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Standard Deviation question
Back to the same expected value question from yesterday, but this time I want to calculate my standard deviation.
To refresh on the details: I am going to deposit $250. They will match my deposit. I can bet with my $500 bankroll. When I withdraw, $250 is deducted from my cashout. So if I play 1 hand of baccarat and bet on banker until there is not a tie: 50.68% -> balance = $975 -> cashout = $725 49.32% -> balance = 0 -> cashout = $0 So expected value = 117.48 0.506852033______475______240.7547157 0.493150343_____-250_____-123.2875858 __________________________117.4671299 What is the standard deviation of this? When I calculate the standard deviation bu taking 240.75 and -123.29 as the other I get 257.42 which I think is wrong since the probabilities of each result occuring isn't 50/50 but 50.68%/49.32%. So I lined up in excel 5069 cells containing 240.75 and then 4931 cells containing -123.28 and then found the standard deviation of this to be 182.01. This I feel should be correct since the sample size now represents the expected ratio of results. Is the second way the proper way to do this? Thanks. |
#2
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Re: Standard Deviation question
[ QUOTE ]
Back to the same expected value question from yesterday, but this time I want to calculate my standard deviation. To refresh on the details: I am going to deposit $250. They will match my deposit. I can bet with my $500 bankroll. When I withdraw, $250 is deducted from my cashout. So if I play 1 hand of baccarat and bet on banker until there is not a tie: 50.68% -> balance = $975 -> cashout = $725 49.32% -> balance = 0 -> cashout = $0 So expected value = 117.48 0.506852033______475______240.7547157 0.493150343_____-250_____-123.2875858 __________________________117.4671299 What is the standard deviation of this? When I calculate the standard deviation bu taking 240.75 and -123.29 as the other I get 257.42 which I think is wrong since the probabilities of each result occuring isn't 50/50 but 50.68%/49.32%. So I lined up in excel 5069 cells containing 240.75 and then 4931 cells containing -123.28 and then found the standard deviation of this to be 182.01. This I feel should be correct since the sample size now represents the expected ratio of results. Is the second way the proper way to do this? Thanks. [/ QUOTE ] No, doing it that way you would have to enter your results of 475 and -250. You can use this formula, where E(x) is the expected value of x: variance(x) = E(x^2) - [E(x)]^2 variance = 0.506852033 * (475)^2 + 0.493150343 * (-250)^2 - (117.47)^2 standard deviation = sqrt(variance) = 362.47 |
#3
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Re: Standard Deviation question
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[ QUOTE ] snip I am going to deposit $250. They will match my deposit. I can bet with my $500 bankroll. When I withdraw, $250 is deducted from my cashout. So if I play 1 hand of baccarat and bet on banker until there is not a tie: 50.68% -> balance = $975 -> cashout = $725 49.32% -> balance = 0 -> cashout = $0 So expected value = 117.48 0.506852033______475______240.7547157 0.493150343_____-250_____-123.2875858 __________________________117.4671299 What is the standard deviation of this? snip [/ QUOTE ] No, doing it that way you would have to enter your results of 475 and -250. You can use this formula, where E(x) is the expected value of x: variance(x) = E(x^2) - [E(x)]^2 variance = 0.506852033 * (475)^2 + 0.493150343 * (-250)^2 - (117.47)^2 standard deviation = sqrt(variance) = 362.47 [/ QUOTE ] Or the equally correct (but usually less practical) alternative formula: Var[X] = Sum ((X - E[X])^2 * P(x)) Here Var[X] = (475-117.47)^2*.50685 + (-250-117.47)^2*.49315 Var[X]= 131381 and SDev[X]=362.47 as above. |
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