In "The Casino Gambler's Guide," Allan Wilson provided a
mathematical proof of the fallacy that a progression can overcome a negative
expectation in a game with even payoffs. This article expands on Wilson's
Proof and provides the proof that progression systems cannot overcome a
negative expectation even if the game provide uneven payoffs.
Let bk = the size of the kth bet.
Mk = the size of the payoff on the kth bet.
pk = the probability that the series terminates with a win on the
kth bet, having been preceeded by k-1 losses in a row.
n-1 = the greatest number of losses in a row that a player can handle, given
the size of the player's bankroll. In other words, the nth bet
must be won, otherwise the player's entire bankroll will be lost.
Let's now define Bn = bn * Mn
The expected value for any series is:
Eseries = p1B1 + p2(B2-b1)
+ p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2-
. . . -b2-b1) + (1-p1-p2- . . .
-pn) * (-bn-bn-1-bn-2- . . . -b2-b1)
If we let
Eseries = A + B where,
A = p1B1 + p2(B2-b1)
+ p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2-
. . . -b2-b1)
and
B = (1-p1-p2- . . . -pn) * (-bn-bn-1-bn-2-
. . . -b2-b1)
then it is easier to see that "A" represents the probability
that the series will end with a win multiplied by the bet size at the nth
term in the series and "B" is the probability that the series ends
in a loss multiplied by the net loss.
Now let's rearrange the terms in "A."
A = p1B1 + p2B2 - p2b1
+ p3B3 - p3b2 - p3b1
+ . . . + pnBn - pnbn-1 - pnbn-2
- . . . - pnb2 - pnb1
A = p1B1 + p2B2 + . . . + pnBn
+ b1(- p2 - p3 - . . . - pn) + b2(-
p3 - . . . - pn) + bn-2(- pn-1 -
pn) + bn-1(- pn)
And for "B" we get
B = -bn(1 - p1 - p2 - . . . - pn)
- bn-1(1 - p1 - p2 - . . . - pn)
- . . . - b2(1 - p1 - p2 - . . . - pn)
- b1(1 - p1 - p2 - . . . - pn)
Now if we combine A and B again, we get,
Eseries = A + B
Eseries = p1B1 + p2B2
+ . . . + pnBn - b1(1 - p1) - b2(1
- p1 - p2) - . . . - bn-1(1 - p1
- p2 - . . . - pn-1) - bn(1 - p1
- p2 - . . . - pn)
Eseries = p1B1 + p2B2
+ . . . + pnBn + p1b1 + (p2
+ p1)b2 + (p3 + p2 + p1)b3
+ . . . + (pn + pn-1 + . . . + p2 + p1)bn
- (b1 + b2 + . . . + bn)
Wilson points out that to get rid of the subscripts, all we have to do is
realize that pk = (1-p)k-1p, where p is the
probability of a win on an individual play and 1-p is the probability of a
loss. If we think about it, it makes sense that the probability of a series
terminating in a win at the kth level is the product of the
probability of k-1 losses in a row multiplied by the probability of win on
the kth trial.
So how do we use this information? Well, let's try substituting this
expression for each pk and see what we get.
Eseries = (1-p)1-1pB1 + (1-p)2-1pB2
+ . . . + (1-p)n-1pBn + (1-p)1-1pb1
+ ((1-p)2-1p + (1-p)1-1p)b2 + ((1-p)3-1p
+ (1-p)2-1p + (1-p)1-1p)b3 + . . . + ((1-p)n-1p
+ (1-p)n-1-1p + . . . + (1-p)2-1p + (1-p)1-1p)bn
- (b1 + b2 + . . . + bn)
Simplifying, we get
Eseries = (p(1-p)0B1 + (1-p)1pB2
+ . . . + (1-p)n-1pBn + p(1-p)0b1
+ ((1-p)1p + (1-p)0p)b2 + ((1-p)2p
+ (1-p)1p + (1-p)0p)b3 + . . . + ((1-p)n-1p
+ (1-p)n-2p + . . . + (1-p)1p + (1-p)0p)bn
- (b1 + b2 + . . . + bn)
If we factor p out of the first parts of the equation and look closely,
we can see that the kth term T can be written as:
T = p[(1-p)k-1]Bk + p[(1-p)k-1 + (1-p)k-2
+ . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
or rephrased for Bk = bk * Mk we get
T = p[(1-p)k-1Mk + (1-p)k-1 + (1-p)k-2
+ . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
If we substitute
C = (1-p)k-1 + (1-p)k-2 + . . . + (1-p)2
+ (1-p)1 + (1-p)0
and if we multiply C by (1-p) and call this D
D = (1-p)C = (1-p)k + (1-p)k-1 + . . . + (1-p)3
+ (1-p)2 + (1-p)1
Now, if we subtract C from D, we get
D - C = (1-p)C - C = (1-p)k - (1-p)0
[(1-p) - 1]C = (1-p)k - (1-p)0
C = [(1-p)k - 1]/[(1-p) - 1] or
C = [(1-p)k - 1]/-p]
Now if we substitute C back into T, we get
T = p[(1-p)k-1Mk + [(1-p)k - 1]/-p]bk
T = p(1-p)k-1Mk + 1 - (1-p)k]bk
T = p(1-p)k-1Mk + 1 - (1-p)(1-p)k-1]bk
T = [[pMk - (1-p)](1-p)k-1 + 1]bk This now
allows us to write the equations in terms of summations. We therefore get
Eseries = sum {[[pMk - (1-p)](1-p)k-1]bk}
+ sum {bk} - sum {bk}, for k = 1 to n
The last two terms cancel, so we are left with:
Eseries = sum {[pMk - (1-p)](1-p)k-1]bk},
for k = 1 to n
Eseries = sum {[(1+Mk)p - 1](1-p)k-1]]bk},
for k = 1 to n
If we now look closely at this equation, we can make several
observations. First, the sign of Eseries depends solely on the
resulting sign of [(1+Mk)p - 1]. To make things a little easier
to follow, let's say we're dealing with a game that has even payoffs. This
means that Mk = 1 and therefore
Eseries = sum {[2p - 1](1-p)k-1]]bk}, for k
= 1 to n
Eseries = [2p - 1]sum {(1-p)k-1]]bk}, for k
= 1 to n
Now it is a little easier to see what is going on. For example, if we are
in an unfair game, then p < 0.5 and we can easily see that 2p-1 will be a
negative value. For example, if our chance of winning is only 49%, then p =
0.49 and 2p-1 = -0.02. In an even game, p = 0.5 and we see that 2*0.5-1 = 0.
In this case, the equation is telling us that in an even game the expected
value is zero just as we would expect it should. If we are playing a game
with an advantage, then p > 0.5 and 2p-1 will be positive.
The general formula for uneven payoffs work just as well, but is more
complicated to understand. Suffice to say that if [(1+Mk)p - 1]
is negative, then regardless of the progression, the game will eventually
result in a loss for the player.
Hopefully, this post will provide definitive proof of the fallacy of
trying to overcome a negative expectation by using any type of progression
whether it be the martingale or some other modern progression.
