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LostInParadise's avatar

How do you explain the math solution from an intuitive point of view?

Asked by LostInParadise (23480points) May 28th, 2012

I wish I could remember where I heard this problem. I worked out the solution, but I still can’t explain at an intuitive level why it should work out the way that it does. Here is the problem.

You are at a casino playing the roulette wheel. Someone comes over to you and complains about the house having the advantage. He suggests that you make a side bet with him and keep the money between the two of you. He proposes that you each choose a permutation of length 3 of the colors red and black and the winner will be the one whose choice comes up first on consecutive spins of the roulette wheel. To make things fair, you will be allowed to choose first.

I trust that if this happened to you, you would sense, apart from any mathematical odds, that you were being hustled. Why is this complete stranger making this offer? You would be absolutely right, but on the face of it, if seems reasonable. If there were eight of you, each choosing one of the possible orderings of 3 choices of red and black, one of you would win in 3 spins with equal likelihood.

Consider the most extreme case. Suppose you had chosen red, red, red. What would the other person have chosen? If you do not win in the first 3 spins (ignoring 0 and 00), you are guaranteed to eventually lose. Then I would appreciate it if you could explain to me in simple terms why, for every choice, there is another that is more likely to occur first.

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10 Answers

whitenoise's avatar

I think that you cannot proof any particular permutation to be better than another.

Why do you say “If you do not win in the first 3 spins (ignoring 0 and 00), you are guaranteed to eventually lose.”?

(You didn’t state that the choices need to be different, but I think that is implied.)

Bill1939's avatar

I cannot follow the ‘roulette wheel’ scenario, but then I am math challenged. However, I believe that scientists, such as Einstein, have sometimes first intuited a notion before seeking the mathematics to support it.

PhiNotPi's avatar

I believe that in your example the other person would choose black red red. Here’s how I think about it:

If the combination of red red red ever shows up in the string of red and black spins, then you know that the combination of ???? red red must have shown up directly before it. If the first player (the one being scammed) is to win, this sequence must also be red red red. If it were black red red then the scammer would have won before before the other person. This circle of logic repeats itself until you reach the very first sequence, win no spins coming before it. By the same logic, if the first player were to ever win, he must win on the first three spins.

LostInParadise's avatar

@PhiNotPi , Nicely done. The nearest I can figure in the general case is that the second person always chooses a combination whose last two colors match the first two colors of the first person. In a sense, you “cut off” the other person’s chances of winning.

CWOTUS's avatar

Assuming that we can ignore the non-red and non-black results; that is, pretend those results never even occurred, and assuming that the order of the results doesn’t matter (in cases of mixed black-and-red endings), then you have a 1 in 8 chance of all-red or all-black results, and a 3 in 8 chance of two-red and one-black or two-black and one-red.

So obviously choosing any three-of-a-kind strategy is a total loser, right off the bat. No one who thinks about it for more than a few seconds would make that bet.

So you, being smart, pick one of the choices that gets you a 3/8 chance of winning the bet.

If there are just two of you in the room, then your chances of winning are exactly equal to his chances; it’s a fair gamble. But there aren’t just two people in the room, are there? He’s playing the same game against any number of other players. He’ll give them whatever choice they want, and chances are that half of the players will take the two-reds-and-a-black bet, and the other half will take the two-blacks-and-a-red bet.

He can take the three-reds or three-blacks bet. This way, he always has the benefit of a 5/8 bet against (most of) the rest of the room.

PhiNotPi's avatar

(In my above answer “win no spins coming” -> “with no spins coming”)

If the first person chose red black red, then I think that the second person should choose either red red black or black red black. He should go with red red black because otherwise the two sequences would be direct opposites, so there would not be an advantage.

In general, the second person should choose a sequence such that the last two colors match the first two colors of the first sequence, but so that the last two colors of the first sequence do not match the first two colors of his sequence.

Here’s a more general method for how the problem will be solved:

In order for the first player to win, then somewhere in the chain of spins the shared sequence must pop up for the first time, in this case red black. Since the second player chose red red black, then there is already a fifty-fifty chance that the game would end immediately at this point with the second player winning. In the 50% chance that this is not the case, then there would be a second fifty-fifty chance of the second player winning. If neither of those are the case, then there would be percentage chance that that whole mini-sequence would have to be ignored and the players must wait for the next time the sequence red black is seen again.

Since there is at least a 50% of the second player winning, and a nonzero chance that player two gets a second try, then he must have an advantage.

LostInParadise's avatar

@CWOTUS , Sorry if I did not make clear that the order does matter. That is what makes the solution seem at first to be so non-intuitive.

It is possible with just a little work to calculate the probabilities. Here is the transition state diagram for red,black red, (RBR) played against RRB. I simplified things by making R the initial state, since the sequences of both players stat with red. You can think of the R state as representing zero or more blacks in a row followed by a red. There is a formal analysis that can be done using Markov chain matrices. I have forgotten most of the little I knew of these, but there is a much simpler analysis.

Here is how to calculate the probability of winning. Each transition is a probability of one half. If you take the path from R to RB to RBR, that is two links with a winning probability of ½ * ½ = ¼. If you take the path from R to and RB back to R, the chance of that leading to a win has probability of ¼ times the original probability. If w is the probability of winning, we get w = ¼ + ¼ w . That works out to a winning probability of ⅓.

CWOTUS's avatar

Okay, then the combination / sequences are:

Black – Black – Black
Black – Black – Red
Black – Red – Black
Black – Red – Red
Red – Black – Black
Red – Black – Red
Red – Red – Black
Red – Red – Red

Given a fair wheel and ignoring non-red and non-black outcomes, there is an exactly 1/8 chance of each of those sequences: ½ * ½ * ½.

If you make a bet, you have a 1/8 chance of winning on the first three spins. Your opponent has the same 1/8 chance of winning on the first three spins. That’s a fair bet. If your opponent makes any other bet he has a 50–50 chance of winning against you if you both wait until one of your two sequences hits. That’s still a fair bet. But if he’s also betting against many others, each of whom has a 1/8 chance of winning on the first three spins, then… then I’m not so sure. It seems to me that he has an advantage playing against a group of players, but I haven’t worked it out yet.

PhiNotPi's avatar

@CWOTUS The bets won’t always be fair- If player one chooses RRR and player two chooses BRR, then player two has a massive 7/8 chance of winning. Test this if you want by flipping a coin and doing an experiment.

mattbrowne's avatar

It’s not always possible. Take the birthday paradox as an example. Intuition can fool people. I’m not sure whether this is the case for your roulette wheel situation.

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