equity which is 11/61, that gives me an equity of 4*(25/36)^2*(11/61)

1), 2) and 3) sum up to 0,229, better although not by much that the 0.18 cubeless equity. Well, if I did not make any mistake. Did I ?

Of course, the argument can be repeated to show that doubling is correct at any point, at least until you are reaching the bottom of your pockets !

[Zorba]

Yes, there is a mistake, you are turning the cube after the roll instead of before it. Assuming you own the cube (even if it's on 1) and double, and opponent doubles too, then you don't double anymore, the game starts with double/take so the cube is already on 2 when you roll. So you get:

2*(396/1296)-4*(275/1296)+ 4*(625/1296)*(11/61) = 0.11

which is less than the 0.18 cubeless equity, so I still believe it's a no double!