100xOdds
Joined: Feb 5, 2012
• Posts: 3481
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
Joined: Jun 22, 2011
• Posts: 5418
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
100xOdds
Joined: Feb 5, 2012
• Posts: 3481
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

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

Joined: Jun 22, 2010
• Posts: 302
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?

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

Joined: Jun 22, 2010
• Posts: 302
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
Joined: Feb 5, 2012
• Posts: 3481
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)

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

Joined: Jun 22, 2010
• Posts: 302
Thanks for this post from:
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?

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
Joined: Oct 2, 2017
• Posts: 1689
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

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

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
It’s all about making that GTA
Ace2
Joined: Oct 2, 2017
• Posts: 1689
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?

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
Joined: Oct 2, 2017