The Roulette Conundrum

[Farsight]

I don't think it's a contradiction Simon. You're forgetting about the intervening zeros. To back up what you were saying earlier:

Suppose I'm betting when the next zero will come up, and I'm wondering whether it'll come up on the first spin or the second.

Like you said there's a 1/37th chance it'll come up on the first spin. Which means the chances of it NOT coming up on the first spin are 36/37th.

The chances of it coming up on the second spin are 1/37th, but the chances of it NOT coming up on the first spin AND coming up on the second spin are 36/37th * 1/37th.

Which means I'm better off betting on the first spin, regardless of what came up last time. If I'm foolish enough to bet it'll come up next on the 37th spin the odds are 36/37th^36 * 1/37. Which is probably your 1/99.

[Qfwfq]

Originally Posted by Simon

It seems clear that for each bet, the chance of a zero is the same for both players - 1/37.

Nevetheless, the following propositon is also true.

The probability of spinning a zero immediately is 1/37
The probabilty of spinning exactly 36 consecutive non-zero spins followed by a zero is about 1/99.

This appears to be a contradiction.

But is it?

It is not a contradiction at all!

The probability of the next spin giving a zero is indeed 1/37 but this is a different thing than "the probabilty of spinning exactly 36 consecutive non-zero spins followed by a zero", which does have a lower probability, slightly less than 1/99.

The difference between the two is that your player 2 is really betting on a single spin, the fact that he waits for 36 in a row to be non-zero has no bearing on the probability of the following result. If you're willing to bet against a result that is already known to have occurred, let me know and I'll finally get rich!

The formula that Ron gave is true, based on the fact that the results of each run are independent of each other, and is a calculation of the probability that, in a set of n + 1 specified spins, only a specified one of them will be zero. That isn't what player 2 is achieving if he waits until 36 in a row have given no zero. If he only bets at that point, the house will never grant him a ratio greater than 37 (indeed they would grant the customary 36).

[Qfwfq]

Note:

The computation is valid even if the word "specified" is understood to mean as you go along, providing the spin is specified before the result, along with either zero or not zero. The player could agree with the house that, before one spin, he can say "zero this time" and the opposite before each of n other spins, at his choice. Neither order nor even consecutivity make any difference to the computation of probability, as long as the player specifies spins before the result is predictable. They could even agree to the player going from wheel to wheel or coming back another day, no difference.

[Simon]

Agree with all of the above.

Incidentally, I did have "none" in mind when I first created the problem. Ron was the first person to choose that answer without prior influence. Others needed some serious persuasion. Well done Ron!

There's another twist to the two player scenario.

We're agreed that if probability is played out, the casino will win from both players. The House pays only 36-to-1 on a 1/37 bet.

The question is which player is likely to lose more?

Both players will be betting £1 on zero.

Player 1 will only bet after a zero is spun.
Player 2 will only bet after a zero is spun + at least 36 other numbers.

The chance of getting Player 1's conditions is 1/37
The chance of getting Player 2's conditions is 1/99

Player 1 will bet £1 almost three times more often than Player 2.

So Player 1 is likely to lose nearly three times as much as Player 2.

On the other hand, if the House paid in the player's favour (e.g 38-to-1), Player 1 will win nearly three times more than Player 2.

This illustrates why there is no contradiction. Win or lose, Player 1 is likely to get more betting opportunities than Player 2. This is simply because an unconditional zero will occur nearly three times more than a zero + at least 36 other numbers.

By the same ratio, two consecutive zeros is nearly three times more likely than than two zeros separated by exactly 36 other numbers.

This is where we came in.

Simon

[LaurieAG]

Originally Posted by Simon

By the same ratio, two consecutive zeros is nearly three times more likely than than two zeros separated by exactly 36 other numbers.

And three consecutive zeros is much more likely than three zeros separated by exactly 36 other numbers etc.

[Simon]

Originally Posted by LaurieAG

And three consecutive zeros is much more likely than three zeros separated by exactly 36 other numbers etc.

Actually, I'm not sure how you define three zeros separated by exactly 36 other numbers. Do you mean exactly 36 other numbers on both sides of the middle zero? Or exactly 36 other numbers between the first and third zero, with the middle zero occuring somewhere between?