Math question on roulette strategy

Factual Questions
1

The math of Roulette is driving me crazy. Please settle a discussion with a friend of mine. We were discussing American roulette, with 38 numbers to bet on and a payoff of 35 to 1 if you bet on a single number and win, with the odds of the wheel landing on any particular number being 37 to 1.

My friend and I were arguing over whether or not it is better strategy, mathematically, to bet on a single number on the board at a time or bet on two different numbers at the same time. When I say two different numbers, I mean putting a token on say, #2 and putting another token on say, #18, I don’t mean the situation where you put a token between two numbers.

I think that the payoff is better when you bet on one number at a time, because you are betting against yourself if you make two simultaneous bets, and because this makes the payoff per win less, because every time you win you also lose. I think this is demonstrated by the extreme example of placing 38 bets at the same time on every number on the board, where you are sure to lose each round.

Another way of asking this question is, suppose I have the option of betting ten dollars on a single number of the roulette board each round or betting 5 dollars on two nonadjacent numbers of the roulette board each round. Is one strategy mathematically superior to the other?

2

I don’t know anything about roulette, so what I say is based on what I gleam from your OP: one of 38 numbers is chosen at uniform random by the wheel. You place X dollars on a number on the wheel, and if the wheel lands on that, you are paid 35*X dollars.

In this case, the expected profit (i.e., the arithmetic mean profit) from putting X dollars on the board is entirely independent of how you choose to distribute those X dollars across the board. Put them all on one number, split them evenly between two adjacent numbers, split them evenly between two non-adjacent numbers, split it across the board, evenly, unevenly, whatever; the expected result is that you’ll win back 35/38*X dollars (i.e., if you play for a long time, with high probability, the ratio of the amount you’ve been paid to the amount you’ve paid will be very close to 35/38).

3

You can win at roulette but there is only one system that works. Without getting into math, when your number hits, whether it is one or two numbers, you leave the table. That’s it!

The only way to win mathematically is to bet that the odds will hit your number before you get wiped out by the odds that favor the house. If you can’t walk away, you can’t win (mathematically).

4

This entirely depends on what your purpose of playing is. At issue here are two independent factors, expected value and variance. I’m going to take a little detour to explain, and then return to your question.

Expected value is the probability of winning multiplied by the win amount, minus the probability of losing multiplied by the amount lost, all divided by the amount wagered. For roulette, every bet on the board except for one (the 5-square bet) has the exact same EV. Note that EV doesn’t depend on how much you wager - it’s expressed as a percentage loss (if negative) or gain (positive).

$10 on a 1-square bet: ((1/38)*350 - (37/38)*10)/10 = -0.053
$5 on 2 1-square bets: ((2/38)*170 - (36/38)*10)/10 = -0.053

Neither system will make you more or less money in the long run.

However, the other thing to consider is variance. Variance is a measure of how often your actual result is close to your average result. Things with high variance often deviate from your EV, things with low variance will frequently be close to your EV.

In your example, putting $10 on a 1-square bet has a much higher variance than putting 2 $5 bets down. In the first case, 1/38th of the time you will be up (but way up), 37/38ths of the time you will be down. In the second case, you’ll be twice as likely to be up, but you’ll be less rich if you hit. To go further, let’s say you put 10 $1 1-square bets down. Your EV is still the same! ((10/38)*26-(28/38)*10)/10 = -0.053 But now, you’ve got a 26% chance of being in the black the next round.

Getting back to your original question, let’s figure out what your goal in playing is.

(a) You want to turn $10 into $360 with no regard to how long you play. Your best bet now is to plunk down all $10 on one number and pray. Remember - every bet in roulette has a negative EV, so you basically want to ride the variance. You’ll meet your goal about 2.6% of the time.

(b) You want to turn $10 into $70 with no regard to how long you play. Your best bet now is to spread out and make 5 $2 bets so that if any one of them hits, you’ll make your $70. You’ll meet your goal about 13% of the time.

(c) You want to play as long as possible on $10. This is the goal of most gambling. In this case, you now want to lower your variance as much as possible. As a matter of fact, you shouldn’t be betting on single numbers at all - you want to bet as little as possible so you get the most bets in, and cover as many numbers as you can with each bet. For example, if the table is a $5 minimum, you want to make $5 bets on one of the following: red, black, 1st half, 2nd half. You have a 70% chance of making it past 2 rounds.

(d) You want to have the highest EV possible. Take your money and deposit it in an FDIC-insured bank. 99.99999% guaranteed.

5

Wouldn’t that be 38 to 1?

Nope, they are identical. The outcome of either bet is not influenced by the presence or absense of the other bet on any given spin.

6

37 against to 1 for.

7

The only way to win at roulette, consistently, reliably, over time, is…be the house.

8

There is only one bet at the roulette wheel that has a different expected value than all the others. If you bet on the intersection that covers 0, 00, 1, 2, and 3, you do not get a payout that provides a long term result the same as the other bets: it’s actually worse.

Otherwise, every other bet has a difference between true odds and payout odds of 2/38, or 5.26%.

9

Speaking of which, it appears that in my post above I misunderstood the wording explaining how the payouts worked. Change all my 35s to 36s. However, also, feel free to ignore me and read aptronym’s excellent post (discussing the pertinent issue of variance, which I was too lazy to do) instead.

10

No, I think you’re correct.

37-to-1 means there’s a 1/38 chance. Think about it this way, if you have a 1-to-1 chance of something, your probability is 1/2.

Also, roulette pays 36 for 1 on your winning bet - that means that you lose your 1, and you get 36 in return. So you have 36 units, but you won 35 units. (This is not the standard way of expressing payouts and differs from every other game in the casino where you are paid 1-to-1, or 2-for-1, for even-money bets.)