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Jul 29, 1998, 3:00:00 AM7/29/98

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Does anyone have any idea roughly what fraction of doubles should be

taken or dropped? I know this is a pretty vague question, so I'll

try to explain what I mean.

taken or dropped? I know this is a pretty vague question, so I'll

try to explain what I mean.

Given that large swings in equity can occur in a few rolls, it is

(as we all know) possible to go from not being good enough to double,

through to a double/drop position in just one roll per player. In

what fraction of games does this occur? I am asking because I have

been analysing my results recently,and I am looking for evidence that

I am on average doubling too early (more doubles taken by my opponents,

less dropped) or too late (more doubles dropped) on average. Do any

of our resident experts have any feeling for what these numbers ought

to be?

Phill

Jul 29, 1998, 3:00:00 AM7/29/98

to

!remove!this!ph...@sun.leeds.ac.uk (Phill Skelton) writes:

> Does anyone have any idea roughly what fraction of doubles should be

> taken or dropped? I know this is a pretty vague question, so I'll

> try to explain what I mean.

> Does anyone have any idea roughly what fraction of doubles should be

> taken or dropped? I know this is a pretty vague question, so I'll

> try to explain what I mean.

Actually I think it's a nice and precise question -- especially compared

to a few of the others that have been asked here recently :-)

(WARNING: Boring mathematical content included. Persons of a

boredom-sensitive or non-mathematically inclined nature should

skip to the last paragraph. Discontinue reading my articles if

pain persists.)

> Given that large swings in equity can occur in a few rolls, it is

> (as we all know) possible to go from not being good enough to double,

> through to a double/drop position in just one roll per player. In

> what fraction of games does this occur? I am asking because I have

> been analysing my results recently,and I am looking for evidence that

> I am on average doubling too early (more doubles taken by my opponents,

> less dropped) or too late (more doubles dropped) on average. Do any

> of our resident experts have any feeling for what these numbers ought

> to be?

I'm more of a resident than an expert, but I'm going to inflict my

feeling on the newsgroup anyway :-) I was thinking about this problem

earlier, in trying to present a reasonable model of the distribution

of points won per game. In the end I decided it was too hard to

reason about exactly, gave up, and was going to assume an empirically

derived take/drop ratio and see whether that assumption led to a

reasonable agreement with observations of points won in real games.

I don't have a complete answer, but here's some thoughts that might

help you. Assume money games, no Jacoby rule, and ignore last-roll

situations (they're `easy' to decide anyway, simply by enumerating all

36 or 1296 dice rolls over the next exchange). Ignore gammons, to

keep things simple. Assume both players are `perfect'. Assume you're

making an initial double, not a redouble (you're more reluctant to

double if you already own the cube, so P would be higher in that case).

The question we now ask is: Assume that it is correct to double this

turn. What is the probability P that it is also correct to drop?

Since you have the opportunity to double, we know you didn't double

last turn. And since we're also assuming perfect play, we know it

would have been _incorrect_ to double then. Since we're not considering

gammons, the only reason it wouldn't have been correct to double

was that your position was not good enough. Therefore, the position

last turn was no double/take, and this turn is double/(we don't know).

To be a take last turn, we know your equity E then must have been less

than your opponent's drop point D. In practice D will vary slightly

from roll to roll (since it depends on the recube vig, which in turn

depends on the cube efficiency, which varies...) but let's ignore that

for now.

To be a double this turn, we know your cubeful equity after doubling

immediately must be higher that your cubeful equity from holding the

cube for one more turn and then deciding. This is loosely equivalent

to claiming that the cost of doubling now (and then rolling junk and

not losing your market after all) is less than the cost of _not_

doubling (and then rolling well and losing your market). Calculating

these costs is a complicated question in itself, but let's go with

the conventional rule-of-thumb for now and assume (given volatility V)

you double if E + V > D, and your opponent drops if E > D. The question

we previously asked (given that it wasn't a double last turn, and it

is a double this turn, what is the probability it is now a drop?)

becomes: given E + V <= D, and E' + V' > D, what is the probability

that E' > D? (E and V are last turn's equity and volatility; E' and

V' are this turn's). E and E' are obviously related: E' = E + delta,

where delta is assumed to be normally distributed with mean 0 and

standard deviation V. There are too many inequalities in there for

me, so let's simplify things by making yet another assumption. Assume

that last roll, it was _almost_ a double (E + V = D). It's obviously

easier to lose your market from that position than from being nowhere

near a double, so any conclusion we make will be an upper bound on P.

