100xOdds
100xOdds
Joined: Feb 5, 2012
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May 8th, 2022 at 3:51:16 PM permalink
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?
Craps is paradise (Pair of dice). Lets hear it for the SpeedCount Mathletes :)
ThatDonGuy
ThatDonGuy
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May 8th, 2022 at 4:15:18 PM permalink
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
100xOdds
100xOdds
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May 8th, 2022 at 9:21:45 PM permalink
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

thx!

So if there was no house edge (ie: flipping a coin), then variance is 0?
Craps is paradise (Pair of dice). Lets hear it for the SpeedCount Mathletes :)
camapl
camapl 
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May 9th, 2022 at 12:53:04 AM permalink
Quote: 100xOdds

thx!

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



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…
* Actual results may vary.
camapl
camapl 
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May 9th, 2022 at 1:15:18 AM permalink
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)
* Actual results may vary.
100xOdds
100xOdds
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May 9th, 2022 at 3:19:09 AM permalink
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)
link to original post

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?
Craps is paradise (Pair of dice). Lets hear it for the SpeedCount Mathletes :)
camapl
camapl 
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Thanks for this post from:
100xOdds
May 9th, 2022 at 12:29:59 PM permalink
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!
* Actual results may vary.
Ace2
Ace2
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May 9th, 2022 at 1:12:14 PM permalink
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
Last edited by: Ace2 on May 9, 2022
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Ace2
Ace2
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May 9th, 2022 at 1:24:16 PM permalink
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
It’s all about making that GTA
Ace2
Ace2
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May 9th, 2022 at 1:35:51 PM permalink
Quote: camapl

The coin only has 2 outcomes - much less than the roulette wheel!
link to original post

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
It’s all about making that GTA

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