# Michael Hall: Risk of Ruin

Risk of Ruin
Last Revised 1/15/92
Permission to copy for own use

This is really seven articles in one:

I.   What is the risk of ruin for a given bankroll and given win goal?
II.  How many units of bankroll for a given risk of ruin?
III. How long will it take to go broke or win a goal amount?
IV.  What is the chance of taking a loss after N hands?
V.   George C. on the Ruin Formula
VI.  Mason Malmuth on the Ruin Formula
VII. How about some C code?

Although blackjack is frequently used as an example here, the
results can be applied to other gambling games.

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I.   What is the risk of ruin for a given bankroll and given win goal?

This can be estimated using the standard ruin formula,
with the bankroll adjusted to reflect the standard deviation, as
shown by Griffin in "Theory of Blackjack".  The average squared
win is very nearly the variance (i.e. the standard deviation
squared), more precisely it is the variance plus the square
of the win per hand.  You can approximate the game of blackjack
by betting the square root of the average squared wager
on a biased coin with P(heads) = 0.5 + wph/2sqrt(asw), where asw is
the average squared wager and wph is the expected win.  So,
here P(heads)= 0.5+.0151/(2sqrt(3.887)) = 0.5038.  The ruin formula is:

R = (1 - S^b)/(1 - S^(a+b))

where  R    is the probability of ruin from 0 (none) to 1 (always)
a    is a'/sqrt(asw), the coin toss bankroll
b    is b'/sqrt(asw), the amount to be won in coin toss units
S    is P/(1-P), the ratio of a coin winning to losing
P    is 0.5 + wph/2asw, the bias of the coin
wph  is the win per hand
asw  is s^2 + wph^2, the average squared win

Suppose we wish to double a bankroll of 300 basic units. The bet size
effectively reduces this to a=b=300/sqrt(3.887)=152.16 units.
S=0.5038/(1-0.5038)= 1.0153.  So R=(1-1.0153^152.16)/(1-1.0153^(152.16*2))
=9.0%.

And here's the same formula, fully expanded:

/sqrt(s^2 + wph^2) + wph\(b'/sqrt(s^2 + wph^2))
1 - | ----------------------- |
\sqrt(s^2 + wph^2) - wph/
R = -------------------------------
/sqrt(s^2 + wph^2) + wph\((a'+b')/sqrt(s^2 + wph^2))
1 - | ----------------------- |
\sqrt(s^2 + wph^2) - wph/

where  R    is the chance of ruin from 0 (none) to 1 (always)
s^2  is the variance (i.e. square of the standard deviation s)
wph  is the win per hand in units
a'   is the blackjack bank size in units
b'   is the number of blackjack units we want to win (usually = a')

If S = 1 (i.e. the game is even), then use this formula instead:

R = b / (a + b)

where  R    is the chance of ruin from 0 (none) to 1 (always)
a    is a'/sqrt(asw)
b    is b'/sqrt(asw)

Here's some more examples, with the wph and s^2 as above. Here are
the probabilities of ruin before doubling for breaking your bankroll
into various numbers of units:

CHANCE RUIN BEFORE DOUBLE
BANKROLL      wph=0.2265        wph=0.151        wph=0.0755
--------    ---------------   --------------    ---------------
100            23.8                31.5              40.4
200             8.9                17.5              31.5
300             2.9                 8.9              23.8
400             0.9                 4.3              17.5
500             0.2                 2.0              12.5
600             0.09                0.9               8.9
700             0.03                0.4               6.2
800             0.009               0.1               4.3
900             0.003               0.09              2.9
1000             0.0009              0.04              2.0

Note: above probabilities do not apply to Frank "Almost Never Lose 40
Units" Irwin ;-)  They also do not apply to Kelly Criterion bettors,
but Kelly Criterion (betting proportional to current bankroll and
advantage) is not practical to apply exactly in practice.

==========================================================================
II.  How many units of bankroll for a given risk of ruin?

