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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 provides 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.
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Progresja nie działa przy ujemnej marży. Matematyczne dowody przeprowadza się na przedmiocie zwanym analiza matematyczna (buntownik z wyboru przeprowadzał takie dowody i Pani prezes bet365 też zajmowała się tym na studiach - na ekonometrii ma się analizę matematyczną). Jeżeli ktoś uważa, że progresja działa to... po prostu tak uważa i tyle... Jeżeli ktoś uważa, że wszystko sobie wyliczył i wie co robi kiedy gra progresją to po prostu sobie tak uważa. Jeżeli ktoś jest w stanie przeprowadzić matematyczny dowód na działanie progresji to pewnie dostanie matematycznego Nobla a ktoś kto po prostu uważa, że progresja działa i szerzy takie teorie to może kandydować do antynobla
