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  #11  
Old 06-14-2005, 04:20 AM
pzhon pzhon is offline
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Posts: 66
Default Re: Standard deviation in roullet

[ QUOTE ]

Then I thought a little more about it, and remembered the other principle of negative expectation bets: to maximize your probability of reaching a fixed target, variance is your friend; play the bets with the highest SD-to-EV ratio.


[/ QUOTE ]
That's close, but you want to play with the highest variance:EV = SD^2:EV ratio. (This explains why you prefer to make larger wagers.) The reason is that the expected total variance of your bets is roughly independent of your strategy, so minimizing the house advantage you pay per unit of variance also minimizes the expected house advantage paid.

A quick way to see that betting $10k on an even money wager is not optimal is to compare it with the following strategy: Bet $5k on something that pays 2:1, then if you miss, bet the remaining $5k on something that pays 3:1. That has a chance to win while only applying the house advantage to $5k, and you never have to bet more than $10k, so the average amount wagered is lower than $10k.
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  #12  
Old 06-14-2005, 08:40 PM
PairTheBoard PairTheBoard is offline
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Default Re: Standard deviation in roullet

Aaron -
"The standard deviation of a fair coin flip for $5 is $5."

I believe the standard deviation for a $5 bet on a fair coin flip is $2.50

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  #13  
Old 06-14-2005, 09:17 PM
BruceZ BruceZ is offline
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Default Re: Standard deviation in roullet

[ QUOTE ]
Aaron -
"The standard deviation of a fair coin flip for $5 is $5."

I believe the standard deviation for a $5 bet on a fair coin flip is $2.50

[/ QUOTE ]

It's $5.

Variance(x) = E(x^2) - [E(x)]^2

= (1/2)*5^2 + (1/2)*(-5)^2 - (0)^2

= 25

standard deviation = sqrt(variance) = 5.
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  #14  
Old 06-15-2005, 05:50 AM
PairTheBoard PairTheBoard is offline
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Default Re: Standard deviation in roullet

[ QUOTE ]
[ QUOTE ]
Aaron -
"The standard deviation of a fair coin flip for $5 is $5."

I believe the standard deviation for a $5 bet on a fair coin flip is $2.50

[/ QUOTE ]

It's $5.

Variance(x) = E(x^2) - [E(x)]^2

= (1/2)*5^2 + (1/2)*(-5)^2 - (0)^2

= 25

standard deviation = sqrt(variance) = 5.

[/ QUOTE ]

ok. I was thinking 5*SQRT(.5*.5*1)

Nevermind.


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