A special case of the ruin formula given before is the following
for when we wish to *double* the bankroll:

1
R = -------
1 + S^n

where R is the risk of ruin
S is the ratio of winning to losing
n is the units of coin toss bankroll

We can solve this in terms of n, to answer the above question, and we get:

log((1/R) - 1)
n = --------------
log(S)

Expanding this out to head off possible confusion:

log((1/R) - 1)
a'= ----------------------------(sqrt(s^2 + wph^2))
/sqrt(s^2 + wph^2) + wph\
log| ---------------------- |
\sqrt(s^2 + wph^2) - wph/

where R is the risk of ruin
wph is the win per hand
s^2 is the variance of wph
a' is the necessary units of blackjack bankroll

For example, plugging in R=.09, S^2=3.88681, wph=.0151, then
a'~= 300.  In other words, if you are willing to risk a 9%
risk of ruin and use a 1-4 spread under conditions similar to
my computer simulation, then you should use 300 units of bankroll.

I don't believe there is any published blackjack book that accurately
explains the bankroll requirements of blackjack.  In fact, most of them
are just plain wrong when it comes to this.  This is confirmed by
the full double-or-nothing simulations of myself and also independently
on a different simulator by a rec.gambler who is no longer with us.  For
example, a 400 unit bankroll with the above single deck situation has
(only) about a 80% chance of doubling before ruin, according to the
simulations, which is about what we'd expect from using the ruin formula.

==========================================================================
III. How long will it take to go broke or win a goal amount?

From Epstein's _Theory of Gambling and Statistical Logic_, page 66:
the expected number of plays before ruin or win limit of a gambler is:

z         a[1-(q/p)^z]
E(n) = ---------- - ---------------------                 for p<>q
(q-p)(1-r)   (q-p)(1-r)[1-(q/p)^a]

where E(n) is the expected number of plays before ruin or win limit
n is the number of plays
z is the initial capital
a is the desired final capital (a > z, z-a is the desired win)
p is the probability of a win
q is the probability of a loss
r is the probability of a tie

Peter Griffin claims we can approximate blackjack by choosing
p=0.5 + wph/2sqrt(asw), where asw is the average squared wager and wph
is the expected win, with q=1-p and r=0.  With the numbers wph=.0086 and
asw=7.63 (for an 8 deck shoe, AC rules, abandoning counts of -1 or worse)
then p=.5015567. Suppose z=100, and a=200, and so E(n)=9689.  Thus, it
would take 9689 hands on average, around 97 hours, to double or nothing
100 units on this shoe game.

There are other formulas that are sometimes useful.

For an even game use:

z(a-z)
E(n) = ------     for p = q
1 - r

For an infinite win limit "a", use:

z
E(n) = ----------     for p < q
(q-p)(1-r)

E(n) = undefined      for p > q

E(n) = infinity       for p = q

For intelligent card counters, p > q, and so it doesn't make much
sense to talk about the average time to ruin or winning an infinite
amount.  Although sometimes a card counter will go broke with a
fixed unit size, sometimes a card counter *will* win an infinite
amount if left to play forever.  p < q represents the case of
most gamblers, such as craps players, who are doomed to lose if
they play long enough and so it's only a matter of how long it will
take.

==========================================================================
IV.  What is the chance of taking a loss after N hands?

Assuming the risk of ruin is low and the number of hands is
more than 100 or so, the Central Limit Theorem makes the
distribution normal.  The mean and standard deviation of this
normal distribution can be computed from the expected win per
hand and the standard deviation per hand.  z = (x-u)/s*sqrt(N),
where z is the standardized normal variable, x is the point
of interest (zero, if we want to know the chance of taking a
loss), u is the expected win, s is the standard deviation,
and N is the number of hands.

Let's say N=40, then z=(-50-(40*.0151))/(sqrt(3.887*40))=-4.0583.  Don't
have my tables handy, but I'm sure that's pretty much off the scale, close
to 0%. That seems reasonable, since it is pretty hard to lose 50 units
in just 40 hands, even if you are spreading 1-4.  I would, however,
expect the "actual" figure to be a bit higher (due to the 4 unit bets
having an extra large influence in the space of only 40 hands.)

==========================================================================
V.   George C. on the Ruin Formula

From "The Ruin Formula", by George C., Blackjack Forum, September 1988...

