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**8 members have voted**

To simplify the math, let's think of it as being based on a race, as opposed to backgammon, where one roll can have huge swings in the the probability of winning. The goal of each player will be to win as much as possible with utility in proportion to the amount won. In other words, you're 2x as happy to win $2 than $1.

The question for the poll is which statements do you agree with. Multiple votes allowed.

I've never ever used the doubling cube

I've watched a lot of backgammon

I've never seen a doubling cube used.

I'm the type of person that wants to play the whole game out and see who wins. Playing a game is about the fun factor of just playing and making lots of decisions after rolling the dice and see who wins at the end of the game. Taking money off a friend is not a fun factor for me. Simply beating a friend is all I need to add to the fun factor.

Backgammon is a great game in itself without a doubling cube

You can really use a doubling cube in any game if you want to allow it.

Maybe it would make some sports more exciting if they could use a doubling cube

Score 1st in a football game and go to the doubling cube 😀

Mahomes scores and Andy Reid throws out the doubling cube to the refs. How will the other team respond?

How about a doubling cube in chess

How will computer chess players use that against a human

Go to doubling cube after 1st move

Pretty bold but probably the best strategy :-)

Quote:TigerWuI've rarely played Backgammon, but never used the doubling cube. I've always thought it was a stupid idea.

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For most people

Getting a backgammon game with the doubling die is akin to getting a deck of cards with jokers. Just extra stuff not needed to play the game

The question I pose to the forum is what should be the probability of winning to double or accept a double, both for the initial double and a redouble.

on the cube usage.

I do however have a question is cube usage prevalent in professional tournament play?

I understand the doubling cube, but I have never used it. The proper use of the doubling cube requires that a player be able to make good estimates of the probability of winning the match and be able to update this probability after each roll. This where the odds are near 3:1 in favor of one or the other player. It is hard enough to find someone matched in ability to play against. Whichever player has more ability to move the checkers well is probably the one who can use the doubling cube more effectively. This would tend to make play more lopsided and less fun.Quote:WizardThe purpose of this thread will be to discuss when to make an initial double, when to accept/decline an initial double, and when to accept/decline when there has already been at least one double.

To simplify the math, let's think of it as being based on a race, as opposed to backgammon, where one roll can have huge swings in the the probability of winning. The goal of each player will be to win as much as possible with utility in proportion to the amount won. In other words, you're 2x as happy to win $2 than $1.

The question for the poll is which statements do you agree with. Multiple votes allowed.

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I think you can accept a double with less than a 25% chance of winning because then you own the cube. But, I am not sure how much that is worth.

When playing for money, the doubling cube hastens the end of matches and gets more money into play. It seems like pro players would hate to play without it. I realized long ago that I prefer to take money from faceless corporations, not from people across the table from me. I like losing to folks across the table even less.

I believe doubling is equivalent to a half pot bet in poker. Folding is equivalent to dropping/declining the cube.

The doubling cube has even spiced up a regular cribbage game I play. It adds a lot to the game and I would expect proper doubling in cribbage to be very nuanced when allowed during the play of the hand.

Quote:WizardI think the doubling cube is an integral part of backgammon and will almost not play without it.

The question I pose to the forum is what should be the probability of winning to double or accept a double, both for the initial double and a redouble.

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If you were playing me

I would assume you would double at the slightest advantage.

Me not playing since high school, not knowing optimum backgammon strategy and simply playing on gaming intuition.

I guess I would fold

At my skill level, it's the best move to lose a little rather then a lot.

Only in town another week.

Would love to play you without a doubling cube lol

Anyplace, anywhere except during Dead Sphere shows 😀

Rather play chess. Not good at that but more fun in that its a more complicated game and more moves to consider and pure strategy with no luck

Quote:terapinedIf you were playing me

I would assume you would double at the slightest advantage.

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Let me review the doubling cube rules.

Initially, the doubling cube is, shall we say, in nobody's possession. Either player may double before he rolls. However, if the double is accepted, that player has control of it. It is powerful to "own the cube." Only the owner can re-double. So, I would not double with a 51% chance of winning because you would then have the power of the cube. The game could swing your way and you double it back to 4x when you have an 80% chance of winning. Then I would probably lose 4 units. It's better to wait to make an initial double until your situation is more sure.

Make an initial double with 75% chance of winning or more.

Re-double with 80% chance of winning or more.

I have not worked out at what points you should accept if your opponent doubles, but I tend to think they are 25% and 20% respectively.

