lilredrooster
lilredrooster
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May 19th, 2019 at 2:15:01 PM permalink
Quote: ssho88

what 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
"𝘦𝘷𝘦𝘳𝘺𝘣𝘰𝘥𝘺 𝘩𝘢𝘴 𝘢 𝘱𝘭𝘢𝘯 𝘶𝘯𝘵𝘪𝘭 𝘵𝘩𝘦𝘺 𝘨𝘦𝘵 𝘩𝘪𝘵 𝘪𝘯 𝘵𝘩𝘦 𝘧𝘢𝘤𝘦" .......ᴍɪᴋᴇ ᴛyꜱᴏɴ
Ace2
Ace2
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May 19th, 2019 at 2:39:14 PM permalink
Quote: ssho88

How 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 ?

See my latest post. Adding ties into the count, my calculation now jives with your simulation. 28.703 / 0.904544 = 31.732.
ThatDonGuy
ThatDonGuy
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May 19th, 2019 at 3:33:58 PM permalink
Quote: Ace2

Google ďgeometric seriesĒ and youíll see this and other formulas. Iíve 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 7ís 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...

Quote: ThatDonGuy


E0 p4 = 1 + p + p2 + p3
E0 = (p4 - 1) / (p4 (p - 1))

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 + E2 + E3 + ... + EN, where E is the expected number of trials to get one result of that type" came from.
Ace2
Ace2
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May 19th, 2019 at 4:57:10 PM permalink
Quote: ThatDonGuy

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 + E2 + E3 + ... + EN, 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 Iíd be shocked if no one ever saw this pattern before. For now Iíll call it the Ace-Onacci series.

When I first saw ďconsecutive winĒ 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 itís 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
Last edited by: Ace2 on May 19, 2019
ssho88
ssho88
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May 19th, 2019 at 5:06:30 PM permalink
Quote: lilredrooster

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




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 . . . .
Last edited by: ssho88 on May 19, 2019
lilredrooster
lilredrooster
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May 20th, 2019 at 1:00:14 AM permalink
Quote: ssho88

The 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
"𝘦𝘷𝘦𝘳𝘺𝘣𝘰𝘥𝘺 𝘩𝘢𝘴 𝘢 𝘱𝘭𝘢𝘯 𝘶𝘯𝘵𝘪𝘭 𝘵𝘩𝘦𝘺 𝘨𝘦𝘵 𝘩𝘪𝘵 𝘪𝘯 𝘵𝘩𝘦 𝘧𝘢𝘤𝘦" .......ᴍɪᴋᴇ ᴛyꜱᴏɴ
ssho88
ssho88
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May 20th, 2019 at 1:11:49 AM permalink
Quote: lilredrooster

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




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 !
lilredrooster
lilredrooster
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May 20th, 2019 at 1:15:44 AM permalink
Quote: ssho88

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 !



okay, my bad - I missed that - sorry
"𝘦𝘷𝘦𝘳𝘺𝘣𝘰𝘥𝘺 𝘩𝘢𝘴 𝘢 𝘱𝘭𝘢𝘯 𝘶𝘯𝘵𝘪𝘭 𝘵𝘩𝘦𝘺 𝘨𝘦𝘵 𝘩𝘪𝘵 𝘪𝘯 𝘵𝘩𝘦 𝘧𝘢𝘤𝘦" .......ᴍɪᴋᴇ ᴛyꜱᴏɴ
ThatDonGuy
ThatDonGuy
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May 20th, 2019 at 4:20:35 PM permalink
Quote: Ace2

Quote: ThatDonGuy

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 + E2 + E3 + ... + EN, 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 Iíd be shocked if no one ever saw this pattern before. For now Iíll call it the Ace-Onacci series.


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 EK 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.
EK = 1 + q E0 + p EK+1

EN = 0
EN-1 = 1 + q E0 + p EN = 1 + q E0
Assume EN-K = (1 + q E0) (1 + p + p2 + ... + pK-1);
EN-(K+1) = 1 + q E0 + p EN-K
= 1 + q E0 + p (1 + q E0) (1 + p + p2 + ... + pK-1)
= 1 + q E0 + (1 + q E0) (p + p2 + ... + pK-1 + pK)
= (1 + q E0) (1 + p + p2 + ... + pK-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:
E0 = EN-N = (1 + q E0) (1 + p + p2 + ... + pN-1)
= (1 + q E0) (1 - pN) / (1 - p)
= (1 - pN) / (1 - p) + q E0) (1 - pN) / q (since q = 1 - p)
= (1 - pN) / (1 - p) + E0) (1 - pN)
E0 (1 - (1 - pN)) = (1 - pN) / (1 - p)
E0 = (1 - pN) / (pN (1 - p))
= (1 - pN) / (pN - pN+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 q2 p + 4 q3 p + ...
= p (1 + 2 q + 3 q p + 4 q2 p + ...)
= p (1 + q + q2 + q3 + ...)2
= p / (1 - q)2 (since -1 < q < 1)
= p / p2 = 1 / p

Finally:
E0 = (1 - pN) / (pN - pN+1)
= 1 / pN x (1 - pN) / (1 - p)
= (1 + p + p2 + p3 + ... +pN-1) / pN
= 1 / pN + p / pN + p2 / pN + ... + pN-1 / pN
= 1 / pN + 1 / pN-1 + 1 / pN-2 + ... + 1 / p
= (1 / p)N + (1 / p)N-1 + (1 / p)N-2 + ... + (1 / p)
= EN + EN-1 + EN-2 + ... + E
where E = 1/p is the expected number of trials to get a result once

Ace2
Ace2
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May 20th, 2019 at 4:36:40 PM permalink
Nice proof. I never did one, I only recognized the pattern.

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