The Gambler’s Puzzle
The following is a quotation from the short novel The Gambler by Fyodor Dostoyevsky:
I was a gambler myself; I realized it at that moment. My arms and legs were trembling and my head throbbed. It was, of course, a rare happening for zero to come up three times out of some ten or so; but there was nothing particularly astonishing about it. I had myself seen zero turn up three times running two days before, and on that occasion one of the players, zealously recording all the coups on a piece of paper, had remarked aloud that no earlier than the previous day that same zero had come out exactly once in twenty four hours.
The probability of obtaining a zero on a (fair) Roulette wheel of the European variety is 1/37. Assuming that such a wheel is spun exactly 370 times in a day, determine the probability of obtaining at most one zero in twenty four hours as described in the quotation. Give your answer to three significant figures.
Answers through the comments box please!
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In the Dark 
December 17, 2013 at 12:35 pm
(36/37)³⁷⁰ + 370×(1/37)×(36/37)³⁶⁹ = 0·000446
December 18, 2013 at 9:45 am
Yes, that’s right. It’s more fun to do it without a calculator – which you can do to 3SF with a couple if nifty approximations…
December 17, 2013 at 12:37 pm
0.000446
December 17, 2013 at 1:37 pm
What did Dostoevsky mean by “that same zero”? Did he mean that same sequence of three zeros? Three consecutive zeros will occur once every 137 days. Eddington would have been pleased to see the fine structure constant emerge from numerology. And perhaps that means that God does play dice (copyright A Einstein), or at least roulette.
December 17, 2013 at 2:17 pm
I think he meant the same slot on the same wheel..
December 17, 2013 at 2:33 pm
I think this is more fun as an order-of-magnitude question (to be answered without a calculator or computer, of course).
December 17, 2013 at 3:44 pm
If you have a whole day to observe 370 spins with at most one zero in the last 24 hours, don’t you have 740 spins to observe 369 consecutive non-zero results on the roulette wheel?
I scribbled down the recursion relations and realised you can probably approximate one of the terms as small pretty quickly, but didn’t go through with the whole calculation, as I’m not sure this is the question Peter wants answering…
December 17, 2013 at 3:45 pm
I found the same answer as Paul, in the same way, but then thought about using the Poisson distribution. The zero coming up is a random occurrence, and we expect 10 zeros in our “experiment” (1/37 * 370). The probability of observing 0 or 1 zeros is then
exp(-10) + 10*exp(-10) = 0.000499
Why is this not the same as the first result? I suspect I am missing something trivial…
December 17, 2013 at 4:16 pm
The Poisson distribution is only an approximation to the Binomial distribution. It corresponds to n -> infty and p -> 0 in such a way that np -> lambda. Here n is large and p is small but the error is finite.
December 17, 2013 at 6:14 pm
Yes, that makes perfect sense – thanks.
December 17, 2013 at 9:42 pm
Doesn’t the answer require a frequentist approach?
Is that allowed here?
December 17, 2013 at 10:02 pm
No, it doesn’t and no it isn’t.
December 18, 2013 at 12:38 pm
I’d have been interested in Bryn’s attempt.
December 18, 2013 at 2:02 pm
Anton, I haven’t tried it yet, either from a frequentist or Bayesian perspective!