Quote:WizardHere are the results of my initial math, which could be wrong.

Make an initial double with 75% chance of winning or more.

Re-double with 80% chance of winning or more.

I have not worked out at what points you should accept if your opponent doubles, but I tend to think they are 25% and 20% respectively.

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75% seems right. 80% seems way too high except in a situation where the cube will have no value to you if you accept it again. Robertie has a relevant example in Advanced Backgammon

Quote:WizardQuote:terapinedIf you were playing me

I would assume you would double at the slightest advantage.

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Let me review the doubling cube rules.

Initially, the doubling cube is, shall we say, in nobody's possession. Either player may double before he rolls. However, if the double is accepted, that player has control of it. It is powerful to "own the cube." Only the owner can re-double. So, I would not double with a 51% chance of winning because you would then have the power of the cube. The game could swing your way and you double it back to 4x when you have an 80% chance of winning. Then I would probably lose 4 units. It's better to wait to make an initial double until your situation is more sure.

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Thanks

Wasn't familiar with the doubling rules

As usual, I don't know Jack and simplified it when of course its more nuanced and complicated.

Yes, I see that control of the cube is a big strategy concept now. There is a lot of luck involved so I see the logic of waiting for a bigger advantage instead of giving up ownership

Quote:WizardHere are the results of my initial math, which could be wrong.

Make an initial double with 75% chance of winning or more.

Re-double with 80% chance of winning or more.

I have not worked out at what points you should accept if your opponent doubles, but I tend to think they are 25% and 20% respectively.

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This ignores the natural doubling effect of getting a gammon or backgammon. There are circumstances when your chance of winning exceeds 75% where you shouldn’t offer a double because of that.

It’s also a cash game analysis. In match plays there are times when the player behind should offer the double at the first opportunity.

There are also end game scenarios where you should double with much less than 75% chance of winning.

Quote:WizardI think the doubling cube is an integral part of backgammon and will almost not play without it.

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As polite disagreement, I had family members who would play socially for about an hour a day, never using the cube.

Their goal was to socialize and try to play out every position, so the doubling cube would have been counterproductive. (They played enough that they had a stack of worn out backgammon briefcases in the garage; the checkers and cups from an early set were the ones they liked playing with, even though they had to keep replacing the board because the felt and points wore off.)

Doubling is advantageous for cash or tournament play. Unfortunately, it shortens the game, and removes some social appeal.

But playing for money the doubling cube is essential, and is part of the ongoing calculation of odds to win. Look at it something like "surrender" in blackjack. And even at 25 cents a point which is a typical stake for low end players such as college students, it can get up there, faster than you'd think.

There might be many moments of truth in a game - if the doubling cube is used.

Quote:RideTheEdgeQuote:WizardHere are the results of my initial math, which could be wrong.

Make an initial double with 75% chance of winning or more.

Re-double with 80% chance of winning or more.

I have not worked out at what points you should accept if your opponent doubles, but I tend to think they are 25% and 20% respectively.

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75% seems right. 80% seems way too high except in a situation where the cube will have no value to you if you accept it again. Robertie has a relevant example in Advanced Backgammon

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Okay, it's been a while since I've studied backgammon and I think my answer is a blunder. Or at least I couldn't find the example I thought existed in Advanced Backgammon where someone could take a cube with less than a 25% chance because of the efficiency of their redouble.

In any case, some thoughts (maybe wrong, again it's been a while since I've thought this stuff through.)

First off, Wizard's premise that we're just analyzing a race and not actual backgammon is critical. Maybe clarify by saying the race is arbitrarily long and that the edge changes very gradually without abrupt turns of fortune. Further note that there is no opportunity to double in which your opponent cannot double back at some later time.

If your opponent can double you back and you can also redouble (and so on), then s/he should accept your offer to double with any chance of winning over 25%. At 25% the take/drop is break even. There is no incentive to double with less than a 75% chance if the edge changes minutely at each doubling opportunity, so double when the edge inches up to just under 75% double in this theoretical race and the opponent is nearly indifferent to take/drop. In real games, the edge changes more abruptly so the cube should be offered at point of last take, which is under 75% when there is a chance of going over 75% on the next turn and losing our market. No need to wait until your edge is 80% before redoubling. At 24.9% chance of winning I think the take is a mistake. There are situations where you can take with less than 25% chance to win, but I think these only occur when you can return the cube without fear of getting it back. Sorry I couldn't find the example. I'll post it if further searching is successful.

