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Undefined Equity - Part 1

by Douglas Zare
2 July 2004

Douglas Zare

"I have intuition about quantum mechanics.
The problem is that my intuition is all wrong."
-- John Preskill, a professor of physics at Caltech.

The doubling cube may escalate so rapidly that the theoretical average value of a position (equity) may not be defined. Backgammon is the only 2-player zero-sum game allowing this possibility. This is a problem for any rigorous theoretical analysis of the whole game of backgammon, since it is unclear whether equity is defined from the initial position. It's also fun to think about it

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Subject: Re: Undefined Equity - Part 1

From: Casper

Date: 03 Jul 2004 05:40 EST

Very interesting as usual Douglas!

What is wrong with the following logic in the Paradox Position: Suppose Red redoubles. You have already shown that Redoubling gains if White fails to redouble if Red dances. So only if White's recube after Red's dance is correct mat Red lose out by redoubling the original position. However, since Red's and White's decisions are identical, White's recube can only be correct if Red's recube is also correct. So the maniac strategy is correct.

Subject: Re: Undefined Equity - Part 1

From: Casper

Date: 03 Jul 2004 05:57 EST

And under the conditions you specify (rolling a 6 is worth 2 with the cube on 1), I'd guess Red's expectation after doubling is 22/86 points times the value of the cube. However, to get that answer I used algebra, and I must confess not understanding your example (especially the "E").

Subject: Re: Undefined Equity - Part 1

From: zare

Date: 04 Jul 2004 05:37 EST

Suppose I assume that my opponent is a maniac, but I am forbidden to use the maniac strategy. In that case, it is wrong to redouble. In some ways this resembles real life, since I won't redouble to stakes I can't cover.

I think real life may be modelled better by part of the n- brave versus m-brave table. You (as first roller) get to choose the row, while your opponent chooses the column. If a player will not double to stakes that can't be covered, this limits the choices to the first few rows or columns.

If you are willing to redouble up to 3 times (to 4, 16, and 64), and your opponent is willing to redouble up to 3 times (up to 128), then the restricted table appears as follows:

Second Roller 0 1 2 3 First 0 .361 .361 .361 .361 1 .721 .220 .220 .220 2 .721 .916 -.050 -.050 3 .721 .916 1.292 -.572

In this restricted game, the last column dominates the others. The entries in the last column are the lowest in each row. So, the second player should be willing to redouble 3 times, and the last column is the only relevant one. The highest entry is in the first row, so you should not redouble.

However, If your opponent is not able to redouble the third time, this is the payoff table:

Second Roller 0 1 2 First 0 .361 .361 .361 1 .721 .220 .220 2 .721 .916 -.050 3 .721 .916 1.292

Now the bottom row dominates the other rows, so you should be willing to redouble 3 times, while your opponent should not redouble.

Before you test your opponent, you might not know whether you are in the former or latter restricted game, and this information determines the best move.

You might say that rather than the first and second player's choices being identical, they are exactly opposite. If the first player is willing to redouble more than the second, the first player should be willing to redouble, but the second player should refrain from redoubling. After redouble/take and a miss, the position is not the same, because the roller has changed, and who is willing to redouble more has switched from roller to nonroller or from nonroller to roller.

Douglas Zare

Subject: Re: Undefined Equity - Part 1

From: zare

Date: 04 Jul 2004 06:01 EST

If you try to find the expected value, the sum you get is a nice geometric series,

A (1 + R + R^2 + R^3 + ...)

where A = -2 (11/36) C, and R = -2(25/36).

If r were between -1 and 1, then the geometric series would converge to A/(1-R). Here, r<-1, so the series diverges. The partial sums oscillate off to +-infinity. The algebra makes it look like the sum is still A/(1-R), but as far as I know, this is not meaningful.

Douglas Zare

Subject: Re: Undefined Equity - Part 1

From: Chris Yep

Date: 08 Jul 2004 21:56 EST

I like the presentation. I'm looking forward to your second part (table stakes and match play).

If I remember right, a sufficient condition for the 1- animal (beavers allowed, raccoons not allowed) table to be valid is (1) the cubeless equity (of the player on roll) is greater than 0 and less than 0.5, (2) correct checker play is idential to money-cubeless checker play, and (3) the cube has no value to either player once someone rolls 6-x from the bar. [Equivalent to (1) is "the value of the first 6 is 61/22 = 2.77 times the cube"]. These are very robust assumptions. "Clearly" (1) is true by a comfortable margin and (2) and (3) are almost true, so overall the 1- animal table is almost surely valid, which is a "nice" result in my opinion. I'm looking forward to using this knowledge, but unfortunately in my chouettes I've yet to reach this Paradox position... Still, it's fun to think about the results...

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