George shows how to compute variance, like so:

6 Decks Strip Rules w/DDAS, 75% Penetration

Advantage  Hands/Hour   Bet Squared $Product$
---------  ----------   -------------  ---------
-3.4%       1.0               10000      10000
-2.9%       2.0               10000      20000
-2.4%       3.0               10000      30000
-1.9%       4.0               10000      40000
-1.4%       8.0               10000      80000
-0.9%      13.0               10000     130000
-0.4%      35.5               10000     355000
0.1%      13.0               40000     520000
0.6%       8.0              250000    2000000
1.1%       4.0              562500    2250000
1.6%       3.0             2250000    6750000
2.1%       2.0             2250000    4500000
2.6%       2.0             2250000    4500000
3.1%       1.0             2250000    2250000
3.6%       0.5             2250000    1125000

Sum of Products      24560000
Sq Root of Products      4956
Times 1.1                5451 = Hourly standard deviation in $Note: When bet size goes to$1000, it is assumed player will
play two hands of $500 each, then 2 hands of$1000 each
[treated above as a bet of .75*2*bet].

The above is a reasonable method of computing variance from
frequency distributions.  George C. does his statistics
in "hourly" units, i.e. 100 hands.  Using the above chart, you
can obviously come up with things like the hourly win rate too.

Then George comes up with this version of the Ruin Formula:

h
/1 + w/v\
| ------- |  -  1
\1 - w/v/
R = --------------------
s+h
/1 + w/v\
| ------- |  -  1
\1 - w/v/

where R is the probability of ruin
w is the hourly expected win (in units)
v is the hourly variance
h is the hours of play
s is the stake (or size of bank)

He wants to figure out what his chance of halving his bank of
$150,000 before doubling it is... George C.: "Since I should average$392/hour, it will take me 383
hours to double my bank of $150,000 (i.e., divide the bank size,$150k, by the hourly expectation, $392, to get 383 hours.) I want to know what my chances are of halving my bank so I divide$75,000 (half of my bank) by $500 to obtain how many units of bank I have. That equals 150 units." [George calls$500
his unit size, even though he goes down to $100 bets in bad counts, but that's okay, so long as we know what he means.] George substitutes the values w=.783,v=119,h=383,s=150 to get .14, or a 14% chance of halving his bank of 300$500 units
before doubling it. Nrrrrr.... but thanks for playing George C.,
and we have some nice consolation gifts for you as well as
Chambliss and Rogenski.  You just can't use whatever units
you feel like in the ruin formula.

My computations show a 11.2% chance of losing *half* the bankroll
before doubling the whole bankroll.

Note: general_ruin(1500.0, 1500.0, 27.016, .0392) = .013, and
general_ruin( 750.0, 1500.0, 27.016, .0392) = .112,
$100 units. 27.016 is the variance in$100 units per hand (squared).

George C.:
"The answer you should get is .14 or about a 1 in 7
chance of losing half of your bank.  Most high stake pros
that I know play double the bet size I recommend here,
considering the size of my bankroll..."

I'm a bit puzzled by the "s+h" term in George's formula, and
I tend to think it's totally bogus. s is the initial bankroll,
while h is the hours. Whoh, ERROR type conflict.

We can see the problem in George's approach by switching to
a different unit size for the time, which shouldn't change the
risk of ruin. Let's say instead of hours (i.e. 100 hands),
how about 100 hours (i.e. 10,000 hands)?  This divides
the necessary units of time by 100, and multiples the variance
and win by 100.  Georges formula then spits out 2.463%, whereas
with hourly units the answer was 14%.  Clearly something wrong
here.  Could be me, but I think it's George.

George's formula will work if you enter the data in terms of
hands and bet units.  Instead of hours, for "h" use the number of
units you wish to win.

==========================================================================
VI.  Mason Malmuth on the Ruin Formula

From a letter to the editor, by Mason Malmuth, Blackjack Forum,
December 1988...

Mason Malmuth:
"George C's article on The Ruin Formula' (BJ VIII #3) was
very well done, [NOT!] but if you use the method give in my
book, Gambling Theory and Other Topics', you can extend
these ideas even further and produce what I think are
even more interesting numbers.  Specifically, consider
the following equation:

LL = (WR)(N)-(3)(SD)sqrt(N)
where LL is your lower limit,
WR is your win rate per hour,
N is the number of hours you play, and
SD is the standard deviation.

For all practical purposes, this is the equation that gives
the lower limit (at three standard deviations) for how you
would do for some period of time.  That is, for practical
purposes, this equation is your worst possible result."