I don't think probability of winning is the right way to think about the difference between owning the cube and having the cube in the middle. I think it's more about volatility in the position. By keeping ownership of the cube, you protect yourself from getting doubled when you change from being a favorite to being a dog (say after missing a double shot when far behind in the race if you expect more chances to hit later.) I think the situations where it is correct to offer an initial double but not redouble (and give up ownership of the cube) occur when the assumption that it is possible to redouble ad infinitum and the lack of abrupt changes is relaxed. In this case the person accepting the cube has ownership that is more valuable since they eventually are in a situation where they can give the cube but you can't accept and turn the game around to give it back. There cube ownership is then more valuable and you'll wish to deny that value by not offering it to them.

Regarding probability to win required to double, if you are at the point of last take and cannot be doubled back (e.g. the last roll of the game) then anything over 50% chance of winning is sufficient. This is a real world case and violates the assumption that the probability of winning changes very gradually.

Quote:DieterDoubling is advantageous for cash or tournament play. Unfortunately, it shortens the game, and removes some social appeal.

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Point taken. To me, backgammon is a gambling game like blackjack or poker. For some reason, I don't see the fun in playing it if money is not at stake.

With all due respect to Magriel, whom I knew before he passed, I will shortly post on why I think those percentages should be 20% and 80% for a redouble.

Quote:WizardIn chapter 22 on Doubling Theory in Backgammon by Paul Magriel, he says to accept a double with better than 25% chance of winning and to double with more than 75% chance. He doesn't distinguish between the first double or a redouble.

With all due respect to Magriel, whom I knew before he passed, I will shortly post on why I think those percentages should be 20% and 80% for a redouble.

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That result cannot be correct. Once I accept a cube, it does not matter if it came from the center or the other side of the board. If I have a 22% chance of winning a particular position, it cannot be correct to drop an initial double but accept a redouble. In both cases I accept the double if my equity owning a doubled cube is greater than -1*(undoubled cube)

Quote:RideTheEdgeThat result cannot be correct. Once I accept a cube, it does not matter if it came from the center or the other side of the board. If I have a 22% chance of winning a particular position, it cannot be correct to drop an initial double but accept a redouble. In both cases I accept the double if my equity owning a doubled cube is greater than -1*(undoubled cube)

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I think the situation is different. If the cube is on the center, there is more of an incentive to double, lest your opponent beats you to it. Once you control the cube, you can bide your time with it and wait for a better opportunity to strike.

Let:

p = Your probability of winning.

x = Some probability of you winning that is so low that your opponent will double if he has the cube. At this point you will be indifferent to dropping and accepting, so to keep things simple, assume you lose by dropping if your probability gets to x.

By reflective logic, if your probability of winning gets to 1-x, you would double and your opponent would be put to an indifference point. Again, to make the logic simple, say he drops.

Let's say the game started with a bet of 0.5 units and your opponent doubled it to 1. You know own the cube.

We need to find a point such that your expected win is 1 unit whether your opponent drops of accepts. If he accepts, and your probability of winning drops to x, then he will double, you will drops and lose 2 units.

So the probability of you winning is (1-2x)/(1-x). Likewise, the probability of losing is x/(1-x).

2*(pr(win) - pr(lose)) = 1

2*((1-2x)/(1-x) - 1/(1-x) = 1

2*(1-3x) = 1-x

2-6x = 1-x

1 = 5x

x = 1/5

So, you will drop if your probability of winning gets to 20%. By reflection, you should double at 80%.

How is the probability p determined? At some point late in the game, an AI program can calculate the probability of a win assuming random dice and both players playing optimally. However, I don't know whether the probabilities are the same with and without considering the doubling cube.Quote:WizardHere is my basic logic for my double at 80% advice.

Let:

p = Your probability of winning.

x = Some probability of you winning that is so low that your opponent will double if he has the cube. At this point you will be indifferent to dropping and accepting, so to keep things simple, assume you lose by dropping if your probability gets to x.

By reflective logic, if your probability of winning gets to 1-x, you would double and your opponent would be put to an indifference point. Again, to make the logic simple, say he drops.

Let's say the game started with a bet of 0.5 units and your opponent doubled it to 1. You know own the cube.

We need to find a point such that your expected win is 1 unit whether your opponent drops of accepts. If he accepts, and your probability of winning drops to x, then he will double, you will drops and lose 2 units.

