crash,

Its simple maths really.

1. If you have a strike rate of x% you are going to need to earn at least 100/x to break even.

I have shown using the staking plan that you can increase a 33% strike rate to have a 75% success rate for producing a profit. Providing the win is more than $1.33 (for 75%) then I am going to be in profit. Do you agree or disagree ?

As Angry Pixie put it .. Think of 1 bank as 1 bet. Therefore I have a 75% success rate at picking winners. I make double the bank each time for that 75%. Which means I make an average winner of $2.00 which straight away makes a return of $1.50 for every $1 bet.

That is maths proving my point.

I am yet to see the counter maths....

Like I said earlier, life is too short to debate the possibility of turning a negative expectation sum game [LOT] into positive expectation [POT] by a 'clever' loss chasing staking plan.

Continue to convince yourself and maybe some of the [mathematically challenged] crowd who visit here, but 'smoke and mirrors' arguments leave me cold when it comes to risking cold hard cash.

There always has been and always will be, a queue to buy Snake-oil. Sell away! :-))

Lets put this silly baby to bed:

Mathematical Proof that Progressions cannot overcome Expectation.
by Richard Reid
--------------------------------------------------------------------------------


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.
--------------------------------------------------

And from a long ago previous post of mine on the subject:

To expand on progressions a bit further, it might be worth thinking about the punter who is exploring the use of, or is trying a progression for the first time. Mostly it is those who are loosing punters, as a winning punter netting profit flat stakes win/place or whatever, is as happy as a pig in mud with what they are already doing.

The loosing punter exploring progressions is searching for a way to turn loss into profit, which is a bit like trying to spend more than you earn without getting into debt [how our society and it's economy works]. In other words the punter interested in progressions is mostly chasing losses though they will rarely admit it [even to themselves].

The only progression that will never get you into trouble is the winning progression. Increasing the bet amount by a progression line after a win and not a loss. Long term of course you will be no better off than sticking to flat stakes betting, but it will give you the 'impression' of being better off. That might be worth something.

crash,

Thank you for posting that and I agree that the maths of using 1 bank doesn't work thats why I use multiple banks.

I am going to run the results for as long as it takes to either proves this works or doesn't work.

Todays selections where Res of 1 means a win and Res of 0 means a loss.

Odds Res Level Stakes Running Profit
1000 1000
2.1 0 990 990
2.38 0 980 978
1.85 0 970 964
1.56 0 960 948
1.65 0 950 929
1.76 1 957.6 945.72
2.66 0 947.6 926.72
1.48 1 952.4 937.28
3 1 972.4 977.28
3.7 0 962.4 963.28
1.82 0 952.4 947.28
2.96 0 942.4 928.28
2.56 0 932.4 906.28
1.96 0 922.4 880.28
3.25 0 912.4 850.28
2.46 0 902.4 820.28
2.88 0 892.4 790.28
2.9 1 911.4 847.28
1.98 0 901.4 812.28
3.4 0 891.4 771.28
2.52 1 906.6 844.24
3.45 0 896.6 808.24
2.96 0 886.6 766.24
1.79 1 894.5 804.95
5.7 1 941.5 1007.05
4 0 931.5 997.05
3.7 0 921.5 985.05
1.6 0 911.5 971.05
2.1 0 901.5 955.05
3.4 0 891.5 936.05
3.85 1 920 998.75
2.18 1 931.8 1011.73
2.74 0 921.8 1001.73
3 0 911.8 989.73
2.7 0 901.8 975.73
4.8 0 891.8 959.73
3.55 0 881.8 940.73
3.75 1 909.3 1001.23


Laying the same selections:

