Help me crunch some numbers

I feel like I should be able to do this myself - turns out I am not:

A shooter in Craps has a probability of 20% of rolling a seven. In 100 (200, 300) rolls, what is the probability of that shooter to have rolled more sevens than a random shooter at 16.67%

Thanks a lot! Wouldn't mind learning how you calculated this as well...

What?

Do you mean what's the chance a random shooter throwing 20% (or higher) sevens in X rolls?

No, I would like to know how likley it would be for a DI shooter that can roll a 7 20% of the time to roll more 7s than a random shooter within 100 rolls.

I don't think there's an "easy" way of calculating this. The one way I can think of doing it is this:

Let F(n) be the probability of rolling exactly N 7s out of 100 with the 20% probability:
F(n) = (100)C(n) x (1/5)ⁿ x (1 - 1/5)^100-n
where (100)C(n) is the number of combinations of n items out of 100
Let G(n) be the probability of rolling exactly N 7s out of 100 with the normal (1/6) probability:
F(n) = (100)C(n) x (1/6)ⁿ x (1 - 1/6)^100-n

The probability of rolling more 7s in 100 rolls is:
F(0) x (G(1) + G(2) + G(3) + ... + G(100))
+ F(1) x (G(2) + G(3) + G(4) + ... + G(100))
+ F(2) x (G(3) + G(4) + G(5) + ... + G(100))
+ ...
+ F(98) x (G(99) + G(100))
+ F(99) x G(100)

This probably can be calculated easiest in an Excel spreadsheet.

Thx Don, makes me feel a little better that it actually is complicated. In that caseI guess I will think of a less complicated wager.

Meet me in the SRR thread everybody :)

I get 69.81% after 100 rolls, 78.76% after 200 rolls, and 84.23% after 300 rolls

Cool, thanks again Don! That is less likely than I would have assumed.

I guess that goes to show how difficult DI were to prove if it were real. Even after 300 rolls, there is a reasonable chance the random shooter would still come out on top.

You can use the binomial distribution to get probabilities of each occurrence and then multiply probabilities.

So, populate a table in excel:

Probability
Rolls 100
Prob 0.166666667 0.2
0 1.20747E-08 2.03704E-10 1.20747E-08
1 2.41493E-07 5.09259E-09 2.41493E-07
2 2.39079E-06 6.30208E-08 2.39079E-06
=a6+1

Column B = BINOMDIST($A4|$B$2|B$3|FALSE)
Column C = BINOMDIST($A4|$B$2|C$3|FALSE)
Column D = B4*SUM(C5:$C$...) -- the probability of outcome 1 (shooter with 1/6th probability of rolling a 7) x the probability of rolling more 7s than the column 1 (probability is the sum of all results in column C > #7s in column B).

Then sum up column D to get 69.81% on 100 rolls, 78.77% on 200 rolls, 84.23% on 300 rolls, 87.98% on 400 rolls, 90.70% on 500 rolls

You don't get to 2 sigma until about 800 rolls which explains why it is very difficult to tell if one shooter is actually better than another in real world conditions.

Quote: boymimbo

You can use the binomial distribution to get probabilities of each occurrence and then multiply probabilities.

So, populate a table in excel:

Probability
Rolls 100
Prob 0.166666667 0.2
0 1.20747E-08 2.03704E-10 1.20747E-08
1 2.41493E-07 5.09259E-09 2.41493E-07
2 2.39079E-06 6.30208E-08 2.39079E-06
=a6+1

Column B = BINOMDIST($A4|$B$2|B$3|FALSE)
Column C = BINOMDIST($A4|$B$2|C$3|FALSE)
Column D = B4*SUM(C5:$C$...) -- the probability of outcome 1 (shooter with 1/6th probability of rolling a 7) x the probability of rolling more 7s than the column 1 (probability is the sum of all results in column C > #7s in column B).

Then sum up column D to get 69.81% on 100 rolls, 78.77% on 200 rolls, 84.23% on 300 rolls, 87.98% on 400 rolls, 90.70% on 500 rolls

You don't get to 2 sigma until about 800 rolls which explains why it is very difficult to tell if one shooter is actually better than another in real world conditions.

Thanks! I learned something today...

You could get an approximation that is a little less computational.

If we let A be a normal distributed random variable for the number of sevens rolled by the 0.2 shooter and let B be the variable for the 0.1667 shooter then…

A ~ N(mean = n/5, var = (1/5)(4/5)n = 4n/25)
B ~ N(mean = n/6, var = (1/6)(5/6)n = 5n/36)

Where n is the number of rolls

Then we can say C = A – B. Therefore,
C ~ N(mean = n/30, var = 269n/900)

We want to know the probability that C is greater than 0.5. So we can find a Z-score (sigma) by taking the mean minus 0.5 and then dividing it by the standard deviation (or var^0.5). So therefore…

Sigma = (n-15) / (269n)^0.5

You can look up the sigmas…or use Excel to get the probabilities by using:
=NORMSDIST(sigma)

And again, it’s just approximation. The higher the n the better the approximation.

Quote: PeeMcGee

You could get an approximation that is a little less computational.

...
And again, it’s just approximation. The higher the n the better the approximation.

Cool. I wonder if there's some magical formula that the Wizard can come up that's exact and better than both. Since these are in a series, I am thinking the answer might be "Yes".