So the probability of you winning is (1-2x)/(1-x). Likewise, the probability of losing is x/(1-x).

2*(pr(win) - pr(lose)) = 1

2*((1-2x)/(1-x) - 1/(1-x) = 1

2*(1-3x) = 1-x

2-6x = 1-x

1 = 5x

x = 1/5

So, you will drop if your probability of winning gets to 20%. By reflection, you should double at 80%.

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For example, a safer move might reduce the chance that p goes below x and lose immediately, but it also reduces the chance that p gets above 1-x and win immediately. If optimum tactics are not independent of the cube, and cube strategy is dependent on x, then there is a feedback loop that makes it harder to come to a definitive answer on the overall problem.

In any case, I just assumed that the question you pose could be verified by MC simulations for simple late game positions. So I assumed this is a solved problem. I would have thought this was only an open question earlier in the game where there is more uncertainty about the exact current value of p, and where there is more time for the cube to pass back and forth.

In a real game where calculators are not allowed, the uncertainty in humans estimating p might be more than the uncertainty in x. In any case, I agree that x would be less than 0.25. I would have to spend a lot of time working with a calculator before I could estimate p for a position +/- 0.05.

At any rate, I appreciate you starting a thread that is based on probability theory as relates to a real-life pro gambling situation. Not many threads here are directed at figuring out how to make more money at a real-world AP game.

When both players will have a live cube, Mr. Ward's analysis probably mirrors the Wizard's since he concludes the minimum take point is 20%. Ward cites Danny Kleinman though so if we're going to chase cites to give credit then the article entitled "The Cube Equations" on pages 237 to 240 of "Vision Laughs at Counting with Advice to the Dicelorn Volume 2 (c) 1980 is the source. After dragging Kleinman off my bookshelf I note that Kleinman also demonstrates the 20% minimum take point (and 18.75% special case mentioned above) when the positions are symmetric in a continuous approximation to backgammon where equities change gradually. Kudos to the Wizard for reaching this same conclusion.

I would only point out that the take point for accepting an initial double must also be 20%, not 25% per the arguments in my prior post.

I'm not so sure how much cube ownership matters in this theoretical continuous approximation. It may only matter when the probability of winning jumps around discontinuously. Kleinman discusses the difference between an initial double and redouble after relaxing the continuous approximation and analyzing a discontinuous model. His analysis focusses on the need to double prior to gradually approaching the 80% optimum and the cost of doubling too early with the cube in the middle vs. the cube owned. In any case, the question is how far from 80% to deviate and he does not derive a 75/25 figure.

Ward's discussion is introductory in a book that focusses on reference positions for practical use over the board. Kleinman is great at the theoretical stuff and has the meatier discussion.

Quote:MentalHow is the probability p determined?

<clipped>

In a real game where calculators are not allowed, the uncertainty in humans estimating p might be more than the uncertainty in x. In any case, I agree that x would be less than 0.25. I would have to spend a lot of time working with a calculator before I could estimate p for a position +/- 0.05.

At any rate, I appreciate you starting a thread that is based on probability theory as relates to a real-life pro gambling situation. Not many threads here are directed at figuring out how to make more money at a real-world AP game.

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As a practical matter in cash games 25% is used as a take point and doubles are rationally offered at a range of probabilities above 50%.

In tournaments the take point is highly dependent on the score and is all over the map. The better players memorize the tournament take points and have various heuristics to get a pretty good estimate over the board. There are a variety of formulas for calculating races. The probability of hitting/escaping is easily calculated. The probability for a lot of reference positions is known and you can estimate a deviation from this reference. Based on a reference position you might know a formula for how much each pip for checkers outside your board increases your chance of getting gammoned, or if you've borne off some checkers and are hit in a backgame what your probability of winning is based on the number of checkers closed out on the bar. And so on.

Nobody seems to agree with me, but I think those markers are different for an initial double. As I've stated before, I think the player should be in more of a rush to double initially, lest things swing and the opponent do so first.

That said, if we accept my 20% and 80% markers, then there is only a space of 0.6 where the game is played. At some distance x from these markers, will be the indifference points to accepting and rejecting a double. Let's look at when the opponent is indifferent to accepting and double and dropping:

1 = 2*[(0.6-x)/0.6 - x/0.6]

1/2 = (0.6-x)/0.6 - x/0.6

0.3 = 0.6 - 2x

2x = 0.3

x = 0.15

Remember, this is a distance from 0.2 and 0.8.

