Quote:ssho88what I am doing and I can't disclose it here.

you ask for assistance in very difficult and complicated math equations and then you say you can't disclose what you're doing here

those that answered you did a lot of work for you

what did you do for them?

zilch

See my latest post. Adding ties into the count, my calculation now jives with your simulation. 28.703 / 0.904544 = 31.732.Quote:ssho88How you got that simple formula ?

If a Tie does not break a streak, my simulation results is 31.74, which is slightly higher compare to your 28.703.

I guess you NOT add in the rounds for TIE ?

Quote:Ace2Google “geometric series” and you’ll see this and other formulas. I’ve used it many times since this is such a common problem (consecutives). If, for instance, someone asks the expected number of rolls to get 3 consecutive 7’s with 2 die (1 in 6 chance each roll) the answer is 6 + 6^2 + 6^3 = 258 which can also be calculated as (6^(3+1) - 6) / (6 - 1).

There are many formulas for series that can make a calculation much easier and in some cases make a calculation possible.

Gee, I'll have to remember that one...

I think what ssho88 (and myself, for that matter) wants to know is, where "the expected number of trials to get N consecutive results of a particular type is E + EQuote:ThatDonGuy

E0 p4 = 1 + p + p2 + p3

E0 = (p4 - 1) / (p4 (p - 1))

^{2}+ E

^{3}+ ... + E

^{N}, where E is the expected number of trials to get one result of that type" came from.

I guess that series is my discovery, though I’d be shocked if no one ever saw this pattern before. For now I’ll call it the Ace-Onacci series.Quote:ThatDonGuyGee, I'll have to remember that one...

I think what ssho88 (and myself, for that matter) wants to know is, where "the expected number of trials to get N consecutive results of a particular type is E + E^{2}+ E^{3}+ ... + E^{N}, where E is the expected number of trials to get one result of that type" came from.

When I first saw “consecutive win” problems a few years ago, it was immediately obvious there was a pattern. For instance, in the case of p = 1/6 the answers are 6, 42 and 258 for 1, 2, and 3 consecutive wins. The first value is clearly 6, then it’s easy to see the pattern: the second value is the first value plus 6^2, then the third value is the second value plus 6^3 and so on. This works for any p value and any number of consecutive wins. Once the wins get over about 3, it gets much easier to use the geometric series formula than to sum up the power series.

I realize you know what a geometric series is. I was answering the OP and thought he/she was asking about that

Quote:lilredroosteryou ask for assistance in very difficult and complicated math equations and then you say you can't disclose what you're doing here

those that answered you did a lot of work for you

what did you do for them?

zilch

Did I promise anything when I ask for help to verify my simulation results ? I may share it ONLY to those(ThatDonGuy, Chesterdog, 7craps) that helped me upon request but definitely not in the public forum. Fair ?

The funny thing here is main contributors here did not ask anything from me but you are trying . . . .

Quote:ssho88The funny thing here is main contributors here did not ask anything from me but you are trying . . . .

I didn't ask you for anything and you don't have to worry - I never will

the other guys didn't ask because that is not in their nature

just thought after they did all that work for you, you might have considered saying thank you

Quote:lilredroosterI didn't ask you for anything and you don't have to worry - I never will

the other guys didn't ask because that is not in their nature

just thought after they did all that work for you, you might have considered saying thank you

Please read carefully my early replies( post #3 and # 9), I have said "Tq" to ChesterDog and ThatDonGuy, and yet I shared my advantage play info with 7craps !

Quote:ssho88Please read carefully my early replies( post #3 and # 9), I have said "Tq" to ChesterDog and ThatDonGuy, and yet I shared my advantage play info with 7craps !

okay, my bad - I missed that - sorry

Quote:Ace2I guess that series is my discovery, though I’d be shocked if no one ever saw this pattern before. For now I’ll call it the Ace-Onacci series.Quote:ThatDonGuyI think what ssho88 (and myself, for that matter) wants to know is, where "the expected number of trials to get N consecutive results of a particular type is E + E

^{2}+ E^{3}+ ... + E^{N}, where E is the expected number of trials to get one result of that type" came from.

Since I have nothing better to do (and I don't), here's a proof of the Ace-onacci Series:

Back to the "state" method - let E

_{K}be the expected number of trials needed to get to N consecutive results of a particular type given that you are currently at K consecutive, where p = the probability of the event happening and q = 1 - p.

E

_{K}= 1 + q E

_{0}+ p E

_{K+1}

E

_{N}= 0

E

_{N-1}= 1 + q E

_{0}+ p E

_{N}= 1 + q E

_{0}

Assume E

_{N-K}= (1 + q E

_{0}) (1 + p + p

^{2}+ ... + p

^{K-1});

E

_{N-(K+1)}= 1 + q E

_{0}+ p E

_{N-K}

= 1 + q E

_{0}+ p (1 + q E

_{0}) (1 + p + p

^{2}+ ... + p

^{K-1})

= 1 + q E

_{0}+ (1 + q E

_{0}) (p + p

^{2}+ ... + p

^{K-1}+ p

^{K})

= (1 + q E

_{0}) (1 + p + p

^{2}+ ... + p

^{K-1}+ p

^{(K+1)-1})

Since it is true for K = 1 and it is true for K + 1 if it is true for K, it is true for all positive integers K by induction

Let K = N:

E

_{0}= E

_{N-N}= (1 + q E

_{0}) (1 + p + p

^{2}+ ... + p

^{N-1})

= (1 + q E

_{0}) (1 - p

^{N}) / (1 - p)

= (1 - p

^{N}) / (1 - p) + q E

_{0}) (1 - p

^{N}) / q (since q = 1 - p)

= (1 - p

^{N}) / (1 - p) + E

_{0}) (1 - p

^{N})

E

_{0}(1 - (1 - p

^{N})) = (1 - p

^{N}) / (1 - p)

E

_{0}= (1 - p

^{N}) / (p

^{N}(1 - p))

= (1 - p

^{N}) / (p

^{N}- p

^{N+1})

Now, a quick proof that the expected number of trials for an event with probability p is 1 / p:

Let E be the expected number

E = p + 2 q p + 3 q

^{2}p + 4 q

^{3}p + ...

= p (1 + 2 q + 3 q p + 4 q

^{2}p + ...)

= p (1 + q + q

^{2}+ q

^{3}+ ...)

^{2}

= p / (1 - q)

^{2}(since -1 < q < 1)

= p / p

^{2}= 1 / p

Finally:

E

_{0}= (1 - p

^{N}) / (p

^{N}- p

^{N+1})

= 1 / p

^{N}x (1 - p

^{N}) / (1 - p)

= (1 + p + p

^{2}+ p

^{3}+ ... +p

^{N-1}) / p

^{N}

= 1 / p

^{N}+ p / p

^{N}+ p

^{2}/ p

^{N}+ ... + p

^{N-1}/ p

^{N}

= 1 / p

^{N}+ 1 / p

^{N-1}+ 1 / p

^{N-2}+ ... + 1 / p

= (1 / p)

^{N}+ (1 / p)

^{N-1}+ (1 / p)

^{N-2}+ ... + (1 / p)

= E

^{N}+ E

^{N-1}+ E

^{N-2}+ ... + E

where E = 1/p is the expected number of trials to get a result once