And if he uses a staking plan....
 

Mathematical Proof that Progressions cannot overcome Expectation.




by Richard Reid

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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 b_k = the size of the k^th bet.
M_k = the size of the payoff on the k^th bet.
p_k = the probability that the series terminates with a win on the k^th 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 n^th bet must be won, otherwise the player's entire bankroll will be lost.

Let's now define B_n = b_n * M_n

The expected value for any series is:

E_series = p₁B₁ + p₂(B₂-b₁) + p₃(B₃-b₂-b₁) + . . . + p_n(B_n-b_n-1-b_n-2- . . . -b₂-b₁) + (1-p₁-p₂- . . . -p_n) * (-b_n-b_n-1-b_n-2- . . . -b₂-b₁)

If we let

E_series = A + B where,

A = p₁B₁ + p₂(B₂-b₁) + p₃(B₃-b₂-b₁) + . . . + p_n(B_n-b_n-1-b_n-2- . . . -b₂-b₁)

and

B = (1-p₁-p₂- . . . -p_n) * (-b_n-b_n-1-b_n-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 n^th 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 + . . . + p_nB_n - p_nb_n-1 - p_nb_n-2 - . . . - p_nb₂ - p_nb₁
A = p₁B₁ + p₂B₂ + . . . + p_nB_n + b₁(- p₂ - p₃ - . . . - p_n) + b₂(- p₃ - . . . - p_n) + b_n-2(- p_n-1 - p_n) + b_n-1(- p_n)

And for "B" we get

B = -b_n(1 - p₁ - p₂ - . . . - p_n) - b_n-1(1 - p₁ - p₂ - . . . - p_n) - . . . - b₂(1 - p₁ - p₂ - . . . - p_n) - b₁(1 - p₁ - p₂ - . . . - p_n)

Now if we combine A and B again, we get,

E_series = A + B
E_series = p₁B₁ + p₂B₂ + . . . + p_nB_n - b₁(1 - p₁) - b₂(1 - p₁ - p₂) - . . . - b_n-1(1 - p₁ - p₂ - . . . - p_n-1) - b_n(1 - p₁ - p₂ - . . . - p_n)
E_series = p₁B₁ + p₂B₂ + . . . + p_nB_n + p₁b₁ + (p₂ + p₁)b₂ + (p₃ + p₂ + p₁)b₃ + . . . + (p_n + p_n-1 + . . . + p₂ + p₁)b_n - (b₁ + b₂ + . . . + b_n)

Wilson points out that to get rid of the subscripts, all we have to do is realize that p_k = (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 k^th level is the product of the probability of k-1 losses in a row multiplied by the probability of win on the k^th trial.

So how do we use this information? Well, let's try substituting this expression for each p_k and see what we get.

E_series = (1-p)^1-1pB₁ + (1-p)^2-1pB₂ + . . . + (1-p)^n-1pB_n + (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)b_n - (b₁ + b₂ + . . . + b_n)

Simplifying, we get

E_series = (p(1-p)⁰B₁ + (1-p)¹pB₂ + . . . + (1-p)^n-1pB_n + 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)b_n - (b₁ + b₂ + . . . + b_n)

If we factor p out of the first parts of the equation and look closely, we can see that the k^th term T can be written as:

T = p[(1-p)^k-1]B_k + p[(1-p)^k-1 + (1-p)^k-2 + . . . + (1-p)² + (1-p)¹ + (1-p)⁰]b_k

or rephrased for B_k = b_k * M_k we get

T = p[(1-p)^k-1M_k + (1-p)^k-1 + (1-p)^k-2 + . . . + (1-p)² + (1-p)¹ + (1-p)⁰]b_k

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-1M_k + [(1-p)^k - 1]/-p]b_k
T = p(1-p)^k-1M_k + 1 - (1-p)^k]b_k
T = p(1-p)^k-1M_k + 1 - (1-p)(1-p)^k-1]b_k
T = [[pM_k - (1-p)](1-p)^k-1 + 1]b_k This now allows us to write the equations in terms of summations. We therefore get

E_series = sum ****[[pM_k - (1-p)](1-p)^k-1]b_k**** + sum {b_k**** - sum {b_k****, for k = 1 to n