So, the sweet spot to make an initial double is 0.8 - 0.15 = 0.65. Likewise to accept/drop a double is 0.35.

Even if all this is right, it still only gives us the sweet spots, or the perfect places to double. I have not looked at when it is good enough to double and not bad enough to accept a double.

Quote:RideTheEdgeI found a couple discussions of when the cube can be accepted with less than a 25% chance of winning, actually the theoretical minimum is 18.75%. ...

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Thank you for an outstanding post. I've done some research and it seems this book is well out of print. I can't find it on Ebay or Amazon.

I am almost to the point of running simulations to find the "good enough" points. I know from playing thousands of games that doubles are often accepted. I would compare it to having a weak hand in poker and you're confronted with a moderate raise. Often calling has a higher EV than folding, although both situations are negative.

Quote:WizardQuote:RideTheEdgeI found a couple discussions of when the cube can be accepted with less than a 25% chance of winning, actually the theoretical minimum is 18.75%. ...

link to original post

Thank you for an outstanding post. I've done some research and it seems this book is well out of print. I can't find it on Ebay or Amazon.

I am almost to the point of running simulations to find the "good enough" points. I know from playing thousands of games that doubles are often accepted. I would compare it to having a weak hand in poker and you're confronted with a moderate raise. Often calling has a higher EV than folding, although both situations are negative.

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I'm not sure of the copyrigjt implications if I photographed this article and either posted it or sent it to you. If legal I'll do so gladly

Quote:WizardThis post is a follow up to this one, where I said the player should make a redouble at 80% and accept it at 20%.

Nobody seems to agree with me, but I think those markers are different for an initial double. As I've stated before, I think the player should be in more of a rush to double initially, lest things swing and the opponent do so first.

That said, if we accept my 20% and 80% markers, then there is only a space of 0.6 where the game is played. At some distance x from these markers, will be the indifference points to accepting and rejecting a double. Let's look at when the opponent is indifferent to accepting and double and dropping:

1 = 2*[(0.6-x)/0.6 - x/0.6]

1/2 = (0.6-x)/0.6 - x/0.6

0.3 = 0.6 - 2x

2x = 0.3

x = 0.15

Remember, this is a distance from 0.2 and 0.8.

So, the sweet spot to make an initial double is 0.8 - 0.15 = 0.65. Likewise to accept/drop a double is 0.35.

Even if all this is right, it still only gives us the sweet spots, or the perfect places to double. I have not looked at when it is good enough to double and not bad enough to accept a double.

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If I accept a dead cube (meaning I cannot profitably redouble) then I'm indifferent to taking with a p=25% chance to win. Cube at C, dropping is worth -C. Taking is worth 2C*p - 2C*(1-p)

-C = 2C*p - 2C*(1-p). ;at what p is dropping the same as taking

-0.5 = p - (1-p) ;divide by 2C

-0.5 = 2p-1

2p = 0.5

p = .25

I think accepting your 20/80 numbers means I'm indifferent to take/drop with 20% probability of winning and a live cube. I'm indifferent to take/drop with an 18.75% chance of winning if my cube is live and yours is dead if I'm lucky enough to give it back.

The sweet spot to make the initial double is when my opponent is indifferent to take/drop, theoretically 80% in the "continuous" game with live cubes.

All of my calculations assume infinite redoubling is allowed.

Quote:WizardThis post is a follow up to this one, where I said the player should make a redouble at 80% and accept it at 20%.

Nobody seems to agree with me, but I think those markers are different for an initial double. As I've stated before, I think the player should be in more of a rush to double initially, lest things swing and the opponent do so first.

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This matches my (anecdotal) experience against software opponents.

I double fairly aggressively, where I expect that the software is a conservative approximation of a book theory. I think I do OK.

Quote:WizardJust so we're clear, what is the definition of "live" and "dead" cube.

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My cube is dead if I cannot increase my equity by doubling. For example, when winning is a certainty.

My cube is completely alive if I double my equity by doubling.

This happens in the theoretical game we're modeling in this thread where we can double with an 80% chance of winning.

Except for some very specific cases the value of the cube is somewhere in between.

Quote:WizardAlso, what is the proper term for a cube in middle?

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Cube in the middle is fine

Quote:WizardI've done some research and it seems this book is well out of print. I can't find it on Ebay or Amazon.

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I'm sure the Jeff Ward book is obsolete now. We're in the age of computer perfect play and there are better resources.