From the definition of E', P( E' > D ) is P( E + delta > D ), and since

we assume E + V = D, that's the same as P( delta > V ). That's just

the plain old normal distribution P( z > 1 ), which is about 0.16.

However, we haven't yet taken into account that we also assumed that

E' + V' > D. Since E + V = D, for E' + V' to exceed D then either

E' must be greater than E; V' must be greater than V; or both. Measuring

the change in volatility gets pretty messy, so let's just assume that

E' must exceed E. Under a normal distribution this will happen exactly

half the time, so although the probability of losing your market is

only 0.16, the probability of losing your market _given that your position

improved_ is twice that, or 0.32. Remember the assumption we made in

the middle of this paragraph will tend to overestimate P, so we

conclude that the probability that any particular double should be

dropped is no more than 32%. The other assumption we made was to

ignore the change in volatility -- unfortunately there's little you

can say about this in general (it's too hard for me to think about,

anyway! We measure the equity, and call the variance in equity the

volatility. What do we call the variance in volatility?) It depends

on the position -- for instance, in prime-vs-prime games the

volatility is very high, and it is often correct to lose your market

without doubling (ie. P will be higher). In long races, the

volatility is very low, and with correct play you should get the

opportunity to make an efficient double (ie. P will be lower). The

overall P depends on the relative probability of these situations

occurring. Our conclusion must be that the _average_ P is under 32%,

but P will vary from position to position.

Well, that's quite a long-winded non-answer :-) But I hope the

reasoning behind it helps in any case. I feel kind of cruel making

you read all that without giving you some sort of independent

verification that I'm vaguely on the right track, so here goes: a few

years ago, Stig Eide posted an article with statistics from 1000 games

of Jellyfish playing itself. Among the results was that 35% of games

resulted in a single point win -- assuming the only way you can win a

single point is for your opponent to drop an initial double, that

would indicate that 35% of doubles ought to be dropped. This is a

little higher than our theoretical P of 32%, so it looks as if

some of our assumptions led us to underestimate P slightly, or

Jellyfish drops slightly more doubles than it should, or it waits

slightly too long before doubling, or that sometimes it is correct to

win a single point without ever turning the cube, or the difference

can be attributed to sampling error in the experiment.

(NB: Boring mathematical content ends.)

Whether you believe the wobbly acalephoid computerised fish or me, it

looks as if P ~= 1/3 (2 takes to every drop) is in the right ballpark.

Cheers,

Gary.

--

Gary Wong, Department of Computer Science, University of Arizona

ga...@cs.arizona.edu http://www.cs.arizona.edu/~gary/

Jul 30, 1998, 3:00:00 AM7/30/98

to

There are several questions here:

What percent of offered doubles are dropped and what percent are

dropped? (This can be answered by watching matches/games and counting.)

What percent of doubles offered are correct drops and what percentage

are correct takes? (The answer will depend on whether it's money or

match play, and requires watching matches/games, recording the results

and determining what the correct cube action was.)

What percentage of correctly offered doubles are drops vs. takes? (as

above, but factor out incorrectly offered doubles.)

With perfectly offered doubles, what percent are drops vs. takes? (This

requires playing a lot of matches perfectly and then counting. Until we

can play perfectly, that's hard to do!)

Gary Wong gave some theoretical analysis, so I'll give some impirical

data. The data is from 100 money games of jellyfish 5 playing itself

and the first 100 non-trivial cube actions in 11-point matches.

