A Roulette Challenge

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Simmy
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A Roulette Challenge

Postby Simmy » Mon Nov 26, 2007 2:59 am UTC

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?

GreedyAlgorithm
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Re: A Roulette Challenge

Postby GreedyAlgorithm » Mon Nov 26, 2007 4:10 am UTC

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
Spoiler:
bet on the first spin, no matter what.
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quintopia
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Re: A Roulette Challenge

Postby quintopia » Mon Nov 26, 2007 4:51 am UTC

are we looking at 37 numbers on the wheel (0-36) or 36 (0-35)?

And what can we assume about the other players? Are they perfect logicians? Or do they bet randomly?

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Beyondthewall
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Re: A Roulette Challenge

Postby Beyondthewall » Mon Nov 26, 2007 5:13 am UTC

It's 36 numbers total, unless I misread the rules.
And, as stated above, the highest chance is
Spoiler:
always the first spin, because it will always occur. The others become less likely to occur as it progresses but do not increase their own chances of getting a zero. So the second spin has only a 35/362 chance of winning, I think, whereas the first has 1/36. The nth spin has (1-1/36)n-1*(1/36) probability, again methinks.


Also, this assumes the players don't follow this logic. If they do, bet on the
Spoiler:
second spin
. 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.
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.

Simmy
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Re: A Roulette Challenge

Postby Simmy » Mon Nov 26, 2007 6:14 am UTC

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)

Buttons
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Re: A Roulette Challenge

Postby Buttons » Mon Nov 26, 2007 6:34 am UTC

Simmy wrote:"What is the most probable number of spins it will take to get a zero?"

Spoiler:
The answer to this question is always 1, assuming you're talking about independent events with the same probability. It's a simple fact about geometric distributions, and it requires absolutely no further information about the game. The payout could be 17 million-to-one, and there could be 2006 numbers. It still wouldn't change anything, because the probability of an event with probability p happening after exactly n steps is p*(1-p)^(n-1). This number always greatest when n = 1.


You could make this a more interesting question by talking about how to earn the most money.
Spoiler:
In that case, you might want to bet on less populated numbers so that your share of the pot (in the event that you do win) is greater. For instance, it's not that much more likely to hit zero on the first try than it is on the second, so if far more people were on 1 than on 2, it would be wiser to bet the latter.

Playing with perfect logicians, and assuming everyone must make their bets simultaneously, I guess you'd want to choose your bet randomly, weighted more towards the lower ones. The exact optimum mixed strategy is probably a messy function, though.

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Re: A Roulette Challenge

Postby Cosmologicon » Mon Nov 26, 2007 8:14 am UTC

Buttons wrote:
Spoiler:
Playing with perfect logicians, and assuming everyone must make their bets simultaneously, I guess you'd want to choose your bet randomly, weighted more towards the lower ones. The exact optimum mixed strategy is probably a messy function, though.

Spoiler:
It's not bad for an infinitely large number of players. I believe the optimum strategy would have the expected payout equal for all possible bets. That is, the probability pn of winning on spin n, times the winnings wn is a constant. We have wn = max(37, 1/fn), where fn is the fraction of players playing that bet. Normalization gives fn = pn, and the constant is 1.

That means that your probability of choosing any particular spin should be equal to its probability of winning, and your expected winnings are equal to your bet (i.e. it's a zero-sum game).

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quintopia
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Re: A Roulette Challenge

Postby quintopia » Mon Nov 26, 2007 8:25 am UTC

Spoiler:
Does that mean bet on turn n with prob 1/37 for turns 1..37?

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Cosmologicon
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Re: A Roulette Challenge

Postby Cosmologicon » Mon Nov 26, 2007 8:37 am UTC

Spoiler:
Not quite. As per Buttons's post, you should bet on spin n with probability (1/37)(36/37)n-1.

DragonWrangler
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Re: A Roulette Challenge

Postby DragonWrangler » Mon Nov 26, 2007 1:18 pm UTC

Can you bet on multiple spins? If so,
Spoiler:
bet a large amount of money on each.
Last edited by DragonWrangler on Wed Nov 28, 2007 2:00 pm UTC, edited 1 time in total.

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JamesCFraser
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Re: A Roulette Challenge

Postby JamesCFraser » Mon Nov 26, 2007 3:45 pm UTC

Spoiler:
The probability of a 0 occuring on any particular roll is 1/37. Therefore, the odds of not getting a 0 is 36/37 {1 - 1/37}.

Given that each event is independent, the probability of receiving a win for a bet on the nth spin is (1/37)*(36/37)^(n-1). Obviously, given that n is a member of the natural numbers, the highest probability we can get is achieved when n is 1. Given that we can only bet on one particular spin, if you want to bet on multiple occasions, it only ever makes sense to bet on the first spin. This means that we always have a probability of 1/37 of winning.

However, I am assuming that seeing as the minimum we can win is 37 times what we originally bet, and the maximum probability of winning just happens to be the reciprocal of this figure, then we receive a payout of our bet multiplied by the reciprocal of the probability we would win. This is pretty boring, as we have an expected value of winnings of exactly what we bet for any n.

This means that if we play the game, the expected amount to be won is the same as the amount we bet. So, you may as well not play the game. The safest strategy would probably to be to bet for the zero to be the first roll, as that will occur most frequently. This means you minimise the amount of time you have to stay at the table to get back to where you started from :D

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quintopia
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Re: A Roulette Challenge

Postby quintopia » Mon Nov 26, 2007 4:47 pm UTC

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

Postby GreedyAlgorithm » Tue Nov 27, 2007 1:29 am UTC

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. :D

Spoiler:
If the question is really what you stated then "safe strategy" means nothing. Yes, certainly, if you want to answer a different question like "how do I make the most money" or "how do I maximize the probability that I make money" (two phenomenally different questions) then all these other considerations like number of games, other players' bets, etc. might be important. Otherwise? Nope, nada, zilch. Not relevant at all.