Fav Price Res Return Level Stakes Running Profit
1000 1000
2.1 1 0 1009.5 1009.5
2.38 1 0 1019 1019
1.85 1 0 1028.5 1028.5
1.56 1 0 1038 1038
1.65 1 0 1047.5 1047.5
1.76 0 1 1039.9 1039.9
2.66 1 0 1049.4 1051.3
1.48 0 1 1044.6 1046.02
3 0 1 1024.6 1024.02
3.7 1 0 1034.1 1038.27
1.82 1 0 1043.6 1050.62
2.96 1 0 1053.1 1061.07
2.56 1 0 1062.6 1071.52
1.96 1 0 1072.1 1081.97
3.25 1 0 1081.6 1092.42
2.46 1 0 1091.1 1102.87
2.88 1 0 1100.6 1113.32
2.9 0 1 1081.6 1092.42
1.98 1 0 1091.1 1106.67
3.4 1 0 1100.6 1118.07
2.52 0 1 1085.4 1101.35
3.45 1 0 1094.9 1114.65
2.96 1 0 1104.4 1126.05
1.79 0 1 1096.5 1117.36
5.7 0 1 1049.5 1056.26
4 1 0 1059 1078.11
3.7 1 0 1068.5 1096.16
1.6 1 0 1078 1111.36
2.1 1 0 1087.5 1124.66
3.4 1 0 1097 1135.11
3.85 0 1 1068.5 1103.76
2.18 0 1 1056.7 1083.7
2.74 1 0 1066.2 1102.7
3 1 0 1075.7 1118.85
2.7 1 0 1085.2 1132.15
4.8 1 0 1094.7 1143.55
3.55 1 0 1104.2 1154
3.75 0 1 1076.7 1121

Using Level Stakes
Backing All Favs : SHOWING A LOSS
Laying All Favs : IN PROFIT

Using Progressive Stakes:
Backing All Favs : IN PROFIT
Laying All Favs : IN PROFIT

For this test I will assume I have 10 banks that can be busted before I have lost everything (starting fund of 10K).

Good Luck.

One question to crash. Is it your maths or something you read in a book?
Do your own research, then you can present a valid argument, not someone else's.

isn't this exciting hee hee, thanks for starting this one wesmip1, I think I can smell burning flesh here (fingertips) that's the problem!

For very good reasons I don't do my own major electrical work when it's needed in our house. I call on a qualified professional.
I don't have to stick a knife into a switched on toaster to prove electricity can kill [obviously some people do] and would argue strongly with anyone suggesting otherwise. If necessary I'll call on professional opinion [that's valid isn't it?] and in 99% of cases most people would take an Electrician's word on the subject !

Likewise, I don't need to be a mathematician to disagree with the original[IMHO] silly maths. proposition that was presented in this thread, but quite validly and if necessary, present professional opinion on the matter. If that's unfair or unacceptable then this discussion is clearly over.

If Reid's maths. are wrong, then there will be other Mathematicians out there who would have said so and presented their maths arguments to prove it. Sorry, I can't find even one that disagrees but plenty who agree.

I like what that Major said on the beach in Apocalypse Now: " I love the smell of Napalm in the morning! "

Lol. lol :-))

Get stuck into them Top rank , I too, smell burning finger tips & I like it.

I beleive in staking plans .
It is very difficult to make sustained profits from level stakes betting , although it is the safest.

The maths chaps are right when it comes to roulett because the prices are fixed, with a max ceiling, but in racing the prices vary race to race where value should be targeted. e.g. bet 2 horses to beat the odds-on pop.
This is the bit that is not factored in by the Maths chaps.
That's why the the debate will continue, because the prices vary in racing.

With most progressions , its a good if your selection method picks a hand full of $4+ shots. Under this amount & things can get tough trying to recover from a bad run.

One of the safest progressional methods is the...
"Ladder staking plan" over 72 bets.
It is desinged to show a profit even with a -30% POT & a 16% SR.
Which, according to the level stakes believers & maths experts , cant be done.
Keep in mind , it needs a strike rate of 16%+
Restart once there is any profit.

For those interested , do a search of this site using the "Search" button above.

Cheers.

__________________
Cheers.

Bagman's Stairway [to heaven] staking plan.

We all know your a born-again staking desperate Bagman!

By the way, the maths wasn't 'Roulette', it was about the futility of trying to turn a loss into a profit by using [any] progressive staking methods. You well know all Casino's, TAB's and Bookies would all go broke if it was possible [think about it].

Your just being mischievous here [as usual].