In the money games, out of 100 offered doubles, 42% were dropped, 58%

were taken. Of the 25 redoubles, 40% were dropped, 60% were taken.

In the match games, I ignored post-crawford doubles, automatic redoubles

(for example when the leader has 2 points to go and doubles, I counted

the double but not the automatic redouble), and doubles at a score of

-2:-2.

Of the 100 doubles, 52% were dropped, 48% were taken. Of the 8

redoubles, 87.5% (7/8) were dropped, 12.5% were taken.

(There were 5 games completed with no doubles, and 2 of the 11 matches

reached a score of -2:-2.)

A few observations:

Initial doubles in money and match play are about the same.

Redoubles in money play are about the same as doubles in money play.

Redoubles are rarer in 11-point matches and a much higher percentage are

dropped. This is consistent with 4-cubes being a lot more volatile than

2-cubes in an 11-point match.

-Michael J. Zehr

Phill Skelton wrote:

>

> Does anyone have any idea roughly what fraction of doubles should be

> taken or dropped? I know this is a pretty vague question, so I'll

> try to explain what I mean.

>

> Given that large swings in equity can occur in a few rolls, it is

> (as we all know) possible to go from not being good enough to double,

> through to a double/drop position in just one roll per player. In

> what fraction of games does this occur? I am asking because I have

> been analysing my results recently,and I am looking for evidence that

> I am on average doubling too early (more doubles taken by my opponents,

> less dropped) or too late (more doubles dropped) on average. Do any

> of our resident experts have any feeling for what these numbers ought

> to be?

>

> Phill

Jul 30, 1998, 3:00:00 AM7/30/98

to

In article <35BF21...@sun.leeds.ac.uk>,

!remove!this!ph...@sun.leeds.ac.uk (Phill Skelton) wrote:

> Does anyone have any idea roughly what fraction of doubles should be

> taken or dropped? I know this is a pretty vague question, so I'll

> try to explain what I mean.

>

> Given that large swings in equity can occur in a few rolls, it is

> (as we all know) possible to go from not being good enough to double,

> through to a double/drop position in just one roll per player. In

> what fraction of games does this occur? I am asking because I have

> been analysing my results recently,and I am looking for evidence that

> I am on average doubling too early (more doubles taken by my opponents,

> less dropped) or too late (more doubles dropped) on average. Do any

> of our resident experts have any feeling for what these numbers ought

> to be?

>

> Phill

>

Hi. I made jellyfish play 1000 games with the cube once, and recorded

the results. About 35% of the games ended with 1 point won (or lost).

That should mean that 35% of the doubles were rejected.

Kit Woolsey replied (after I posted the results) and said that was about

the result he would expect, probably even lower. He also mentioned that

experts rejected a much higher fraction of the doubles.

The article is in Tom Keiths articledatabase (www.bkgm.com).

Stig Eide

!remove!this!ph...@sun.leeds.ac.uk (Phill Skelton) wrote:

> Does anyone have any idea roughly what fraction of doubles should be

> taken or dropped? I know this is a pretty vague question, so I'll

> try to explain what I mean.

>

> Given that large swings in equity can occur in a few rolls, it is

> (as we all know) possible to go from not being good enough to double,

> through to a double/drop position in just one roll per player. In

> what fraction of games does this occur? I am asking because I have

> been analysing my results recently,and I am looking for evidence that

> I am on average doubling too early (more doubles taken by my opponents,

> less dropped) or too late (more doubles dropped) on average. Do any

> of our resident experts have any feeling for what these numbers ought

> to be?

>

> Phill

>

the results. About 35% of the games ended with 1 point won (or lost).

That should mean that 35% of the doubles were rejected.

Kit Woolsey replied (after I posted the results) and said that was about

the result he would expect, probably even lower. He also mentioned that

experts rejected a much higher fraction of the doubles.

The article is in Tom Keiths articledatabase (www.bkgm.com).

Stig Eide

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