Standard deviation in roullet

[Stefan_K]

I'm not posting this because i think that i will win in roulet, I'm just going to play. thinking about playing with 5$ and just put them on either red or black. After my calculations i will lose 13cent ((18/37)*5) per game, but what i do not know is how to calculate how big my standard deviation will be?

And also how big swings will this lead to?

[AaronBrown]

You will actually lose twice that much in roulette with two zeros. You pay $5 and win $10 18 times out of 38, for a return of $4.74, a $0.26 loss per bet.

The standard deviation of a fair coin flip for $5 is $5. The house edge in roulette does not change things much, the standard deviation is $4.99.

To translate that into swings, multiply the expected loss by the number of spins and the standard deviation by the square root of the number of spins. For example, after 100 spins you expect to lose 100 x 0.2632 = $26.32. Your standard deviation is 10 x 4.993 = $49.93.

About 1 time in 44 you will down more than two standard deviations, that is down more than $126.18. 1 time in 44 you will be up more than two standard deviations, up more than $73.55. About 1 time in 3 you will be down more than expectation but less than one standard deviation below it, that is between down $26.32 and $76.25. Another 1 time in 3 you will be above expectation but within one standard deviation of it, that is between down $26.32 and up $23.61. The rest of the time you will be between one and two standard deviations from your expectation. 29.9% of the time you will make money.
23.61491042

[Stefan_K]

[ QUOTE ]
You will actually lose twice that much in roulette with two zeros. You pay $5 and win $10 18 times out of 38, for a return of $4.74, a $0.26 loss per bet.

The standard deviation of a fair coin flip for $5 is $5. The house edge in roulette does not change things much, the standard deviation is $4.99.

To translate that into swings, multiply the expected loss by the number of spins and the standard deviation by the square root of the number of spins. For example, after 100 spins you expect to lose 100 x 0.2632 = $26.32. Your standard deviation is 10 x 4.993 = $49.93.

About 1 time in 44 you will down more than two standard deviations, that is down more than $126.18. 1 time in 44 you will be up more than two standard deviations, up more than $73.55. About 1 time in 3 you will be down more than expectation but less than one standard deviation below it, that is between down $26.32 and $76.25. Another 1 time in 3 you will be above expectation but within one standard deviation of it, that is between down $26.32 and up $23.61. The rest of the time you will be between one and two standard deviations from your expectation. 29.9% of the time you will make money.
23.61491042

[/ QUOTE ]
But isn't it diffrent between american an european roullet, i thought that it was only 1 zero in the europeean vertion am i wrong?

Thank you a lot for your help

[Jazza]

i apologize for the thread hijack, but here is a quiz:

you have $10,000, and you want to double it playing roulette, what method give you the highest chance of success?

[AaronBrown]

Buy your own wheel and take bets from other people.

[AaronBrown]

You are correct (although I have seen double zero wheels in Europe, and heard there are some single zero wheels in Las Vegas but never seen one). Also, some casinos let the bet ride when a zero comes up; if you win the next spin you get your bet back with no payout; if you lose, you lose. That cuts the expected loss in half again. I apologize for my provincialism.

The standard deviation is virtually the same, but your expected loss is cut in half.

[pzhon]

[ QUOTE ]

you have $10,000, and you want to double it playing roulette, what method give you the highest chance of success?

[/ QUOTE ]
Because the house advantage is constant over almost all wagers, you want to minimize the expected amount you bet by the time you bust out or hit your target. I believe the proper basic strategy is to bet on one number at a time, either just enough to hit your target, or all of the money you have left.

[smartalecc5]

[ QUOTE ]
i apologize for the thread hijack, but here is a quiz:

you have $10,000, and you want to double it playing roulette, what method give you the highest chance of success?

[/ QUOTE ]
I give up, wahts teh answer.

[Jazza]

what phzon said

[Siegmund]

[ QUOTE ]

Because the house advantage is constant over almost all wagers, you want to minimize the expected amount you bet by the time you bust out or hit your target.


[/ QUOTE ]

Very true.

[ QUOTE ]

I believe the proper basic strategy is to bet on one number at a time, either just enough to hit your target, or all of the money you have left.

[/ QUOTE ]

I almost fired off a fast response saying "no, bet it all on an even money bet right away, because sometimes your strategy requires you to win twice or more and cover the same ground repeatedly." (Making the largest bet possible is the correct strategy when you can choose the bet size, but don't have a choice of the payoff table.)

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.

On reflection, I agree with pzhon. Did a quick simulation of it, too: his strategy yields a success rate of 48.09% on the American wheel (better than 18/38~47.37% of one bet on red) and 49.03% (better than 18/37~48.65%) on the single-zero wheel without the en prison rule.

You do lose a little bit because of the small chance of it taking two wins, but you gain more by exposing a smaller portion of the roll to the juice if you get lucky in your first few spins.

[pzhon]

[ 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.

[PairTheBoard]

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

PairTheBoard

[BruceZ]

[ 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.

[PairTheBoard]

[ 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.


PairTheBoard