Kleinman's stuff is interesting regardless since he approaches the game mathematically and has a sense of humor. I saw a reprint of Vision Laughs at Counting on Amazon, though it is expensive. I don't know if it is a legit reprint or if it has the article I referenced.

Quote:RideTheEdgeQuote:WizardI've done some research and it seems this book is well out of print. I can't find it on Ebay or Amazon.

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I'm sure the Jeff Ward book is obsolete now. We're in the age of computer perfect play and there are better resources.

Kleinman's stuff is interesting regardless since he approaches the game mathematically and has a sense of humor. I saw a reprint of Vision Laughs at Counting on Amazon, though it is expensive. I don't know if it is a legit reprint or if it has the article I referenced.

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Kleinman books and many others can be purchased here: https://www.flintbg.com/boutique.html#anchor6807056

Doubling in Backgammon" here: https://bkgm.com/articles/KeelerSpencer/OptimalDoublingInBackgammon/

Quote:MDawgMy experience is that the doubling cube is handed over pretty early in the game, and in some games, tends to go back and forth quite a lot.

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That sounds like a nice loose game. Juicy.

Quote:WizardTo give an update, under my assumption that the probability of winning changes on a continuous basis, like a race, two risk averse players would never accept a double. I tend to think two skilled players in backgammon will double and accept so much is single rolls can have huge changes in the odds late in the game. In other words a roll can suddenly propel the probability of winning across the 80% line by significant amounts.

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Agreed that in an ideal continuous game risk averse players would never accept. The cube would be offered at exactly 80%, the recipient would be indifferent to take/drop and would drop to minimize variance.

Since a roll can propel the probability of winning well beyond 80% as you note, the favorite must double earlier to avoid losing their market. As the recipient of the cube, the dog will be in a similar position if they turn the game around to redouble, so in general they cannot take with a mere 20% chance of winning because their cube will only be partially live. They'll need more than 20%, but no more than the 25% they would need if the accepted cube were completely dead. So, the real-world take point is somewhere between 20% and 25%

"If you aren't absolutely sure whether the position is a take or a pass, then it is always correct to double."

Player A has the cube with chips on the 1 and 5 spots.

Player B has a single chip on his own 1 spot.

Player A has 13* ways to lose and 23* to win. The EV of that situation is 0.277778. By doubling it is 0.555556. Player B would accept because losing 0.555556

is better than losing 1.

* Corrected

Quote:WizardHere is a situation where it is correct to double and for the other player to accept it.

Player A has the cube with chips on the 1 and 5 spots.

Player B has a single chip on his own 1 spot.

Player A has 7 ways to lose and 29 to win. The EV of that situation is 0.277778. By doubling it is 0.555556. Player B would accept because losing 0.555556

is better than losing 1.

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I think that is not correct. With checkers on the 5 and 1 points, player A misses with 4-3, 4-2, 4-1, 3-2, 3-1, 2-1, 1-1 for a total of 13 misses out of 36, not 7. Count the non-doubles twice. Agreed this is a double and a take though. I did not verify your EVs.

But say player A did have 7 out of 36 misses. This happens if you move the rear checker to the 4 point instead of the 5 point. Now 3-2, 3-1, 2-1, and 1-1 miss. This is a double and a drop. Player A wins 29/36 with the cube at 2 (total +58) and loses 7 when he misses with the cube at just 2 (he'll drop a double when certain to lose, so -14.). 58 minus 14 is 44. Player B is better off dropping all 36 for -36 rather than taking and losing 44 over 36 games.

If player A had 9 misses, that's break even for player B to take or drop. Player A doubles and wins 2*27=54 and loses 2*9= -18. 54-18=36. So player B does just as well to take or drop with 9 winners. But wait, that's only 25%! The reason player B needs 25% to take is that his cube is completely dead. He cannot win any games with a redoubled 4 cube since player A will always drop the redouble. Cube is dead ==> 25% winning chance required to take.

Quote:RideTheEdgeI think that is not correct. With checkers on the 5 and 1 points, player A misses with 4-3, 4-2, 4-1, 3-2, 3-1, 2-1, 1-1 for a total of 13 misses out of 36, not 7. Count the non-doubles twice. Agreed this is a double and a take though. I did not verify your EVs.

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You're right. I had 13 in my spreadsheet for 1-5, but looked at 7 for 1-4 when I made the post. My EV's were right, I believe.