100xOdds
100xOdds
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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?
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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?
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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…
Expectation is the root of all heartache.
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)
Expectation is the root of all heartache.
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?
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camapl
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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!
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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
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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
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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
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camapl
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May 10th, 2022 at 3:31:56 AM permalink
I stand by what I wrote. 100xOdds seemed to be confusing the outcomes on the wheel with the outcomes on the bet, and I was merely trying to differentiate between the two. Wheel has over 30 outcomes, the even money bet has 2.

Thanks anyway for your assistance!
Expectation is the root of all heartache.
Ace2
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May 10th, 2022 at 12:15:22 PM permalink
If you have a wheel with 30 pockets and you win $1 on 1-15, lose $1 on 16-30, the bet is identical to a coin flip. Both scenarios have 2 outcomes: 50% you win, 50% you lose. Variance of 1, zero edge
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camapl
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May 10th, 2022 at 1:10:06 PM permalink
Right, but that doesn’t change the fact that the wheel itself has 30 outcomes regardless of how the bets are designed. The OP’er thought that a wheel with fewer pockets should not create a higher variance, and my post was trying to distinguish between WHEEL outcomes and BET outcomes. While they are related, they are not identical.

Reading and reacting to a single post without reading prior can also create further confusion.

Thanks yet again for your assistance. If you absolutely must have the last word, as I’ve noticed many members here do, rest assured that I won’t stop you…
Expectation is the root of all heartache.
JackSpade
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August 28th, 2022 at 10:38:43 PM permalink
Suppose a player planned on making $300 bets per spin on European roulette and had the singular goal of making his session last as long as possible before depleting his bankroll. He's going to place outside bets only to get the lowest house edge.

He could bet:

$300 on a single outcome - red.
$150 on red and $150 on odd.
$100 on red, $100 on odd, and $100 on 1-18.

Splitting the bet up 3 ways would give him the most wheel coverage and presumably the lowest volatility. Would it help his bankroll last longer? Would he earn more comps as a result (if casinos even reward players for this version of roulette)?
Ace2
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August 28th, 2022 at 10:51:37 PM permalink
Best options for minimum volatility and maximum play time:

$150 on red and $150 on black
$150 on odd and $150 on even
$150 on low and $150 on high

European or American wheel

Play until your bankroll is gone then spend all the comps

Caveat: the lowest volatility play is actually to lay the zero, but it’s not offered in all casinos
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JackSpade
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August 28th, 2022 at 11:09:18 PM permalink
Haha... I had never thought of betting on combinations that had no chance of winning, only pushing. I wonder if anyone actually plays this way.
Last edited by: JackSpade on Aug 29, 2022
Ace2
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teliot
August 28th, 2022 at 11:14:22 PM permalink
Not exactly. You can push or lose $300

I believe there are comp junkies that do this stuff
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MDawg
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August 28th, 2022 at 11:28:25 PM permalink
Quote: JackSpade

Haha... I had never thought of betting on combinations that had no chance of winning, only pushing. I wonder if anyone actually plays this way.
link to original post


At any casino at which I play in Vegas, as far as I know, that sort of thing is not allowed. A single player is not allowed to bet both red and black, pass and don't pass, Player and Bank, at the same time. As well, I've observed a player dump conflicting bets all over a roulette table and heard the pit boss later declare to another pit boss that gave the player credit per spin only for the sum of bets that were at risk of total loss.

Anyway, that's where reality and hypothesis diverge.
Last edited by: MDawg on Aug 29, 2022
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teliot
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August 29th, 2022 at 10:22:03 AM permalink
Quote: Ace2

Not exactly. You can push or lose $300

I believe there are comp junkies that do this stuff
link to original post

Even more on baccarat, playing as a team, one side betting Banker the other Player. Funny that casinos don't like players doing this, not sure why. From the casino side, you either push or beat the player, never lose to the player.
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AlanMendelson
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August 29th, 2022 at 10:49:02 AM permalink
Quote: MDawg

Quote: JackSpade

Haha... I had never thought of betting on combinations that had no chance of winning, only pushing. I wonder if anyone actually plays this way.
link to original post


At any casino at which I play in Vegas, as far as I know, that sort of thing is not allowed. A single player is not allowed to bet both red and black, pass and don't pass, Player and Bank, at the same time. As well, I've observed a player dump conflicting bets all over a roulette table and heard the pit boss later declare to another pit boss that gave the player credit per spin only for the sum of bets that were at risk of total loss.

Anyway, that's where reality and hypothesis diverge.
link to original post



I've written several times about a guy who shoots craps at about 5am at Bellagio on Sundays. He bets both pass and dont pass simultaneously. He makes no other bets.

Theres no heat from the crew, and they dont care.

When he takes a bathroom break we discuss his motive and the best guess is he's practicing his shot OR this is his escape from Sunday morning home life.
Ace2
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August 29th, 2022 at 11:20:07 AM permalink
Quote: AlanMendelson

[
I've written several times about a guy who shoots craps at about 5am at Bellagio on Sundays. He bets both pass and dont pass simultaneously. He makes no other bets.

Theres no heat from the crew, and they dont care.

When he takes a bathroom break we discuss his motive and the best guess is he's practicing his shot OR this is his escape from Sunday morning home life.
link to original post



What’s your strategy when he shoots? I assume your standard procedure is to lay the 12 since the shooter is trying to avoid it (only number that gives him a loss). Laying the 12 could be accomplished by hopping every number except 12
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Ace2
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August 29th, 2022 at 11:51:57 AM permalink
Quote: MDawg


At any casino at which I play in Vegas, as far as I know, that sort of thing is not allowed. A single player is not allowed to bet both red and black, pass and don't pass, Player and Bank, at the same time. As well, I've observed a player dump conflicting bets all over a roulette table and heard the pit boss later declare to another pit boss that gave the player credit per spin only for the sum of bets that were at risk of total loss.

Anyway, that's where reality and hypothesis diverge.
link to original post

I’ve heard this before but it makes no sense. Someone making offsetting bets should be the casino’s favorite customer and should be fully comped on 100% of his bets. For example, the casino still gets their ~1.4% edge on someone simultaneously betting pass/DP, but the variance is reduced by 97% and the casino has no risk of ever losing money.

Is there a reason that casinos frown upon or ban this kind of play?
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MDawg
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August 29th, 2022 at 2:56:16 PM permalink
I talked to a different pit boss - at the exact same casino where I saw a different pit boss disallow a player from betting both Bank and Player, and where that pit boss told me that the casino doesn't allow that sort of contrary simultaneous betting at table games, same casino where a different pit boss told me that doesn't give credit for conflicting bets at roulette,

and

this pit boss said that doesn't care, will give credit for every chip on the table as far as bet tracking,
and
will allow red/black, Player/Bank, etc. simultaneous bets as long as promo chips are not involved.

So...go figure. Maybe at this casino, which is one of the majors on the Strip, it's at the pit boss' discretion.
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JackSpade
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August 31st, 2022 at 12:09:39 PM permalink
MGM Grand's -0- wheel has a $100 minimum and a max bet that can be effortlessly negotiated up to however much cash you put on the table.

$60,000 on black:

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