The mid-streak roulette bet

I don't want to legitimize the silliness that's been going on in the other roulette thread(s), but the original post in one raises an interesting math problem.

You observe 25 roulette spins. Of those, select three numbers A, B and C with frequencies fA, fB, and fC, based on maximizing the sum of the adjusted frequencies fA', fB', fC' where fA' = min(fA, 3); fB' = min(fB, 3); and fC' = min(fC, 3). Then, select three more numbers D, E, and F with frequencies 0.

What is the probability that, after an additional 13 spins, the frequency of D, E, and F will all be >=3 AND the frequency of A, B, and C will not all be >= 3?

I don't know if this is exactly what you want, but I think it has some relevance to your thread.
It was a reply provided by 7craps to a question I posed, in an older thread, about 37 spins.


Is it possible to establish averages for 37 spins of a European wheel?

How many numbers should be hit?

yes,
23.57 unique numbers

How many multiples can be expected?

13.43 numbers NOT hit, 13.43 repeating numbers

What sizes will the multiples be?


0 Time repeat/ 13.43 numbers
1 Time repeat/ 13.80 "
2 Times repeat/ 6.90 "
3 Times repeat/ 2.24 "
4 Times repeat/ 0.53 "
5 Times repeat/ 0.10 "

you can download an Excel worksheet (no macros) HERE

Slightly over 20% of the time, 37 spins will result in 24 unique numbers - 13 total repeats
Sightly under 20% of the time, 37 spins will result in 23 unique numbers - 14 total repeats

The Excel worksheet has the results of 10 million 37 spin trials and the distributions.

"but the original post in one raises an interesting math problem" >>> You seem embarrassed asking this, why? It is a FANTASTIC question that SOME POSTERS do not want their name attached to an answer (not really sure why?). Ken

Quote: MathExtremist

I don't want to legitimize the silliness that's been going on in the other roulette thread(s), but the original post in one raises an interesting math problem.

You observe 25 roulette spins. Of those, select three numbers A, B and C with frequencies fA, fB, and fC, based on maximizing the sum of the adjusted frequencies fA', fB', fC' where fA' = min(fA, 3); fB' = min(fB, 3); and fC' = min(fC, 3). Then, select three more numbers D, E, and F with frequencies 0.

What is the probability that, after an additional 13 spins, the frequency of D, E, and F will all be >=3 AND the frequency of A, B, and C will not all be >= 3?

The math to me seems to be a very difficult question. Simulations may be easier.

The challenge I see is that after 25 spins, there are many outcomes to deal with.

There can be 19 unique #s with 6 repeats with any number of numbers repeating 1, 2 or more times
or 18 unique numbers with 7 repeats and so on.

I have a feeling mrjjj knows how to do the math.

MathExtremist also seems to have the qualifications to solve the question.

Fair enough but AGAIN, people can not have BOTH views. One view is for THIS scenario and one view is for THAT scenario. Ken

Cenario? Are you a pro wrestling fan of John Cena?

I changed it. Please tell me you have more on me other than words that are spelled wrong? KEEP IN MIND. I responded to a post which means, my post count goes up. You guys cant have it both ways. Ken