Standard Deviation question

[mike1212]

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.

[BruceZ]

[ 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

[Shawsy]

[ QUOTE ]
[ 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.