A Roulette Challenge
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A Roulette Challenge
Imagine you're in a casino that is holding a special one off promotion at Roulette.
Here are the rules of the game:
1) It is a single zero roulette wheel with 36 regular numbers (no double zero).
2) The croupier is going to spin the wheel until a zero occurs.
3) Before this starts, each player submits a bet for the croupier to place on zero at a specific time. Each bet must be accompanied by a slip with a number on it. This indicates the exact spin that the player wants to place the bet.
4) Submitted in advance, all bets are final and cannot be cancelled, changed or added once the croupier starts spinning.
5) When a zero comes up the game is over. Those who ordered their bets to be placed for that spin are paid a respective share of the Jackpot, proportional to their bet.
6) The Jackpot consists of all money bet by losing players, which is occasionally vast.
7) The House guarantees that every winning player is paid a minimum of 37 times their original stake (not 36). If the Jackpot can't cover that, the bank always will.
Assume it is an unbiased wheel and that you will have the opportunity to play repeated games. And remember, the House guarantees a minimum payout of of 37 to 1.
Using the safest strategy based on probability, can you decide in advance exactly which spin your Zero bet should be placed to maximise your chance of winning?
Or is it equally likely when Zero will occur?
Here are the rules of the game:
1) It is a single zero roulette wheel with 36 regular numbers (no double zero).
2) The croupier is going to spin the wheel until a zero occurs.
3) Before this starts, each player submits a bet for the croupier to place on zero at a specific time. Each bet must be accompanied by a slip with a number on it. This indicates the exact spin that the player wants to place the bet.
4) Submitted in advance, all bets are final and cannot be cancelled, changed or added once the croupier starts spinning.
5) When a zero comes up the game is over. Those who ordered their bets to be placed for that spin are paid a respective share of the Jackpot, proportional to their bet.
6) The Jackpot consists of all money bet by losing players, which is occasionally vast.
7) The House guarantees that every winning player is paid a minimum of 37 times their original stake (not 36). If the Jackpot can't cover that, the bank always will.
Assume it is an unbiased wheel and that you will have the opportunity to play repeated games. And remember, the House guarantees a minimum payout of of 37 to 1.
Using the safest strategy based on probability, can you decide in advance exactly which spin your Zero bet should be placed to maximise your chance of winning?
Or is it equally likely when Zero will occur?

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Re: A Roulette Challenge
Simmy wrote:you will have the opportunity to play repeated games
Using the safest strategy
maximise your chance of winning?
It sounds like you're confused about something, whether it be the problem itself or communicating the problem to us. What do you want to know here? How to maximize your probability of winning? If so the repeated games don't matter and the phrase "safest strategy" is meaningless. Also the answer in that case is
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Re: A Roulette Challenge
are we looking at 37 numbers on the wheel (036) or 36 (035)?
And what can we assume about the other players? Are they perfect logicians? Or do they bet randomly?
And what can we assume about the other players? Are they perfect logicians? Or do they bet randomly?
 Beyondthewall
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Re: A Roulette Challenge
It's 36 numbers total, unless I misread the rules.
And, as stated above, the highest chance is
Also, this assumes the players don't follow this logic. If they do, bet on the. It has only slightly lower probability and you no longer have to split the pot. Depending on how many other players there are. Actually, this only applies if you ignore rule 7. To determine the other players' logic, just watch a round or two.
And, as stated above, the highest chance is
Spoiler:
Also, this assumes the players don't follow this logic. If they do, bet on the
Spoiler:
I have yet to see any problem, however complicated, which, when you looked at it in the right way, did not become still more complicated.
Re: A Roulette Challenge
I'm not too confused Greedy. The question essentially is: "What is the most probable number of spins it will take to get a zero?"
You've given an answer. I assume you're standing by it.
I acknowledge that repeated games are not required to state the actual probability. And perhaps I should have reduced the problem to those bare essentials.
'Safe strategy' does mean something, however. If you want to actually stand a better chance of breaking even or winning, more than one game is required. Whatever the correct answer, it is more realistic you would break even or win if you played 20 games than if you played one.
I haven't given any information about the pattern of other players bets. Such information could affect the answer. In the absence of any, an even distribution is to be assumed. All you know is that the payout is at least 37 to 1, perhaps more.
And there are 37 numbers on the wheel, including the Zero. (I'm going by a British Roulette wheel)
You've given an answer. I assume you're standing by it.
I acknowledge that repeated games are not required to state the actual probability. And perhaps I should have reduced the problem to those bare essentials.
