Thread: gambling problem/ standard deviation

1. gambling problem/ standard deviation

I'm analyzing some data from online gambling sites.I'm trying to prove they are stealing the customers.
I'm not an expert in math ....just having "common sense" knowledge.

Here is the problem I need to solve:

I have a series of N independent event where I have a chance ( W[i] ) to win some money ( P[i] )

At the end of day my mathematical expectation is :
E= sum after i (W[i]*P[i])

In reality I will have won R dollars after these N independe events.

If N is big then R and E should converge somehow.

Is it possible to apply here some Standard deviation/Chebyshev's inequality/Weighted standard deviation to get some statistical interpretation about this?

I would really like a formula that will say to me something like this:
There is a 80% chance that the possible values of R to be in (E-x,E+x) range.

2. Originally Posted by misu200
I'm analyzing some data from online gambling sites.I'm trying to prove they are stealing the customers.
I'm not an expert in math ....just having "common sense" knowledge.

Here is the problem I need to solve:

I have a series of N independent event where I have a chance ( W[i] ) to win some money ( P[i] )

At the end of day my mathematical expectation is :
E= sum after i (W[i]*P[i])

In reality I will have won R dollars after these N independe events.

If N is big then R and E should converge somehow.

Is it possible to apply here some Standard deviation/Chebyshev's inequality/Weighted standard deviation to get some statistical interpretation about this?

I would really like a formula that will say to me something like this:
There is a 80% chance that the possible values of R to be in (E-x,E+x) range.
E will be approximately normally distributed and you can use that to calculate the range you want. But you have to get the standard deviation right. If you describe some actual wagers in more detail, I can recommend how to estimate the standard deviation.

3. Originally Posted by JakeD
E will be approximately normally distributed and you can use that to calculate the range you want. But you have to get the standard deviation right. If you describe some actual wagers in more detail, I can recommend how to estimate the standard deviation.
sure.

W(i) is between 0 and 1

For example:

Imagine that you are in a casino and:

At t0 time you have a chance of 20% to win 100$. This means W(0) = 0.2 and P(0)=100 At t1 time you have a chance of 70% to win 150$.
This means W(1) = 0.7 and P(1)=150

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,
At t(N-1) time you have a chance W(N-1) = 0.8 to win P(N-1) = 70

On average you expect to win E=W(0)P(0)+W(1)P(1) ........W(N-1)P(N-1)
In reality you win R

I would really like a formula that will say to me something like this:
There is a 80% chance that the possible values of R to be in (E-x,E+x) range.

I've tried to use "weighted standard deviation" ( http://www-minos.phyast.pitt.edu/disdocs/weightsd.pdf ) and "weighted mean" ( http://www-minos.phyast.pitt.edu/disdocs/weigmean.pdf )

with the weights being W(i):
w(i)=W(i)

and x(i) to be the outcome at the moment t(i)
x(i) = {
0, if you loose
P(i),if you won
}

I'm not sure if this is right.

Thanks,

4. Since the player has control over the number of trials the usual analysis
where N is determined beforhand is invalid.

That is we dont know what stopping rule is being applied.

RonL

5. Originally Posted by CaptainBlack
Since the player has control over the number of trials the usual analysis
where N is determined beforhand is invalid.

That is we dont know what stopping rule is being applied.

RonL
I 'm trying to analyze some existing data. ....so I have a finite number of trials (N=450 in my case).

6. Originally Posted by JakeD
E will be approximately normally distributed and you can use that to calculate the range you want. But you have to get the standard deviation right. If you describe some actual wagers in more detail, I can recommend how to estimate the standard deviation.
Originally Posted by misu200
sure.

W(i) is between 0 and 1

For example:

Imagine that you are in a casino and:

At t0 time you have a chance of 20% to win 100$. This means W(0) = 0.2 and P(0)=100 At t1 time you have a chance of 70% to win 150$.
This means W(1) = 0.7 and P(1)=150

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,
At t(N-1) time you have a chance W(N-1) = 0.8 to win P(N-1) = 70

On average you expect to win E=W(0)P(0)+W(1)P(1) ........W(N-1)P(N-1)
In reality you win R

I would really like a formula that will say to me something like this:
There is a 80% chance that the possible values of R to be in (E-x,E+x) range.

