Page 1 of 3 123 LastLast
Results 1 to 15 of 31

Math Help - gambling problem/ standard deviation

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by misu200 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2007
    Posts
    10
    Quote Originally Posted by JakeD View Post
    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,
    Last edited by misu200; May 23rd 2007 at 11:57 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2007
    Posts
    10
    Quote Originally Posted by CaptainBlack View Post
    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).
    Sorry about not being clear about this.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by JakeD View Post
    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.
    Quote Originally Posted by misu200 View Post
    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}].
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by misu200 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2007
    Posts
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by JakeD View Post
    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
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View 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.
    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
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    May 2007
    Posts
    10
    Quote Originally Posted by CaptainBlack View Post
    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??
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by JakeD View Post
    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.
    Quote Originally Posted by CaptainBlack View 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.
    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.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by JakeD View Post
    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}].
    Quote Originally Posted by misu200 View Post
    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}].
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    May 2007
    Posts
    10
    Quote Originally Posted by JakeD View Post
    [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.
    Last edited by misu200; May 24th 2007 at 10:56 AM.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. Standard Deviation Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 18th 2010, 10:22 PM
  2. Standard Deviation problem/help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 22nd 2010, 04:14 AM
  3. Problem with mean and standard deviation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 29th 2009, 02:07 PM
  4. Standard deviation problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 21st 2009, 09:10 PM
  5. Standard deviation / mean problem
    Posted in the Statistics Forum
    Replies: 8
    Last Post: January 11th 2009, 08:24 PM

Search Tags


/mathhelpforum @mathhelpforum