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17 votes (50%)
13 votes (38.23%)
5 votes (14.7%)
2 votes (5.88%)
11 votes (32.35%)
3 votes (8.82%)
6 votes (17.64%)
5 votes (14.7%)
10 votes (29.41%)
8 votes (23.52%)

34 members have voted

ThatDonGuy
ThatDonGuy
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July 28th, 2021 at 12:12:12 PM permalink
Quote: Ace2

Iíve never played repeater, but Iím asking what is the chance of rolling (at least) two 2s, three 3s, four 4s etc before rolling a 7.


Did you mean two 2s, three 3s, and so on, through twelve 12s, or just two of each?
Just rolling 12 12s before a 7 is (1/7)^12, or about 1 in 13.84 billion.

What is the over/under on how many posts before somebody's reply mentions "18 Yos"?
Ace2
Ace2
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July 28th, 2021 at 12:14:48 PM permalink
Going by the repeater rules, a 12 is like a 2 (must be repeated twice), an 11 three times etc
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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July 28th, 2021 at 1:20:04 PM permalink
Quote: Ace2

Going by the repeater rules, a 12 is like a 2 (must be repeated twice), an 11 three times etc


I get about 1 in 185,100

Question for the calculus boffins: I know the probability of not rolling a 2 in time t is 1 / e^(t/36), but what is the probability of not rolling a 2 at least twice in time t?
Ace2
Ace2
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July 28th, 2021 at 2:18:52 PM permalink
1 / e^(t/36) + (t/36) / e^(t/36)
Itís all about making that GTA
Ace2
Ace2
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teliot
July 28th, 2021 at 4:28:19 PM permalink
Quote: ThatDonGuy

I get about 1 in 185,100

I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000

Does anyone want more time before I post the method?
Itís all about making that GTA
teliot
teliot
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Ace2
July 28th, 2021 at 4:41:53 PM permalink
These are such small numbers, but I do appreciate multi-line integers. I'm personally a big fan of the fast growing hierarchy. :)
Poetry website: www.totallydisconnected.com
unJon
unJon
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July 28th, 2021 at 8:21:36 PM permalink
Quote: Ace2

I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000

Does anyone want more time before I post the method?



That is a surprising number of zeros at the end of the denominator. Are you sure it didnít round that?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ace2
Ace2
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July 28th, 2021 at 9:26:40 PM permalink
Quote: unJon

That is a surprising number of zeros at the end of the denominator. Are you sure it didnít round that?

Yes Iím pretty sure. Even if it was rounded, thereís still about 300 significant digits
Itís all about making that GTA
Ace2
Ace2
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July 29th, 2021 at 9:49:13 AM permalink
The answer is 1 minus the integral from zero to infinity of:

(1 - ((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 dx

The integral evaluates the probabilities at all time (x) that at least one repeater bet has not been won and a seven has not been rolled. Multiply by 1/6 for the chance that the next roll is a seven, causing the bet to lose.
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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Thanks for this post from:
unJon
July 30th, 2021 at 9:14:41 AM permalink
Quote: Ace2

The answer is 1 minus the integral from zero to infinity of:

(1 - ((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 dx

The integral evaluates the probabilities at all time (x) that at least one repeater bet has not been won and a seven has not been rolled. Multiply by 1/6 for the chance that the next roll is a seven, causing the bet to lose.


I assume that ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36) is the probability of not rolling a 6 at least 6 times in time x. How did you work that out?

Quote: unJon

Quote: Ace2

I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000


That is a surprising number of zeros at the end of the denominator. Are you sure it didnít round that?


My Markov chain result is the same, right down to the same number of zeroes in the denominator.

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