The last two terms cancel, so we are left with:

E_series = sum ****[pM_k - (1-p)](1-p)^k-1]b_k****, for k = 1 to n
E_series = sum ****[(1+M_k)p - 1](1-p)^k-1]]b_k****, for k = 1 to n

If we now look closely at this equation, we can make several observations. First, the sign of E_series depends solely on the resulting sign of [(1+M_k)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 M_k = 1 and therefore
E_series = sum ****[2p - 1](1-p)^k-1]]b_k****, for k = 1 to n
E_series = [2p - 1]sum ****(1-p)^k-1]]b_k****, 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+M_k)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.

wow!!!!!!

I knew that...

Hi Benny,
You have the positive opportunity of refining these selections into a plan of attack so as to create a greater profit here.
Your tipster has had a good run , with a SR of 66% of picking the winner in 3 .
You will see that this will drop down to an average of 45% over time because it is difficult of any tipster to sustain that SR because he is targeting every race .

A couple of ideas to think about .
1)Target only 3 races for each venue say races 4,7,8 only.
These are usually good value races.
If you go over past results , you should find your POT would have stronger just by doing this.
Just for interest , my data program tells me that for some reason , race 4 has had higher value winners than any other race No. based on 20,000 races.
In race 8 , the Fav wins less but pays the highest average div than any other race No. & its loss on turnover is less than any other race No.
Race 7 is the next best for the same reasons as adove.

2) Consider targeting the 2 highest payers of the 3 selections.
That way one is betting 2 horses a race instead of 3 , at good prices.
3) Consider targetting the 2 horses of the 3 , with the best Wt. rating as per TABQ. (100Pters. ect.)
4) Consider using the Neurals std. setting , minus the Time setting (set to 0), to split his first 2 selections only ignoring the 3rd selection.
That way one is betting only one horse per race .
5) Bet 1 or 2% of bank level stakes.

Check this out , I beleive you will increase profits dramaticly by using just one of these ideas .
Put in the homework & see what is revealed.
Remember , Tipsters are not machines , with a guaranteed outcome day in day out , so be prepared for a few bad days in a row .
The damage should be minimised if one restricts their betting to 3 races a venue rather than betting every race.
Or
If one feels one needs more action , delete the first 3 races for any venue , these are usually poor value races with questionable form.
e.g. Maidens, 2YO , lots of first starters & resumers , low divs, ect.

Cheers.

Some good advice there i agree with most of it but imo wouldnt he be better off working out the type of races the tipster performs best in,who's to say this tipster is a good read of the type of race race 4 on the programe usualy is.

Personaly if i was betting 3 horses a race i wouldnt touch a race with less than 10 starters but of course i'd have to check to see how the tipster performs in races with bigger fields.

benny
 

benny, you would almost certainly have a greater chance of success by dumping the tipsters & selecting a couple of youngish trainers within each state that have a decent number of nags in work & following them, esp after a couple of runs up & when there are signs of stable support.

i, like you & a 1,000,000 others would love to be able to wake up each morning (abt 10 ish), give the family jewels a good scratch, stumble down to the local tab, quickly scan the fields (or whatever) for the day, lay the $1k in bets, go home & forget about it all until beer ocklock (abt 4 ish) & go & collect the $1.2k that is just sitting there, waiting for me.

& i am sure that i managed to do that for a few months thru one of my alcoholic delusional phases...

Without checking the forum first for your morning laugh ?????????????????

woops, my bad

Bhagwan,
Long time since I posted, but I'm glad you knew all the those mathematical calculations, I wasn't quite sure so I contacted JFC and He steered me in the right direction.
Seriously though I want to congratulate you for all the time and effort you put in trying to help people on this forum and ask nothing in return.
Not ********ing in your pocket because I have no reason to.

Cheers

Hi Kenchar ,
Tarrs for that .
I was wondering what happened to our friend "JGF".
I reckon he will be having the time of his life on one of those ,
"lets have a fight , sites".


Cheers.

Zlotti,

Don't throw one of those brilliant, if not complex energy sapping equations at me but are saying if you are getting a negative return (loss) on your punting from level stakes, no magical progression method will turn it around into profit.

If that is not what you are saying, can you answer my question anyway.