# Thought for the day

1 view

Jan 16, 2003, 4:17:05â€¯PM1/16/03
to
To have a 99% chance of surviving an n-game money session against an equally
skilled opponent, you need roughly 5*sqrt(n)+20 points of capital.

Jan 16, 2003, 10:51:04â€¯PM1/16/03
to

news:b076ul\$oko\$1...@newsg1.svr.pol.co.uk...

### Christopher Alvino

Jan 17, 2003, 2:05:36â€¯PM1/17/03
to
Ok, I'll bite.

Such formulae are useful for people with more finite bankrolls
(politically correct for poor) so that they know not to go in over their
head. But, without proper justification they are just letters and
numbers.

Can you provide some insight about how you got this? This might help
people interpet it appropriately or modify it.
e.g, assumed a large number of games is probably Gaussian, assumed
something about the variance, you have empirical evidence after lots of
money sessions, etc. I assume you'll have to make some assumptions
problem with this before as long as it's well stated.

I'm particularly interested in where the +10 part comes from. It seems
that the amount of capital should be directly proportional to sqrt(n).
Is this just to make sure the formula works for sessions with a low
number of games, where a Gaussian distribution is a bad assumption?

There's one way that I'd like to modify it. Since I assume most people
want to play against people who are worse than them in the long run, it
might be useful to see where the chance of winning against that opponent
in a single game fits into this formula. When you play against someone
worse than you, the necessary capital goes down, but by how much?

Thanks for the food for thought,
Chris

### rugcutter

Jan 17, 2003, 2:06:09â€¯PM1/17/03
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For those of us who are mathematically impaired, please interpret..

news:b07u19\$5uq\$1...@news5.svr.pol.co.uk...

Jan 17, 2003, 7:17:53â€¯PM1/17/03
to
>...

> For those of us who are mathematically impaired, please interpret..

> >

> > news:b076ul\$oko\$1...@newsg1.svr.pol.co.uk...
> > > To have a 99% chance of surviving an n-game money session against an
> equally

> > > skilled opponent, you need roughly ' 5*sqrt(n)+10 '
points of capital.

Say you agree to play someone 25 games at \$10 a point. This formula
then says that you should have \$350 in order to be fairly confident
that you won't go broke and can handle the normal equity swings.

But, if you are playing someone who say is 1/3 of a point per game
stronger than you... it may take more cash...

Jan 17, 2003, 8:10:44â€¯PM1/17/03
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"Christopher Alvino" <gtg...@prism.gatech.edu> wrote in message
news:3E285400...@prism.gatech.edu...

> Ok, I'll bite.
>
> Such formulae are useful for people with more finite bankrolls
> (politically correct for poor) so that they know not to go in over their
> head. But, without proper justification they are just letters and
> numbers.

Everybody's bankroll is finite, even Bill Gates'. To maximise profit
against a weaker opponent, one must play for the maximum stakes possible
within the bounds of one's personal threshold for acceptable risk of ruin -
i.e going in NEARLY over one's head, but not quite. A similar principle
applies if the opponent is of equal skill level, except one is trying to
minimise the amount of capital required to stay within the ruin threshold
for a given session length.

> Can you provide some insight about how you got this? This might help
> people interpet it appropriately or modify it.
> e.g, assumed a large number of games is probably Gaussian, assumed
> something about the variance, you have empirical evidence after lots of
> money sessions, etc. I assume you'll have to make some assumptions
> problem with this before as long as it's well stated.

I, being a simple type of folk, (I'm not a mathematician) used a simple
model to determine the stated formula. The base formula I came up with
gives capital in direct proportion to sqrt(n), then I adjusted it.

The distribution of net (games won - games lost) as a result of playing n
games will tend towards a Gaussian distribution as n gets larger, i.e. of
the form (p+q)^n. The standard deviation of the net gain from an n-game
session will be sqrt(n). Don't ask me to proove it, it just does for all
the examples I tried :-). A 1% probability threshold in a Gaussian
distribution occurs at about 2.5 standard deviations above/below mean, so
the probability of losing 2.5 or more standard deviations of games is about
1%.

In the base formula, I have subjectively assumed a fixed amount of gain/loss
per game equal to 2.25*(stake per point), a figure which the backgammon
community seems to have deemed reasonable for typical (non-beginner)
players. Therefore, there is about 1% chance of losing (2.5*2.25) = 5.625
or more standard deviations of points in an n-game session. Therefore there
is a 99% chance of not losing 5.625*sqrt(n) points in an n-game session. So
Capital=5.625*sqrt(n) was my base formual.

> I'm particularly interested in where the +10 part comes from. It seems
> that the amount of capital should be directly proportional to sqrt(n).
> Is this just to make sure the formula works for sessions with a low
> number of games, where a Gaussian distribution is a bad assumption?