'Safe strategy' does mean something, however. If you want to actually stand a better chance of breaking even or winning, more than one game is required. Whatever the correct answer, it is more realistic you would break even or win if you played 20 games than if you played one.
I haven't given any information about the pattern of other players bets. Such information could affect the answer. In the absence of any, an even distribution is to be assumed. All you know is that the payout is at least 37 to 1, perhaps more.
And there are 37 numbers on the wheel, including the Zero. (I'm going by a British Roulette wheel)
Re: A Roulette Challenge
Simmy wrote:"What is the most probable number of spins it will take to get a zero?"
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You could make this a more interesting question by talking about how to earn the most money.
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 Cosmologicon
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Re: A Roulette Challenge
Buttons wrote:Spoiler:
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Re: A Roulette Challenge
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 Cosmologicon
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Re: A Roulette Challenge
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Re: A Roulette Challenge
Can you bet on multiple spins? If so,
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Last edited by DragonWrangler on Wed Nov 28, 2007 2:00 pm UTC, edited 1 time in total.
Re: A Roulette Challenge
The real question is why a casino would run this game, considering they cannot possibly make any money on it and stand to lose quite a bit if players bet low or not many players.

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Re: A Roulette Challenge
Simmy wrote:I'm not too confused Greedy. The question essentially is: "What is the most probable number of spins it will take to get a zero?"
You've given an answer. I assume you're standing by it.
I acknowledge that repeated games are not required to state the actual probability. And perhaps I should have reduced the problem to those bare essentials.
'Safe strategy' does mean something, however. If you want to actually stand a better chance of breaking even or winning, more than one game is required. Whatever the correct answer, it is more realistic you would break even or win if you played 20 games than if you played one.
I haven't given any information about the pattern of other players bets. Such information could affect the answer. In the absence of any, an even distribution is to be assumed. All you know is that the payout is at least 37 to 1, perhaps more.
And there are 37 numbers on the wheel, including the Zero. (I'm going by a British Roulette wheel)
You've confirmed that you are confused.
Spoiler:
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Re: A Roulette Challenge
The real question is why a casino would run this game, considering they cannot possibly make any money on it and stand to lose quite a bit if players bet low or not many players.
This is not true. With a high turnout evenly distributed, the House always wins. With a low turnout, the House stands to win over time or break even.
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Re: A Roulette Challenge
Excuse me, the house can win if they automatically bet on bets that no one makes. However, it wasn't said that players couldn't collaborate. If players made sure that they bet 1 each, on, say, 125, with only 25 players, then the house stands to lose big.
Re: A Roulette Challenge
If players made sure that they bet 1 each, on, say, 125, with only 25 players, then the house stands to lose big.
Ok lets examine what happens if 25 players each placed $100 on zero coming up between 125 spins. A total of $2,500 has been bet.
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Convinced?
Re: A Roulette Challenge
Beyondthewall wrote:It's 36 numbers total, unless I misread the rules.
And, as stated above, the highest chance isSpoiler:
Also, this assumes the players don't follow this logic. If they do, bet on the. It has only slightly lower probability and you no longer have to split the pot. Depending on how many other players there are. Actually, this only applies if you ignore rule 7. To determine the other players' logic, just watch a round or two.Spoiler:
Here's the problem, it's a cleverly disguised prisoner's dilema.
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Re: A Roulette Challenge
Yes, but I don't understand intuitively. . .I'd have to see a proof.
Re: A Roulette Challenge
You did.It's 36 numbers total, unless I misread the rules.
That's a zero + 36 regular numbers = 37 numbers total.1) It is a single zero roulette wheel with 36 regular numbers (no double zero).
Re: A Roulette Challenge
I was wrong. My apologies.There is no example you can come up with where the odds would go against the House.
There are betting scenarios where the House stands to lose.
For example:
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Re: A Roulette Challenge
Simmy wrote:For example:Spoiler:
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Re: A Roulette Challenge
Simmy wrote:There are betting scenarios where the House stands to lose.
I spent quite a while trying to come up with this example, but couldn't, which is why I asked for a proof. Good to know my original intuition (that collaborating players could beat the house) was correct. So this leads to a very interesting question: Give a strategy (in terms of n players where the player betting on turn 1 bets p) which will cost the house the most, assuming that each player must bet at least 1 penny and they don't want to bet more than $100 apiece total (they split expenses).
There are definitely better cases than the one you gave: for instance, player 1 bets 123.76 and 24 other players bet .01. Then the house generally pays almost 3 times what it earns.
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