Roulette Probability Example

Since there are 18 red spaces there is an 18/38 probability of winning, with a net gain of $1. There is a 20/38 probability of losing your initial bet of $1. The expected value of this bet in roulette is 1 (18/38) + (-1) (20/38) = -2/38, which is about 5.3 cents. Here the house has a slight edge (as with all casino games). This would normally not be an issue, but American Roulette uses the same payout ratios as European and French without adjusting them for the addition pocket, which complicates things for players. For example, the chance to win a Straight Up bet (single number) is 37:1, but the payout for that is 35:1. As an example if you bet on a single number your chances of winning is 1 in 37 or 1 in 38 (depending on the table). However the casino will only pay out to you 35 to 1 or 36 to 1. A second example would be if you placed a $10 bet on red and win the casino will give back to you $20. To pander to a larger population, we made a video covering basic concepts of probability. To apply these concepts, we give an example using the probability o. Therefore, the odds for winning with a Straight Up bet on 32 Red, for example, would be expressed as Odds for Winning = 1/36 or 1 to 36, because there is only one winning number and 36 numbers that result in a loss. As you can see, probability differs from odds in that it is a likelihood of 1 out of 37 outcomes.

Odds and probability are often used interchangeably to describe your chances of winning at gambling. In fact, they are not the same thing. Let’s find out what they mean.

Probability allows us to predict how often an outcome or win will occur, but does not allow us to predict when exactly it will happen. It’s useful to know the difference in order to make informed decisions about the games you choose to play and to consider as a game strategy if applicable.

In gambling, the concept of odds is different for table games and slot machines. For table games, odds refer to the total amount you are paid if your bet wins, and is typically expressed in ratios. Odds can also be referred to as a ‘payout’. For example, if the payout is 2 to 1 (2:1), this means that for each $1 wager you make, any winning outcome will pay you a total of $2.

Understanding the difference between the odds and probability helps us determine the house advantage. The bigger the difference between the odds and probability, the bigger the house advantage.

An example with roulette

Let’s say you’re playing roulette, and you place a $10 bet that a red number will come up next.
The odds state that the bet pays 1 to 1. If you win, your $10 bet would pay $10.

But the probability of the ball landing on a red number is 47.4 percent. We calculate the probability like this:

• There are 18 red numbers on the roulette wheel.
• And 38 total numbers that could come up: 18 red, 18 black, 0 and 00.
• 18 / 38 = 47.37%.

Over time, you would expect to win 47.4 percent of the time. But you’re being paid as if the probability was 50 percent. The difference between the two helps us determine the house advantage.

The odds for a game are typically listed somewhere close by. In craps, for example, they’re usually written right on the craps table.

For the lottery, odds and probability mean the same thing. They both refer to the mathematical likelihood of winning. For example, the odds of winning a 6/49 jackpot are 1 in 13,983,816. Expressed as a probability, you have a 0.00000715112384 percent chance of winning. In other words, it’s very unlikely.

If gambling is no longer a fun, affordable activity for you or someone close to you, it's time to take a closer look.

Probability Roulette Wheel Math

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Can probability theory save your life? Perhaps not in usual circumstances, but it sure would help if you found yourself playing a game of Russian roulette.

Today’s article covers a couple of variations on the Russian roulette puzzle, which is occasionally asked as an interview brain teaser.

Puzzle 1: Single bullet

I received this puzzle in an email:

Let’s play a game of Russian roulette. You are tied to your chair and can’t get up. Here’s a gun. Here’s the cylinder of the gun, six chambers, all empty. Now watch me as I put a single bullet in the gun. I close the cylinder and spin it. I put a gun to your head and pull the trigger. Click. Lucky you! Now I’m going to pull the trigger one more time. Which would you prefer, that I spin the cylinder first, or that I just pull the trigger?

(I’m not sure of the original source, but the wording is similar to a problem in William Poundstone’s book How Would You Move Mount Fuji?)

The problem can be solved by calculating the probability of survival for the choices.

