ThatDonGuy Joined: Jun 22, 2011
• Posts: 5418
August 12th, 2021 at 4:55:26 PM permalink
Quote: Ace2

I don’t understand the question

Within a time period of duration x:
The probability of rolling at least two 2s is 1 - (e^(-x/36) * (1 + 1/36))
The probability of rolling at least three 3s is 1 - (e^(-x/18) * (1 + 1/18 + (1/18)^2 / 2))
...
The probability of rolling at least three 11s is 1 - (e^(-x/18) * (1 + 1/18 + (1/18)^2 / 2))
The probability of rolling at least two 12s is 1 - (e^(-x/36) * (1 + 1/36))
The probability of not rolling a 7 is e^(-x/6)
You would think the probability of doing all 11, thus winning the "repeater fire bet", would be the integral over x = 0 to positive infinity of the product of these 11 values, but it's not.
Ace2 Joined: Oct 2, 2017
• Posts: 1689
August 13th, 2021 at 7:11:04 PM permalink
For example, the probability of rolling at least three 3s is (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18))

Take the product of such probabilities for all ten points times the probability of zero sevens times 1/6 to get:

((1 - ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36)) * (1 - ((x/9)^4/24 + (x/9)^3/6 + (x/9)^2/2 + (x/9) + 1) * e^(-x/9)) * (1 - ((x/12)^3/6 + (x/12)^2/2 + (x/12) + 1) * e^(-x/12)) * (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18)) * (1 - ((x/36) + 1) * e^(-x/36)))^2 * e^(-x/6) * 1/6

Plug that into the integral calculator to get the exact answer shown below. This formula is slightly more efficient than the one I posted on easy math puzzles since it gets you the answer directly (instead of the complement of the answer). You multiply by 1/6 to avoid overcounting.

228296650211142223842235175926562819565310943733017295518075498704612736041415500474174790373335885790742403815919759376329015547908450301690458558920417797769549693735380351101698849553401942470326707347462797442543279964206576929140810851250499537038847980599155971380821671855091570073172919198633489 /

42257698361772482521904922230217663263013990621490761617970468605585232439497131274911233268994399835192106236921993503595172067323864679436391989136613954326164471151500091945731257099763259926492493587351085501547293599982107240564537220360657324137622979804472478269440000000000000000000000000000000000000

=~ 1 in 185,000
It’s all about making that GTA
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5418
August 13th, 2021 at 8:07:35 PM permalink
The only thing I don't understand about that is, why multiply by 1/6? What is being "overcounted"?
Ace2 Joined: Oct 2, 2017
• Posts: 1689
August 13th, 2021 at 9:34:27 PM permalink
This integral is summing the probabilities for all strings that have won all ten repeater bets and have zero sevens for all time values (x). To avoid overcounting you must end each winning sting with a seven (1/6 chance)

If you didn’t end with a seven, then a winning string at, for instance, x=8 can get counted again at x=9, since some winning strings at x=9 were already in a winning state at x=8 (and maybe at x= 7.2564897567 and at x= 6.2 etc). Ending with a seven ensures a winning string gets counted only once
It’s all about making that GTA
daveyandersen1 Joined: Mar 27, 2015