Right. My base model assumes a fixed amount of 2.25 points per game as the
gain or loss, which is obviously not the case in practice. Due to big cubes
and gammons, the net gains/losses possible for each game will form a
non-rectangular (Gaussian-looking) distribution of their own, (i.e. a
distribution within a distribution) which, for a large n, won't matter, but
for a small n, will give 'fatter tails' to the overall distribution,
particularly relevant to small n values. Therefore, I weighted the formula
in respect of small n values by adding the +10 constant, and reducing the
5.625 factor down to 5.000 to help compensate for the +10 constant, giving
Capital=5*sqrt(n)+10. It's only an approximation, but it's easy to remember
(hey, it even rhymes), and looks very reasonable to me for any realistic n.

> There's one way that I'd like to modify it. Since I assume most people
> want to play against people who are worse than them in the long run, it
> might be useful to see where the chance of winning against that opponent
> in a single game fits into this formula. When you play against someone
> worse than you, the necessary capital goes down, but by how much?

That was my next quest. Unfortunately, since I'm not a mathematician, I
will have to model this using a spreadsheet to provide, say, a (0.52+0.48)^n
distribution, and see empirically what the standard deviations are etc.
Although I'm sure Doug will be delighted to enlighten us all :-).

> Thanks for the food for thought,
> Chris

Yw,

Jan 17, 2003, 8:36:19â€¯PM1/17/03
to

That's true (see my other posting). But if THINK are the stronger player,
you would be prudent to work out the risk of ruin based on the assumption
that you MAY be only, say, of equal skill level, not actually stronger,
since by playing him, you are simply testing your hypothesis that you are
stronger, so you need to know the risk of ruin for the case that your
hypothesis is wrong.

### Christopher Alvino

Jan 18, 2003, 2:02:25â€¯AM1/18/03
to

Ok, then I agree with your formula for bankroll, assuming finite length
sessions and that you are of equal skill with your opponent. I've seen
arguments like this before in poker.

Now, here's another formula to add to your collection, based on similar
assumptions. I think this is well known for poker people and maybe for
BG people too.

If you have an edge on your opponent, and you win on average P points
per game against that opponent, then you need a bankroll of:

BR = (Z^2 * S^2) / (4 * P),

to have some chance (depending on the below table) of not going bust,
where Z is the number of standard deviations away from the mean and S is
the standard deviation (where you used 2.25). Note that the number of
games disappears from this equation since you have an edge on this
opponent. Since you have an edge, we can find the local minimum of the
'bad luck' curve for a certain number of standard deviations away from
the mean.

Table:
Risk of Ruin Z
-----------------------
20% 0.85
10% 1.3
5% 1.7
2% 2.1
1% 2.3 (by the way, I get 2.3 here instead of 2.5 from my Gaussian
random variables reference)
0.5% 2.6
etc.

You can plug is whatever you like depending on what you want your risk
of ruin to be and whatever your edge on your opponent is. For instance,
say I'm Bill Robertie and I have a 0.02 point per game edge on my
opponent, TD-gammon 2.1 [http://www.research.ibm.com/massive/tdl.html].
I want my risk of ruin to be 1% and my standard deviation per game is
2.25 points per game as you stated. Then my necessary bankroll would be,

(2.25^2 * 2.3^2) / (4 * 0.02) = 334.8 points.

Which is about \$1674 if I'm playing \$5 per point (a stake at which I'm
sure Robertie wouldn't bother). This seems reasonable for a game with
that small of an edge and that low of a risk of ruin.

One more thing to note, if P approaches 0 (that is you are equal with
your opponent) then this equation grows without bound (approaches
infinity). This makes sense since I'm assuming an infinite number of
games. Therefore, it's the same as when N approaches infinity in your
formula.

Chris

Jan 18, 2003, 1:14:01â€¯PM1/18/03
to
Chris,

It looks to me at first glance as if your BR formula is of the 'freeze-out'
type - i.e. gives the bankroll necessary to avoid ruin before your opponent
is ruined, during an indefinite number of plays (mine doesn't attempt to do
that, since it assumes that the opponent is always solvent). I'm not quite
sure how the P (p.p.g.) variable would be applied to poker though. Do you

Adam (Mindsports Olympiad No Limit Hold'em Gold medalist, 2002 { i just
winged it lol } )

"Christopher Alvino" <gtg...@prism.gatech.edu> wrote in message

news:3E28FC01...@prism.gatech.edu...

### Christopher Alvino

Jan 18, 2003, 1:52:46â€¯PM1/18/03
to

Actually, it doesn't care what your opponent's bankroll is. It assumes
that the opponent is always there. So it's not of the freeze-out type.
It only assumes that you are a favorite over your opponent. What do you
mean about the opponent being solvent?

You can construct this formula in the following way:
Assume that at time 0, you have a bankroll of BR and you win P points
per game on average. Then your expected assets after N games is BR+N*P.
But of course this differs based on the variance. If you have bad luck
that is Z standard deviations below the mean, then your assets will
follow the curve BR+N*P-Z*S*sqrt(N). Again S is the standard deviation
per game. Keep in mind, this is accurate when the variable becomes
"more Gaussian".

