Roulette: Half the zeros at twice the cost

I used to be really good at calculating house edge and probabilities, but I've gotten rusty. I enjoy roulette, in large part because at a busy table, the game runs very slowly (35 spins per hour, says Ask the Wizard #136), and if I bet the table minimums, I can sit there for hours with a tiny bankroll, even if I'm on the bad side of the variance.

Double-zero is all we've got here in Washington State, but I'll be heading back to Vegas again soon. I typically play five numbers at $1 each, and with a house edge of 5.263%, I expect to lose $0.263 per spin. Assuming a session of 100 spins, I can expect to lose $26.32.

The Riviera has a single zero wheel that I played last year, but the table minimum is $10. If I bet five numbers at $2 each, with a house edge of 2.703%, I'm expected to lose $0.270 per spin. Assuming a session of 100 spins, I can expect to lose $27.03.

I had assumed that doubling the bet and halving the house edge would be a wash, and that playing $5 double-zero tables would have the same -EV as the $10 single-zero tables. I'm seeing that my error was assuming that the house edge was halved, but the 2.703% house edge is more than half of the 5.263% house edge.

Thus, the counter-intuitive result I'm seeing is that the expected loss betting X on a double-zero wheel is less than the expected loss betting 2X on a single-zero wheel.

Care to check my math and let me know if/where I'm mistaken?

Also, the CET blog post linked from this discussion says "Caesars Palace has [single zero] roulette in both its high roller area and on its main casino floor." Any idea what the table limits are on the table outside of the high-roller area?

Quote: travisl

Care to check my math and let me know if/where I'm mistaken?

Looks fine.
You do not think the EVs for both are about equal? I do.
What is not equal is the standard deviation over say 100 bets
The big difference will be in your bankroll swings. Way more with the $10 total bets
You even have slightly higher chance of being even or ahead with the 0wheel

But as you should know, the EV is just the center of the possible ending fortune you can have after 100 bets
For 00wheel you can end up after 100 bets (assuming you can cover every bet) between +$339 and -$392
99.7% of the time and nothing you can do to know where you will end up after your session. Not at -$26.
More likely somewhere else as you know.

And you have the same chance of winning about $95 as you do at losing $148 (assumes your average bet stays the same)

0wheel


FYI
For the unit standard deviation the formula for simple (win or lose) bets is
square root of (P*(1-P))
* (x+1)
In your example P=5/38
x = 31/5 (your payoff - you net $31 from a $5 wager) = 6.2
((5/38)*(33/38))^.5 * 7.2 = 2.434

Take 2.434 * $Bet ($5) * square root of 100 (# of bets) =
~ $121.70 your 1SD value you add and subtract from the EV

Good Luck in Vegas

Don't forget about the outside bets on single-zero roulette. Some operators offer la partage or en prison for even money propositions, either of which will cut the 2.7% edge roughly in half. If you primarily play outside bets, you'll do far, far better on a $10 single-zero table than a $5 double-zero table. I know you say you play 5 inside bets, but maybe that's worth re-examining in view of the cost?

Quote: travisl

Thus, the counter-intuitive result I'm seeing is that the expected loss betting X on a double-zero wheel is less than the expected loss betting 2X on a single-zero wheel.

Care to check my math and let me know if/where I'm mistaken?

Excluding the 5-number bet on double-zero and en prison/half-back rules on single-zero:

The house edge of double-zero roulette is 2/38.
The house edge of single-zero roulette is 1/37.

You are comparing betting 1 unit at 2/38 to betting 2 units at 1/37, and as you have discovered, 2/38 does not equal 2/37.

Thanks to each of you for the insight.

7craps, I know the SD on double-zero roulette makes for some pretty wild swings -- that's one of the things I like about the game -- but I had no idea how to compute it. That's some great data. And I had no idea how much more variable a session at the single-zero wheel would be. I usually only bring a $200 bankroll to a session, and cringe when I peel off the second $100 bill. The variance on the single zero wheel is much more likely to deplete my session bankroll.

MathExtremist, I don't know of any low-limit European roulette games in Vegas. Any of the single-zero games near my bankroll level don't use the en prison rule, according to other threads I've read here. If you know of a $10 or less wheel in Vegas with a 1.35% house edge, I'd love to hear about it. I'll gladly give them many hours taking a shot at my money.

JB, your final sentence is elegant in its conciseness. The assumption that they were equal had been rolling around in my brain for months, and although my math yesterday showed my assumption was wrong, your statement made my understanding concrete.