First, consider the odds of survival if the cylinder is spun. The cylinder is equally likely to stop at any of the six chambers. One of the chambers contains the bullet and is unsafe. The other five chambers are empty and you would survive. Consequently, the probability of survival is 5/6, or about 83 percent.

Next, consider the odds if the cylinder is not spun. As the trigger was already pulled, there are five possible chambers remaining. Additionally, one of these chambers contains the bullet. That leaves four empty or safe chambers out of five. Thus the probability of survival is 4/5, or 80 percent.

Comparing the two options it is evident that you are slightly better off if the cylinder is spun.

But is this always the case? What if there are two bullets in the gun?

Puzzle 2: Two bullets

In the two bullet version, everything is the same as before except the gun is loaded with one more bullet:

Let’s play a game of Russian roulette. You are tied to your chair and can’t get up. Here’s a gun. Here’s the cylinder of the gun, six chambers, all empty. Now watch me as I put two bullets in the gun. I close the cylinder and spin it. I put a gun to your head and pull the trigger. Click. Lucky you! Now I’m going to pull the trigger one more time. Which would you prefer, that I spin the cylinder first, or that I just pull the trigger?

This problem is a bit trickier because there are different ways the two bullets can be loaded-either in adjacent or not adjacent chambers.

Suppose you were lucky and caught a glimpse of how the bullets were loaded. What should you do?

Suppose the bullets were loaded in adjacent chambers

This problem, similarly, can be solved by considering the odds of survival for the different choices.

First, consider the odds when the cylinder is spun. The cylinder is equally likely to stop at at any of the six chambers. Two of the chambers contain bullets and are unsafe. The other four chambers are empty and you would survive. Consequently, the probability of survival is 4/6, or about 67 percent.

Next, consider the odds if the cylinder is not spun. This situation is a bit trickier so careful accounting is helpful. We know the gun is loaded with two bullets so it has two loaded chambers (let’s label them 5 and 6) and four empty chambers (let’s label these 1, 2, 3, and 4). When the game starts, the trigger is pulled and nothing happens, so we know the cylinder started on an empty chamber. Then, the cylinder advances one chamber forward, meaning the cylinder must now be in one of the positions 2, 3, 4, or 5. If you don’t spin, you will survive in three of these four positions (all but 5). Therefore, not spinning the gun has a probability of survival of 3/4, or 75 percent.

It is now better not to spin, unlike the single bullet case.

American Roulette Probability

Now suppose the bullets are not in adjacent chambers

Probability Of Winning Roulette

We proceed as before comparing the two choices.

First, when the cylinder is spun the situation is the same as in the adjacent case. The cylinder is equally likely to stop in any of the six chambers, four of which are empty. Consequently, the probability of survival is 4/6, or about 67 percent.

Next, consider the odds if the cylinder is not spun. This situation again requires careful accounting. We know the gun is loaded with two bullets so it has two loaded chambers (let’s label them 4 and 6) and four empty chambers (let’s label these 1, 2, 3, and 5). When the game starts, the trigger is pulled and nothing happens, so we know the cylinder started on an empty chamber. Then, the cylinder advances one chamber forward, meaning the cylinder must now be in one of the positions 2, 3, 4, or 6. If you don’t spin, you will survive in two of these four cases (4 and 6 are loaded). Therefore, not spinning the gun has a probability of survival of 2/4, or 50 percent. Amazingly this is quite a bit lower than the odds if the bullets were adjacent!

In conclusion, the best choice in the non-adjacent case is to spin.

Possible extensions:

Russian Roulette Probability Calculator

After going through these puzzles, I was thinking about ways this game could be generalized.

What happens in the case of n bullets? It is easy to compute the probability of survival for spinning (=number of empty chambers / number of total chambers) but what about the probability for not spinning? If you don’t know exactly how the bullets are loaded, is it better to spin or not?

Also, these puzzles have been one-stage analyses. What happens if the game continues with each player alternating the shots, like normal Russian roulette? Does a one-stage advantage allow you to survive longer, or can your opponent use a counter strategy to even the odds?

Roulette Probability Statistics Problem

I have a few ideas and I hope to post my thoughts in the comments later. What are your thoughts?