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?
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
thx!Quote: ThatDonGuyAssuming 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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So if there was no house edge (ie: flipping a coin), then variance is 0?
Quote: 100xOddsthx!
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…
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.Quote: camaplTo 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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So without a 0, var is higher?? (.5 vs 0.4736)
How can having 1 less # give more variance?
Quote: 100xOddsahh.. 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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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!
Actually, the variance is the sum of:Quote: ThatDonGuyAssuming 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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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
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" basisQuote: 100xOddsahh.. 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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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
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 1/2Quote: camaplThe coin only has 2 outcomes - much less than the roulette wheel!
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Thanks anyway for your assistance!
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…
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)?
$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
I believe there are comp junkies that do this stuff
Quote: JackSpadeHaha... I had never thought of betting on combinations that had no chance of winning, only pushing. I wonder if anyone actually plays this way.
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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.
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.Quote: Ace2Not exactly. You can push or lose $300
I believe there are comp junkies that do this stuff
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Quote: MDawgQuote: JackSpadeHaha... I had never thought of betting on combinations that had no chance of winning, only pushing. I wonder if anyone actually plays this way.
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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.
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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.
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.
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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
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.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.
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Is there a reason that casinos frown upon or ban this kind of play?
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.