So this curve BR+N*P-Z*S*sqrt(N) has the property that it dips down
because of the bad luck, then starts to go back up because of the N*P
term which eventually overcomes the bad luck term, -Z*S*sqrt(N). So we
can find the number of games M at which this curve is at it's minimum
with some calculus.. and that is:

M = (Z^2 * S^2) / ( 2 * P)

Now, plugging this into N in the formula BR+N*P-Z*S*sqrt(N) will give us
that our assets at the bottom of that curve (which represents the lowest
point of the bad luck streak) are:

BR - (Z^2 * S^2) / (4 * P)

Now, we simply don't want to go bust. So this has to be greater than 0
at all times, which means we want BR >= (Z^2 * S^2) / (4 * P).

In poker, some people chart their hourly win and hourly standard
deviation. This information, once you have enough games for it to be
accurate, can be plugged into this formula in the same way.

Chris

### Christopher Alvino

Jan 18, 2003, 1:57:53â€¯PM1/18/03
to
Ok, my turn for a correction. M should have been,

>
> M = (Z* S)^2 / ( 2 * P)^2
>

Jan 18, 2003, 3:36:51â€¯PM1/18/03
to
(I'm a wee bit too rusty to check right now), shoudn't the minimum function
read M = (Z*S)^2 / (4*(P^2)) ?

"Christopher Alvino" <gtg...@prism.gatech.edu> wrote in message

news:3E29A27E...@prism.gatech.edu...

### Christopher Alvino

Jan 18, 2003, 4:37:07â€¯PM1/18/03
to
Yes, that's the same as,

M = (Z* S)^2 / ( 2 * P)^2

Jan 19, 2003, 11:43:01â€¯AM1/19/03
to
"Christopher Alvino" <gtg...@prism.gatech.edu> wrote in message
news:3E29A27E...@prism.gatech.edu...

>
> Actually, it doesn't care what your opponent's bankroll is. It assumes
> that the opponent is always there. So it's not of the freeze-out type.
> It only assumes that you are a favorite over your opponent. What do you
> mean about the opponent being solvent?
>
> You can construct this formula in the following way:
> Assume that at time 0, you have a bankroll of BR and you win P points
> per game on average. Then your expected assets after N games is BR+N*P.
> But of course this differs based on the variance. If you have bad luck
> that is Z standard deviations below the mean, then your assets will
> follow the curve BR+N*P-Z*S*sqrt(N). Again S is the standard deviation
> per game. Keep in mind, this is accurate when the variable becomes
> "more Gaussian".
>
> So this curve BR+N*P-Z*S*sqrt(N) has the property that it dips down
> because of the bad luck, then starts to go back up because of the N*P
> term which eventually overcomes the bad luck term, -Z*S*sqrt(N). So we
> can find the number of games M at which this curve is at it's minimum
> with some calculus.. and that is:
>
> M = (Z^2 * S^2) / ( 2 * P)

Agreed. Minimum occurs when P - 0.5*Z*S/sqrt(N) = 0, which results in BR
being minimum at N =(Z^2 * S^2) / (2*P )^2, which when plugged into the BR
formula, reduces to

BR >= (Z^2*S^2) / (4*P)

I had to check my differentiation formula was right.

> Now, plugging this into N in the formula BR+N*P-Z*S*sqrt(N) will give us
> that our assets at the bottom of that curve (which represents the lowest
> point of the bad luck streak) are:
>
> BR - (Z^2 * S^2) / (4 * P)
>
> Now, we simply don't want to go bust. So this has to be greater than 0
> at all times, which means we want BR >= (Z^2 * S^2) / (4 * P).
>
> In poker, some people chart their hourly win and hourly standard
> deviation. This information, once you have enough games for it to be
> accurate, can be plugged into this formula in the same way.
>
> Chris

I didn't have a Z reference table to hand at the time, so I interpolated
from manual (p+q)^n calculations to get my 2.5 figure, hence the
discrepancy. Your 2.3 figure is better.

There has been some work previously done to come up with a p.p.g advantage
for a given Elo rating advantage, and the most common figure seems to be
about +0.1ppg per 50 Elo (although I prefer +0.09ppg, so I will use 0.09 for
the minute). And, also in the backgammon case, S will be about 2.25ppg,
(leaving Z as a constant 2.3), we can now say that for an R Elo advantage,

BR = (Z^2 * S^2) / (4*P) = (26.780625) / (4*0.0018*R)

BR = 3719.53125 / R

e.g. Bill Robertie 0.02ppg, R = (0.02/0.09)*50 = +11.1111111 Elo

BR = 3719.53125 / R = 334.76 points

If we use 0.1ppg per 50 Elo instead of 0.09ppg, then BR = 3347.578125 / R,
so a reasonable compromise would be

BR = 3500 / R

Jan 19, 2003, 12:40:24â€¯PM1/19/03
to
My own cube distribution calculations give an approximate probability of a
cube being 8 or more in a money game as being roughly 4%, and with a 20%
gammon rate, the probabilty of losing 16 or more points in a single game
would be roughly 0.004. This is significantly over the 1% Z=2.3 threshold,
and with my using Z=2.5 earlier in error, I think that my original formula
for equal players should be more like BR = 5*sqrt(N)+5, which gives a more
realistic looking BR = 10 for a 1-game session, 12.1 for 2 games, 13.7 for 3
games, 40.4 for 50 games.