Re: Craps question
What is the probability of three boxcars in sucession in a series of 100 rolls?
First I'll give a simple solution and then show that it is really only an approximation in general, but one that works well in this case. Then I'll derive a difference equation for the situation in a method analogous to the one that generated the Fibonacci series for coin flips.
The probability of getting 3 boxcars in a row is 1/46656 per try, but how many tries are there in 100 throws? The average length of a try can be 1, 2 or 3 throws. 1 throw is far more common than anything else, so the average length must be close to 1. It will be exactly 1 long with probability 35/36. It will be 2 long with probability (1/36)*(35/36), and it will be 3 long with probability 1/46656. So the average length of a try is 35/36 + 2(1/36)(35/36) + 3(1/46656) = 1.0262989 throws. So in 100 throws we will average 100/1.0262989 = 97.437504 tries. The probability that we make boxcars 3 times in a row is 1 - (46655/46656)^97.437504 = 0.2% or 2 in 1000.
Now in general the above method is an approximation because it is not generally true that the average number of tries will yield the correct probability. In this case, since almost all the tries are of length 1, the approximation should be good. As a proof that this is an approximation, refer back to the coin flip problem. The probability of getting 2 tails in a row by the 10th flip is the sum of the probabilities of getting it on each flip from 2-10, which we derived is sum[n=2 to 10][F(n-1)/2^n] where F(n) is the nth Fibonacci number. This is equal to 85.9%. If we use the simple method above, we would say that the average length of a try is 1/2 + 2(1/2) = 1.5. So in 10 flips we would have an average of 10/1.5 = 6.67 tries so the probability of 2 in a row would be 1 - (.75)^6.67 = 85.3%, close to the exact answer of 85.9%, but different nonetheless. Perhaps it would be more accurate in this case to say that since we have an average length of 1.5, that we will have a length of 1 half the time and a length of 2 half the time, so we will have 5 tries half the time and 9 tries half the time, since the last try must begin on the 9th flip. Then we would have an average of 7 tries, and 1 - (.75)^7 = 86.7%. Still different from the exact result in the other direction this time.
Now let's derive the difference equation which will yield the exact probability for the dice case. Similar to the coin problem, let b = number of continuing sequences that end in boxcars, k = number of continuing sequences that do not end in boxcars, c = number of continuing sequences, and t = number of terminating sequences. k is equal to 35 times the total number of sequences ending in boxcars (b+t) so we have:
b+k=c
35(b+t)=k
So b = (c-35t)/36
If o = total outcomes, c = o-t so
b = (o-36t)/36
Since t(n) = b(n-1), and o(n) = 36c(n-1), we have the system:
t(n) = [o(n-1)-36t(n-1)]/36
o(n) = 36[o(n-1)-t(n-1)]
This is similar in form to the system for the coin flip case except that 2 is replaced by 36. This comes from the fact that we now have 36 outcomes at each stage instead of 2. With a little algebra this reduces to:
t(n) = 35[t(n-1) + t(n-2)]
t(3) = 1
Except for the factor of 35, this resembles Fibonacci. Note that it is not 35 times Fibonacci, it is very different. Each number is the sum of the preceding two times 35.
probability = sum[n=3 to 100][t(n)/36^n]
Classical renwal methods compute this probability as 1 - t(101) - t(102) - ... where the series can be evaluated to the desired degree of approximation by Laplace transform and partial fraction expansion methods. A classical text which includes this method, Feller, An Introduction to Probability Theory and It's Applications, does not mention that the Fibonacci series comes up in the solution to the problem with p = .5, nor does it take the approach above. This makes me think that this approach and its results may not be so widely known after all.
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