
The Roulette Conundrum
Here is a curious probability puzzle, the answer to which has been hotly disputed elsewhere.
Suppose there is an unbiased roulette wheel with the usual 36 regular numbers plus a single Zero. A Zero has just come up:
"How many nonzero spins are most likely to occur before you hit zero again?"
Simon
Last edited by Simon (20070129 14:00:07)
 Ricky
 Moderator
Re: The Roulette Conundrum
This seems to be a simple binomial probability. Or negative binomial, I always get the two confused.
In any case, at that point, its just a matter of plugging in the numbers. Unless I'm missing something...
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: The Roulette Conundrum
Ok that's cool!
And your answer is?
Re: The Roulette Conundrum
Since each spin is independent of all other spins, it doesn't matter that a 0 just came up. The odds of spinning a 0 are 1 in 37 no matter whether the last 33 spins were all 0 or if a 0 hasn't come up in the last 1000 spins (I believe you when you say the wheel was unbiased!). I think the answer would be the same if a 16 just came up or if you asked how many non32 spins come up before you get a 32.
Randomly walk up to a roulette table and count how many spins in takes to get a zero. Once it hits, walk to another table and do the same over and over again. It may hit on the first spin, it may hit on the 50th spin. I'm going to say on average that you would have to wait 18 or 19 spins (37/2).
Re: The Roulette Conundrum
Interesting observations Pi Man.
You said the average might be 18 or 19 nonzero spins before zero comes up.
I should point out immediately, the question was "how many nonzero spins are most likely to occur?" That is not the same as asking "on average, how many nonzero spins will occur?". The answer to both might or might not be the same. However, I submit that 18 or 19 is the wrong answer in terms of being either the average or the most likely number of spins.
Sorry!
Just to illustrate the difference: a practical test would be to do exactly as you suggested  sample as many wheels as possible. Suppose you tried 38 wheels, just as an example. On 37 wheels you found it took a different number of spins ranging from none to 36. On the 37th table you found it took 30 spins. That means 30 nonzero spins occured twice. The rest in the noneto36range occured once. Although the average was 18 spins, the indications so far suggest that 30 spins might occur more often than any other number of spins.
With a large enough sample  say 10,000 wheels  that is what you would be looking for. You would want the number of spins that occurs the most frequently.
In other words, in any representative sample, the most likely number of nonzero spins to occur is always the Mode rather than the Mean. But feel free to offer the Mean as well. There will probably be less debate about that.
Simon
Last edited by Simon (20070129 17:44:42)
 Ricky
 Moderator
Re: The Roulette Conundrum
A more interesting question than it appeared on inspection. I was right about the negative binomial giving the proper result, but I took a lot of convincing (though this was all done by myself) before I believed what I calculated.
My answer: 0 spins.
Now here's the trick. Anyone want to try to explain? Or shall I?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: The Roulette Conundrum
That is my answer too!
If you have to bet on how many spins it'll take before you get Zero, bet on none. The most probable of all outcomes is that a Zero will occur immediately.
It'll be fun to see how many protests we get before one of us justifies this statement.
Last edited by Simon (20070129 19:26:20)
Re: The Roulette Conundrum
When I first read that question, I thought of 0 before scrolling down. It shouldn't be too hard to prove, although convincing people might be harder.
Basically, on the first roll, the probability is 1/37. Then on the second roll, the probability is 1/37*36/37, because you need to not only get a 0 the second time, but you also need to have not got a 0 the first time.
On the third roll, the probability will be 36/37*36/37*1/37, and so on. The probability will be less for each subsequent roll, because each time you are multiplying the probability by 36/37, and so it decreases. Therefore, it's most likely to happen on the first roll.
Still not likely at all, but more likely than any other option.
Why did the vector cross the road? It wanted to be normal.
 MathsIsFun
 Administrator
Re: The Roulette Conundrum
On first reading the question i went for half of 37 because I interpreted it as pi man did. I though you would then say "but if a zero is likely in half of 37 spins then it should come up twice in 37 spins"
However, for the single likeliest number of spins I am convinced by mathsy's argument of decreasing likelihood.
I like the puzzle, simon!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: The Roulette Conundrum
Thanks.
Incidentally, Pi Man also said that the average should be 18 or 19 spins. Not surprisingly, the average is 36. Most people conclude this must be the most likely number of nonzero spins to occur. This is of course the Mean.
Others find themselves tempted by the Median, which is 25 spins. They calculate, correctly, that this is the 50/50 benchmark. If you took a large sample of roulette tables, on half the wheels it will take more than 25 spins to hit Zero. (On the other half it will take less.) This makes "none" sound like an impossible answer.
And yet "none" is the Mode. If probabilty were played out and narrowed down to a single most likely number of spins, you would hit Zero straight away.
If there were 100,000 roulette wheels, and millions of people paid for lottery tickets in which they guessed the six most recurring numbers of spins it took before getting Zero, those few who chose 0,1,2,3,4, and 5 in that order would almost certainly win and share the jackpot!
Last edited by Simon (20070129 22:45:43)
 Ricky
 Moderator
Re: The Roulette Conundrum
But, it's time for application to turn into theory.
Let p be the probability of success, q = 1  p be the probability of failure
The probability for the rth success to occur *exactly* on the nth trial is known as the negative binomial distribution:
This example (the roulette) is a special case where r = 1, that is, you only allow one "success": Which certainly agrees with the function mathsyperson came up with. But here is the kicker. Any question even remotely like this will have the answer of 0. As long as you are only looking for one "success" (the ball landing on 0), the highest likely candidate will be on the 0 because p and q must be less than one, and thus this function is always decreasing (called monotonely decreasing). But, what if r does not equal 1? Can we find a negative binomial function which is decreasing when r > 1?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