I've tried to use "weighted standard deviation" ( http://www-minos.phyast.pitt.edu/disdocs/weightsd.pdf ) and "weighted mean" ( http://www-minos.phyast.pitt.edu/disdocs/weigmean.pdf )

with the weights being W(i):
w(i)=W(i)

and x(i) to be the outcome at the moment t(i)
x(i) = {
0, if you loose
P(i),if you won
}

I'm not sure if this is right.

Thanks,
You have not described actual wagers. You haven't said how much was wagered, so how much you can lose. Also, for actual casino wagers, there are often multiple payoffs possible. For example: for blackjack if you wager $10, there are many possible payoffs: -$10 for a straight loss, -$20 if you double down and lose, +$15 if you get a blackjack and the dealer doesn't, $0 if both you and the dealer get a blackjack, etc. What games are you wagering on? Are you keeping track of how much you are wagering and your actual net winnings? To accomplish what you want, this is what you need at a minimum: the expected net payoff for each wager and the actual net payoff. So for that$10 blackjack wager, you would need the expected net payoff, say -$.10 (that is -10 cents) if you had a 1% disadvantage to the house, and the actual net payoff, say +$15 if you got a blackjack, -$10 for a straight loss. Then to get your expected net winnings $E$, you take the unweighted sum of the expected net payoffs. To estimate the standard deviation per wager, calculate the unweighted standard deviation $S$ of the actual payoffs. Then after $n$ wagers, your actual net winnings $R$ are approximately normally distributed with mean $E$ and standard deviation $S\sqrt{n}$. (This is from the Central Limit Theorem, BTW.) An 80% confidence interval (range) for $R$ is $[E- 1.28 S\sqrt{n}, E+1.28 S \sqrt{n}].$ 7. Originally Posted by misu200 I 'm trying to analyze some existing data. ....so I have a finite number of trials (N=450 in my case). Sorry about not being clear about this. But where did this data come from? That is do you know that there is no selection effect going on? RonL 8. The game is poker and I consider the ALL IN situations between you and exactly one opponent. I did include your "bet"/wage in P(i) In fact P(i) = your bet + opponent bet + some dead money in the pot from other players that folded or blinds Anyway I can compute the chance for you to win the pot P(i) I've called this chance W(i) So I've simplified the problem by not including my wage . It's just about pots ( P(i) ) that I have a W(i) chance to win them. 9. Originally Posted by CaptainBlack Since the player has control over the number of trials the usual analysis where N is determined beforhand is invalid. That is we dont know what stopping rule is being applied. RonL In most casino games, the outcome of each wager is independent of what happened in the past. So no stopping rule can change the expected winnings per wager. Thus to find the total expected win, you just need the expected win per wager and the number of wagers as I describe in my previous post. 10. Originally Posted by JakeD In most casino games, the outcome of each wager is independent of what happened in the past (blackjack is an exception). So no stopping rule can change the expected winnings per wager. Thus to find the total expected win, you just need the expected win per wager and the number of wagers as I describe in my previous post. Suppose my stopping rule is I stop when I am$10 down. At the end of play I am always $10 down. Now even if the games are as advertised if I collect my results together I will be doing worse that the expectation for the set of wager and stakes. Also if I have some effect on the outcome, like which cards to change/play, or how I play in blackjack (with an infinite number of decks!) If I play sub-optimaly I will come out worse than expectation with optimal play RonL 11. Originally Posted by CaptainBlack Suppose my stopping rule is I stop when I am$10 down. At the end of play I am always $10 down. Now even if the games are as advertised if I collect my results together I will be doing worse that the expectation for the set of wager and stakes. No ignore this one, only if the data is from a single run of play with a stopping rule like this will the result be significantly biased. RonL 12. Originally Posted by CaptainBlack No ignore this one, only if the data is from a single run of play with a stopping rule like this will the result be significantly biased. RonL There is no stopping rule here or wages(bets) and I can loose nothing in this games (the way i have defined it). N is the number of some games I've played and logged on my computer (N=450). In each of these games there was a chance W(i) to win an amount of money P(i) So in each game I can win R(i) dollars: R(i) = { P(i), I win ......W(i) chance to happen 0 , I lost .......1-W(i) chance to happen Theoreticaly after all these games ( N=450 in my case) I should have win somenthing like E = sum after i ( W(i)*P(i) ) with i=1,450 In reality I have won S dollars. My total expectation E was 8500$ and the total outcome S was 7500$. How likely is to have such a big abs(E-S) value?? 13. Originally Posted by JakeD In most casino games, the outcome of each wager is independent of what happened in the past. So no stopping rule can change the expected winnings per wager. Thus to find the total expected win, you just need the expected win per wager and the number of wagers as I describe in my previous post. Originally Posted by CaptainBlack Suppose my stopping rule is I stop when I am$10 down. At the end of play I am always $10 down. Now even if the games are as advertised if I collect my results together I will be doing worse that the expectation for the set of wager and stakes. I am using the theorem that says when wagers are independent trials, no wagering system can influence your expected win rate. If you calculate an expected win rate after N wagers using a stopping rule, you will find you may not have stopped after N wagers, that is, you may not have ever been$10 down. Taking that into account, the expected win rate is independent of the stopping rule.