Okay, let's apply Mason's equation to George's problem, so
WR=392, N=383, SD=5451, producing LL=-169898.00.  Actually,
come to think of it this doesn't really give you much of
an indication as to your risk of ruin during those 383
hours - it is just the worst result you would expect after
383 hours IF YOU HAD AN INFINITE BANKROLL.

But Mason Malmuth continues:
"Going one step further, if we take the above equation and
take the first derivative with respect to N, set it to zero,
and solve for N, we find the number of hours where our bankroll
can by minimized

0 = WR - (3)(1/2)(SD)/(sqrt(N))

2
/ (3)(SD) \
N = | --------  |
\ (2)(WR) /

Now by substituting this value back into the original lower
limit equation, we produce practical bankroll requirements.
In George C's example WR=.783, and SD=10.9.  Thus N=436,

2
/ (3)(10.9) \
436 = |  ---------  |
\ (2)(.783) /

and the required bankroll is 341 units.

-341 = (.783)(436)-(3)(10.9)sqrt(436)"

First, note that 341 units is 341 of George C's $500 units, which is really 1705$100 units, or $170,500. Using my version of the ruin formula, I calculate a 0.7% chance of ruin before doubling 1705$100 units, under George C's conditions.

Mason's technique seems to work okay, but I'd trust the ruin
determining bankroll size for a given risk of ruin.

==========================================================================
VII. How about some C code?

And finally, here's some C code to do the ruin formula above...

Compile by doing something like:

cc -c ruin.c
cc -o ruin ruin.o /usr/lib/libm.a

/*
Gambling ruin formulas. Version 1.1.

Permission to use pretty much however you want, but at your own
risk of ruin :-)
*/

#include <math.h>

#define sqr(X) (X * X)

/*
Just a test.  Should print following (approximately):
0.088595
300.0
*/
main()
{
double general_ruin(), bankroll();

printf("%f\n", general_ruin(300.0, 300.0, 3.886851, 0.0151));
printf("%f\n", bankroll(0.088595, 3.886851, 0.0151));
}

/*
Probability of ruin when trying to win "bb" units starting with "aa"
units of bankroll on a gambling game where the expected win (or
loss) per trial is "wph" and the variance is "s2".  Note that "aa" and
"bb" are doubles, just like the other parameters.
*/
double general_ruin (aa, bb, s2, wph)
double aa, bb, s2, wph;
{
double coin_units, a, b, S, asw, compute_asw(), coin_bet_size(), big_S(),
coin_ruin();

asw = compute_asw(s2, wph);
coin_units = coin_bet_size(asw);
a = aa / coin_units;
b = bb / coin_units;
S = big_S(s2, wph);
return(coin_ruin(a, b, S));
}

/*
Generates proper bankroll size for doubling given a desired risk of
ruin and the other information as above.
*/
double bankroll (R, s2, wph)
double R, s2, wph;
{
double S, coin_size, coin_bet_size(), coin_bankroll();

S = big_S(s2, wph);
coin_size = coin_bet_size(s2, wph);
return(coin_bankroll(R, S) * coin_size);
}

double coin_bankroll(R, S)
double R, S;
{
double log();

return(log((1.0 / R) - 1)/log(S));
}

/*
asw is the average squared win, which is a simple function
of s2 and wph.
*/
double compute_asw(s2, wph)
double s2, wph;
{
return(s2 + sqr(wph));
}

/*
Number of units to bet on a coin tossing game, to get
the same mean and variance as the more general game
with asw average squared wager.
*/
double coin_bet_size(asw)
double asw;
{
double sqrt();

return(sqrt(asw));
}

/*
The big S in the coin ruin formula.
*/
double big_S(asw, wph)
double asw, wph;
{
double sqrt();
return((sqrt(asw) + wph) / (sqrt(asw) - wph));
}

/*
Probability of ruin when trying to win "b" units starting with "a"
units of bankroll on a biased coin tossing game where "S" is the
ratio of the probability of winning to the probability of losing
(i.e. P/(1-P) where P is the probability of winning.
*/
double coin_ruin (a, b, S)
double a, b, S;
{
double pow();

return((1.0 - pow(S,b)) / (1.0 - pow(S,a + b)));
}