Variance of red/black or odd/even European roulette?

European roulette = single 0, and you get 1/2 your bet back on a even money wager (ie: red/black or odd/even #) if the ball lands on 0 (green, not counted as odd or even).

Someone i know lives in England and constantly plays even money wagers on European roulette.

The lower the variance, the more consistent your bankroll will be grinded down by the house edge.
(in this case, losing 1/2 your bet on 0 is 1.43% house edge.)

Variance of multi-deck blackjack is about 1.4.
What's the var of even money wagers on European roulette?

Assuming you lose half of your bet if zero comes up, the mean result is (18/37 x 1) + (18/37 x -1) + (1/37 x -1/2) = -1/74.
The variance is the sum of:
(18/37 x (1 - (-1/74)))^2
(18/37 x (-1 - (-1/74)))^2
(1/37 x (-1/2 - (-1/74)))^2
which is about 0.4736

Quote: ThatDonGuy

Assuming you lose half of your bet if zero comes up, the mean result is (18/37 x 1) + (18/37 x -1) + (1/37 x -1/2) = -1/74.
The variance is the sum of:
(18/37 x (1 - (-1/74)))^2
(18/37 x (-1 - (-1/74)))^2
(1/37 x (-1/2 - (-1/74)))^2
which is about 0.4736
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thx!

So if there was no house edge (ie: flipping a coin), then variance is 0?

Quote: 100xOdds

thx!

So if there was no house edge (ie: flipping a coin), then variance is 0?
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Simply flipping a coin doesn’t constitute making a wager, as no money is changing hands, so there’s no way to answer your question. For what it’s worth, I tried squaring “heads” and “tails” and just came up with more letters…

To answer what I think you’re asking, let’s look at a version of roulette without any zeros (which you will never ever see in a casino for any significant wagering).

EV = (1/2 X 1) + (1/2 X -1) = 0
(no house edge)

Var = (1/2 X (1 - 0))^2 + (1/2 X (-1 - 0))^2 = (1/2)^2 + (-1/2)^2 = 1/4 + 1/4 = 1/2
(non-zero)

Quote: camapl

To answer what I think you’re asking, let’s look at a version of roulette without any zeros (which you will never ever see in a casino for any significant wagering).

EV = (1/2 X 1) + (1/2 X -1) = 0
(no house edge)

Var = (1/2 X (1 - 0))^2 + (1/2 X (-1 - 0))^2 = (1/2)^2 + (-1/2)^2 = 1/4 + 1/4 = 1/2
(non-zero)
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ahh.. I was thinking ev when I said 0.

So without a 0, var is higher?? (.5 vs 0.4736)
How can having 1 less # give more variance?

Quote: 100xOdds

ahh.. I was thinking ev when I said 0.

So without a 0, var is higher?? (.5 vs 0.4736)
How can having 1 less # give more variance?
link to original post

Good question. I think I see the confusion… This is the variance of the BET outcome, not the WHEEL/BALL outcome. The variance of getting either +1 or -1 with equal probability is higher than the variance of getting +1, -1, or -1/2, as in single zero, because of the -1/2 outcome. Think of variance as a (more complicated) measure of distance of all outcomes from the average value. In the even money case, all outcomes are 1 unit away from zero (+ or - makes no difference); whereas, in European roulette, you have one outcome that is closer to the mean than the others, the -1/2 when hitting zero on the wheel. The number of outcomes on the wheel is “less” relevant to the calculation. In other words, 1/36 is not much different from 1/37.

You could assign +1 and -1 to heads and tails, respectively, and it would have the same variance of 1/2 on a fair coin. The coin only has 2 outcomes - much less than the roulette wheel!

Quote: ThatDonGuy

Assuming you lose half of your bet if zero comes up, the mean result is (18/37 x 1) + (18/37 x -1) + (1/37 x -1/2) = -1/74.
The variance is the sum of:
(18/37 x (1 - (-1/74)))^2
(18/37 x (-1 - (-1/74)))^2
(1/37 x (-1/2 - (-1/74)))^2
which is about 0.4736
link to original post

Actually, the variance is the sum of:

18/37 x (1 - (-1/74))^2
18/37 x (-1 - (-1/74))^2
1/37 x (-1/2 - (-1/74))^2

which is about 0.9795

The variance of any even-money game will be around 1 as long as the edge isn't too high

Quote: 100xOdds

ahh.. I was thinking ev when I said 0.

So without a 0, var is higher?? (.5 vs 0.4736)
How can having 1 less # give more variance?
link to original post

Because they are paying you less than a fair payout with a zero on the wheel. For a binary game, variance is pqf^2, where p is the probability of winning, q is the probability is losing and f is the winning payout on a "for-1" basis

For example, the variance of an even money coin flip (no edge) is .5 * .5 * 2^2 = 1. The variance of a coin flip that pays .95 to 1 (2.5% edge) is .5 * .5 * 1.95^2 = 0.9506

Quote: camapl

The coin only has 2 outcomes - much less than the roulette wheel!
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Actually, for a roulette game that does not pay back half your bet when a zero hits, an even money bet has two outcomes...win 1 or lose 1...same as a coin flip. If it pays half back when a zero hits (rare in the USA), then there are three outcomes: win 1, lose 1 or lose 1/2