Roulette Problem: Solving Z Distribution & Win/Loss Probabilities

• dthmnstr
In summary, the player bets $100 on a single number in European roulette, with a 1/37 chance of winning. They continue to play until they win once, at which point they stop playing. The distribution of the number of times they play, Z, is dependent on the probability of losing on the first n-1 plays and winning on the nth play. The probability of winning at least$1000 is impossible since the player quits upon their first win. The probability of losing $1000 is dependent on the player losing 10 times in a row, with a probability of approximately 0.76034. However, this scenario is highly unlikely in the real world since the actual payoff for winning in roulette is 35 to dthmnstr Hey guys long time since I've posted on here, got this random question a friend of mine and myself were wondering over. 1. Homework Statement A player bets 100 dollars continuously on any number besides 0 (european roulette so 1/37 chance to win). He plays until he wins once and stops playing. a) Z is the amount of times he plays. What is the distribution of Z? b) What is the probability of the player winning at least 1000 dollars and the probability of the player losing 1000 dollars?2. The attempt at a solution I'm not so sure as how to go about this as i don't have any background in statistics and am currently studying calculus. But statistics is on my curriculum after the summer so it would be nice to have a little heads up on what's to come. dthmnstr said: Hey guys long time since I've posted on here, got this random question a friend of mine and myself were wondering over. 1. Homework Statement A player bets 100 dollars continuously on any number besides 0 (european roulette so 1/37 chance to win). He plays until he wins once and stops playing. a) Z is the amount of times he plays. What is the distribution of Z? b) What is the probability of the player winning at least 1000 dollars and the probability of the player losing 1000 dollars?2. The attempt at a solution I'm not so sure as how to go about this as i don't have any background in statistics and am currently studying calculus. But statistics is on my curriculum after the summer so it would be nice to have a little heads up on what's to come. Start by figuring out the probability that he loses on the first ##n-1## plays and wins on the ##n##th play. That will give you the distribution. dthmnstr said: Hey guys long time since I've posted on here, got this random question a friend of mine and myself were wondering over. 1. Homework Statement A player bets 100 dollars continuously on any number besides 0 (european roulette so 1/37 chance to win). He plays until he wins once and stops playing. a) Z is the amount of times he plays. What is the distribution of Z? b) What is the probability of the player winning at least 1000 dollars and the probability of the player losing 1000 dollars?2. The attempt at a solution I'm not so sure as how to go about this as i don't have any background in statistics and am currently studying calculus. But statistics is on my curriculum after the summer so it would be nice to have a little heads up on what's to come. You need to tell us what is the payoff if the player wins; otherwise we cannot possibly tell if he/she can even win$1000 at all. Does this $1000 win mean the player quits and goes home$1000 richer than when he/she started (which could happen if he/she wins in game 2 and receives a $1200 payoff). Or, does it just mean that the payoff upon winning is$1000?

The player pays 100 for a single spin and if he wins makes 200 back, so a 100 dollar profit, but each spin costs 100. and this is for any number besides zero so he has 1/37 gone already for the zero and another 1/37 probability for any other number.

S
dthmnstr said:
The player pays 100 for a single spin and if he wins makes 200 back, so a 100 dollar profit, but each spin costs 100. and this is for any number besides zero so he has 1/37 gone already for the zero and another 1/37 probability for any other number.

So, ever winning $1000 is impossible, since he quits upon his first win. Aha that's true for b), didnt see that...as for losing 1000, he has to lose 10 times in a row before quitting, how likely is losing 10 times in a row? dthmnstr said: Aha that's true for b), didnt see that...as for losing 1000, he has to lose 10 times in a row before quitting, how likely is losing 10 times in a row? Respond to post #2 and you will have your answer. I suppose you realize that, in the real world, your "roulette" game is completely absurd. You bet$100 with only 1/37 probability of winning, and in the unlikely event you are lucky, you just get $100 back? So you lose$100 virtually every time you spin and win $100 once every blue moon. Some game. dthmnstr said: Aha that's true for b), didnt see that...as for losing 1000, he has to lose 10 times in a row before quitting, how likely is losing 10 times in a row? Your description does not match the actual way roulette is played. See, eg, http://en.wikipedia.org/wiki/Roulette . In particluar, the section entitled "Bet Odds Table" states that the payoff is ##\frac{1}{n}(36-n) = (36/n) - 1##, where ##n## is the number of squares the player is betting on. In your description you say the player bets on ONE square, so the payoff is 35 to 1; that is, they would receive a payoff of$3500 for their $100 investment, in the lucky event that they win. (Just to be clear, the$3500 is a net gain, so they paid $100 and got back$3600.) Betting on more than 1 square would be needed if they wanted to increase the probability of winning while lowering the possible payoff.

BTW: If we assume they bet on single squares each time, the probability of losing 10 times in a row would be ##(36/37)^{10} \doteq 0.76034##.

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1. What is the Roulette Problem and why is it relevant to Z Distribution?

The Roulette Problem is a mathematical problem that involves determining the probability of winning or losing in a game of roulette. It is relevant to Z Distribution because the outcomes of a game of roulette can be represented by a normal distribution, also known as the Z distribution.

2. How do you calculate the Z distribution for the Roulette Problem?

To calculate the Z distribution for the Roulette Problem, you need to know the mean and standard deviation of the game's outcomes. The mean can be calculated by adding all the possible outcomes and dividing by the number of outcomes. The standard deviation can be calculated using a formula involving the mean and the variance of the outcomes. Once you have these values, you can use a Z table or a statistical software to determine the probabilities of winning and losing.

3. What is the probability of winning or losing in a game of roulette?

The probability of winning or losing in a game of roulette depends on the type of bet you place. If you place a bet on a single number, the probability of winning is 1/38 or approximately 2.63%. If you place a bet on red or black, the probability of winning is 18/38 or approximately 47.37%. The probability of losing is the complement of the probability of winning.

4. How can the Z distribution be used to improve your chances of winning in roulette?

The Z distribution can be used to improve your chances of winning in roulette by helping you make informed betting decisions. By understanding the probabilities of winning and losing for different types of bets, you can choose to place bets with higher probabilities of winning. However, it is important to remember that roulette is a game of chance and the outcomes are ultimately unpredictable.

5. Are there any other factors that can affect the Z distribution in the Roulette Problem?

Yes, there are other factors that can affect the Z distribution in the Roulette Problem. These include the number of players at the table, the speed of the roulette wheel, and any biases in the wheel or table. These factors can potentially alter the expected probabilities and outcomes of the game, making it more difficult to accurately calculate the Z distribution.

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