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  #1  
Old 10-01-2002, 12:38 AM
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Default Craps question

I was in AC last weekend playing craps and Boxcars(6,6) came up 3 times in a row. I'm not sure how to figure out the odds of this happening. I'm guessing 36 to the 3 power or 1 in 46656.
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  #2  
Old 10-01-2002, 07:57 AM
irchans irchans is offline
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Default Re: Craps question

That number looks good to me.

Another related question is:

What is the probability of three boxcars in sucession in a series of 100 rolls?

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  #3  
Old 10-01-2002, 05:04 PM
BruceZ BruceZ is offline
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Default 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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  #4  
Old 10-03-2002, 09:05 AM
irchans irchans is offline
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Default Re: Craps question

I was very surprised to see the Fibonacci like formula

t(n) = 35[t(n-1) + t(n-2)].

I had assumed that since 3 consecutive boxcars were needed that the recurrence relation would have 3 terms.

Have you solve the general problem:

What is the probability of n consecutive successes in a list of m tries if the probability of success for any particular try is p?

Using your approximate method we would get something like

1-(1-p^n)^k

where

k = m/( q + p q + p q^2 + ... + p q^(n-1) ).
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  #5  
Old 10-04-2002, 12:09 PM
BruceZ BruceZ is offline
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Default Re: Craps question

I had assumed that since 3 consecutive boxcars were needed that the recurrence relation would have 3 terms.

It appears that the 3 consecutive only influence where the Fibonacci-like sequence starts, since it can only occur starting on trial 3. The sequence itself only depends on the probability of boxcars on one particular throw. BTW, my sentence about there being 36 possible outcomes should read that there are 36 outcomes per surviving sequence on the previous stage. In the coin flip problem there were 2 outcomes per surviving sequence. This number is the only thing that changes the series.


What is the probability of n consecutive successes in a list of m tries if the probability of success for any particular try is p?

I believe my method generalizes to:

sum[i = n to m]t(i)*p^m

t(i) = (1/p - 1)[t(i-1) + t(i-2)]


BTW, you did post your Markov process matrix solution to the coin flip problem, which can be used here too.


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  #6  
Old 10-04-2002, 01:03 PM
BruceZ BruceZ is offline
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Default Re: Craps question

I believe my method generalizes to:
sum[i = n to m]t(i)*p^m
t(i) = (1/p - 1)[t(i-1) + t(i-2)]


Sorry, make that:

sum[i = n to m]t(i)*p^i
t(i) = (1/p - 1)[t(i-1) + t(i-2)]
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  #7  
Old 10-03-2002, 12:09 AM
Ray Zee Ray Zee is offline
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Default Re: Craps question

well if it seems unusual for you, would you also think the same if you saw ace ace 3 times in a row or hard 8. see what i mean. then you wouldnt really also look, unless you saw it happen once first. so it really is only two times that you are looking for. so when you take it all in, you get to see these not likely occurances happen quite frequently.
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  #8  
Old 10-03-2002, 01:55 AM
BruceZ BruceZ is offline
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Default Re: Craps question

Absolutely. Every hand you play in holdem or stud is a 134 million to 1 shot. That is what the odds are of getting any particular 7 cards. If you look at the order they appeared in you can make that 674 BILLION to 1. If you start taking into account what other players have....well you get the picture. It's just that there isn't anything remarkable enough about most of those hands to warrant a post on twoplustwo, but that is only to our minds. They are all equally remarkable.
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