Also if I have some effect on the outcome, like which cards to change/play, or how I play in
blackjack (with an infinite number of decks!) If I play sub-optimaly I will come out worse
than expectation with optimal play

RonL
There is no doubt that in blackjack, poker, sports betting, etc , you influence your expected win rate and actually you probably don't know what your real expected win rate is.

14. Originally Posted by JakeD
You have not described actual wagers. You haven't said how much was wagered, so how much you can lose. Also, for actual casino wagers, there are often multiple payoffs possible. For example: for blackjack if you wager $10, there are many possible payoffs: -$10 for a straight loss, -$20 if you double down and lose, +$15 if you get a blackjack and the dealer doesn't, $0 if both you and the dealer get a blackjack, etc. What games are you wagering on? Are you keeping track of how much you are wagering and your actual net winnings? To accomplish what you want, this is what you need at a minimum: the expected net payoff for each wager and the actual net payoff. So for that$10 blackjack wager, you would need the expected net payoff, say -$.10 (that is -10 cents) if you had a 1% disadvantage to the house, and the actual net payoff, say +$15 if you got a blackjack, -$10 for a straight loss. Then to get your expected net winnings $E$, you take the unweighted sum of the expected net payoffs. To estimate the standard deviation per wager, calculate the unweighted standard deviation $S$ of the actual payoffs. Then after $n$ wagers, your actual net winnings $R$ are approximately normally distributed with mean $E$ and standard deviation $S\sqrt{n}$. (This is from the Central Limit Theorem, BTW.) An 80% confidence interval (range) for $R$ is $[E- 1.28 S\sqrt{n}, E+1.28 S \sqrt{n}].$ Originally Posted by misu200 There is no stopping rule here or wages(bets) and I can loose nothing in this games (the way i have defined it). N is the number of some games I've played and logged on my computer (N=450). In each of these games there was a chance W(i) to win an amount of money P(i) So in each game I can win R(i) dollars: R(i) = { P(i), I win ......W(i) chance to happen 0 , I lost .......1-W(i) chance to happen Theoreticaly after all these games ( N=450 in my case) I should have win somenthing like E = sum after i ( W(i)*P(i) ) with i=1,450 In reality I have won S dollars. My total expectation E was 8500$ and the total outcome S was 7500\$.
How likely is to have such a big abs(E-S) value??
Going with the method I described above, you need the unweighted standard deviation $S$ of the actual payoffs. To calculate this, use the value for the actual payoff of P(i) if you won and 0 if you lost. Then, using the values you've given, an 80% confidence interval (range) for your actual total win will be $[8500- 1.28 S\sqrt{450}, 8500+1.28 S \sqrt{450}].$

15. Originally Posted by JakeD
$[8500- 1.28 S\sqrt{450}, 8500+1.28 S \sqrt{450}].$

I think you mistype the above formula:
$[8500- 1.28 S / sqrt{450}, 8500+1.28 S /sqrt{450}].$

I've computed the standard deviation as you said:

The mean is M=7500/450

S = sqrt (( sum after i of pow( R[i]-M , 2) ) / (450-1)

The bad think is I got S=14.2
and 1.28*14.2/sqrt(450) = 0.85

I think somenthing is not right here.

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