If you still think they're relevant, ask the right question. "What is the most probable number of spins it will take to get a zero?" is not the same as either of my sample questions and is not the same as a whole slew of questions I currently believe you're really thinking about.

I mean come on, do you think the probability of where a fair wheel ends up will change if three more people bet on 7? (if trillions more people bet then gravity may be a factor...)

ETA: spoiler tags
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Re: A Roulette Challenge

Postby Simmy » Tue Nov 27, 2007 2:41 am UTC

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.

Spoiler:
Remember, the Jackpot is defined as being all players losing bets.

In the games where a zero turns up on a spin which nobody bet on, the House keeps everything. In the games where the Jackpot is more than 37 times the amount bet by winning players, the House keeps keeps the winning players' bet.

In the games where the Jackpot is less than 37 times the amount bet by winning players, the House makes up the difference and so pays less than 37 to 1. In the games where there is no jackpot, the House pays 37 to 1.

The maximum probability of zero occuring is 1/37 on the first spin. For subsequent spins the probability is progressively less.

Therefore, the worst case scenario for the Casino is if everyone always bet on the first spin. Played out over time, the House would break even.

In games where at least one bet is placed on any subsequent spin, the House will win over time. But the bulk of it's profit would come from games with a high turnout of bets spread out over a range of spins - where it is guaranteed to win

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quintopia
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Re: A Roulette Challenge

Postby quintopia » Tue Nov 27, 2007 3:15 am UTC

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, 1-25, with only 25 players, then the house stands to lose big.

Simmy
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Re: A Roulette Challenge

Postby Simmy » Tue Nov 27, 2007 4:52 am UTC

If players made sure that they bet 1 each, on, say, 1-25, 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 1-25 spins. A total of $2,500 has been bet.
Spoiler:
If a Zero occurs in 25 spins or less, one of the players must be paid $3,700. $2,500 is already available. The House must pay $1,200.

If a Zero occurs in 26 spins or more, no-one is paid. The House wins $2,500.

25 spins happens to be the Median benchmark for whether a Zero will occur. In other words, the probability that Zero will occur in less or more than 25 spins is about 50/50.

These are excellent odds for the House, which stands to win more than double what it stands to lose.
There is no example you can come with where the odds would go against the House.

Convinced?

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Macbi
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Re: A Roulette Challenge

Postby Macbi » Tue Nov 27, 2007 6:24 pm UTC

Beyondthewall wrote:It's 36 numbers total, unless I misread the rules.
And, as stated above, the highest chance is
Spoiler:
always the first spin, because it will always occur. The others become less likely to occur as it progresses but do not increase their own chances of getting a zero. So the second spin has only a 35/362 chance of winning, I think, whereas the first has 1/36. The nth spin has (1-1/36)n-1*(1/36) probability, again methinks.


Also, this assumes the players don't follow this logic. If they do, bet on the
Spoiler:
second spin
. 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.

Here's the problem, it's a cleverly disguised prisoner's dilema.
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quintopia
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Re: A Roulette Challenge

Postby quintopia » Tue Nov 27, 2007 8:36 pm UTC

Yes, but I don't understand intuitively. . .I'd have to see a proof.

Simmy
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Re: A Roulette Challenge

Postby Simmy » Tue Nov 27, 2007 10:15 pm UTC

It's 36 numbers total, unless I misread the rules.
You did.
1) It is a single zero roulette wheel with 36 regular numbers (no double zero).
That's a zero + 36 regular numbers = 37 numbers total.

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Re: A Roulette Challenge

Postby Simmy » Wed Nov 28, 2007 12:13 am UTC

There is no example you can come up with where the odds would go against the House.
I was wrong. My apologies.

There are betting scenarios where the House stands to lose.

For example:
Spoiler:
lets take those 25 syndicate players collaborating their bets between 1 and 25 spins.

Player 1 places $100 on the first spin. However, the remaining 24 players each place $1 on spins 2-25. $124 has been bet.

There is a 2.7% chance that Zero will occur on the first spin.
The House pays $3,576 to Player 1.

There is a 47.7% chance that Zero will occur between spin 2-25.
The winning player gets $123. The House has taken $1.

There is a 49.6% chance that Zero will occur above 25 spins.
The House takes $124.

Adding the probabilities, the Casino loses approximately one and a half times more money than it wins.

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Re: A Roulette Challenge

Postby no-genius » Wed Nov 28, 2007 12:34 pm UTC

Simmy wrote:For example:
Spoiler:
lets take those 25 syndicate players collaborating their bets between 1 and 25 spins.

Player 1 places $100 on the first spin. However, the remaining 24 players each place $1 on spins 2-25. $124 has been bet.

There is a 2.7% chance that Zero will occur on the first spin.
The House pays $3,576 to Player 1.

There is a 47.7% chance that Zero will occur between spin 2-25.
The winning player gets $123. The House has taken $1.

There is a 49.6% chance that Zero will occur above 25 spins.
The House takes $124.

Adding the probabilities, the Casino loses approximately one and a half times more money than it wins.

Spoiler:
Or run a syndicate with an infinite number of players :D
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Re: A Roulette Challenge

Postby quintopia » Wed Nov 28, 2007 4:37 pm UTC

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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