Thread Rating:

Poll

16 votes (50%)
12 votes (37.5%)
5 votes (15.62%)
2 votes (6.25%)
10 votes (31.25%)
3 votes (9.37%)
6 votes (18.75%)
5 votes (15.62%)
10 votes (31.25%)
7 votes (21.87%)

32 members have voted

Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1041
June 15th, 2021 at 11:37:34 AM permalink
Quote: miplet

spreadsheet

I saw there was an old thread on this. I'm looking for a calculated (not simulated) answer expressed as a rational number

I disagree with your answer
Itís all about making that GTA
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
June 15th, 2021 at 1:10:17 PM permalink

I get 1 in 344,842,585.

I'll show my method if I'm correct.
Last edited by: Wizard on Jun 28, 2021
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
June 15th, 2021 at 1:10:30 PM permalink
Quote: Ace2

I'm looking for a calculated (not simulated) answer expressed as a rational number



I get:
125,085,337,830,737
,750,040,946,884,515,600,828,826,106,122,701,544,847,430,450
,629,046,605,089,708,493,923,039,495,721,017,570,418,222,323
,016,473,845,293,309,175,319,694,738,875,002,802,771,338,500
,366,299,331,953,447,220,689,670,202,784,565,639,415,289,332
,084,779,481,597,229,081,746,080,364,803,215,921,066,911,536
,541,381,291,811,341,713,825,167,788,888,871,197,475,636,470
,784,515,810,574,611,990,369,837,114,575,201,669,878,148,297
/
43,134,751,192,953,914,210,993
,943,271,802,543,777,980,988,349,922,891,546,230,519,480,217
,557,626,817,255,231,365,456,478,947,627,840,676,555,835,923
,392,957,024,113,969,618,568,826,138,354,283,301,839,972,359
,049,974,938,603,820,014,108,197,177,751,482,827,734,201,244
,571,353,546,913,323,627,624,872,502,898,073,424,297,340,815
,406,874,056,759,075,587,522,737,125,509,897,688,175,170,811
,478,174,333,289,594,416,341,915,851,631,174,536,884,518,568
(which is about 1 / 344,842,584.6)

Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
June 15th, 2021 at 1:29:22 PM permalink
Let the record show I beat Don by 13 seconds.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
June 15th, 2021 at 1:59:32 PM permalink
Quote: Wizard

Let the record show I beat Don by 13 seconds.


But you didn't express it as a rational number, as requested
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
June 15th, 2021 at 2:06:04 PM permalink
Quote: ThatDonGuy

But you didn't express it as a rational number, as requested



Grrrrrrrrrrrrrrrr!
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1041
June 15th, 2021 at 2:31:32 PM permalink
First off, the answer provided by Miplet is correct for the digits provided. But from what I can tell his answer was simulated. Miplet, is this not the case?

Today I have to declare ThatDonGuy as winner since he was the first to provide the answer in rational format as requested. What a close race

I would be interested in seeing everyone's method...I'm guessing the Wizard integrated a Poisson formula like I did.
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
June 15th, 2021 at 2:46:42 PM permalink
Quote: Ace2

I would be interested in seeing everyone's method...I'm guessing the Wizard integrated a Poisson formula like I did.


As usual, I used a brute force Markov chain.
Let a2, a3, ..., a12 = 0 if that point has not been made yet, and 1 if it has.
Let P(a2, a3, ..., a12) be the probability of winning from that state; P(1,1,1,1,1,1,1,1,1,1) = 1.
P(a2, a3, ..., a12) = 1/30 x 1/7 x P(1, a3, a4, ..., a12) + 1/15 x 1/4 x P(a2, 1, a4, ..., a12) + 1/10 x 1/3 x P(a2, a3, 1, a5, ..., a12) + ... + 1/30 x 1/7 x P(a2, a3, ..., a11, 1)
The two fractions are the probability that this is the point (1/30 for 2 and 12, 2/30 = 1/15 for 3 and 11, 3/30 = 1/10 for 4 and 10, 4/30 = 2/15 for 5 and 9, and 5/30 = 1/6 for 6 and 8) and the probability that, if it is the point, the point is made (1/7 for 2 and 12, 2/8 = 1/4 for 3 and 11, 3/9 = 1/3 for 4 and 10, 4/10 = 2/5 for 5 and 9, and 5/11 for 6 and 8)
Work backwards from (1,1,1,...,1,1,0) to (0,0,0,...,0,0,0), which is the solution.
miplet
miplet
Joined: Dec 1, 2009
  • Threads: 5
  • Posts: 1971
June 15th, 2021 at 2:49:56 PM permalink
My spreadsheet answer was calculated, not simulated, using google sheets. Iím sure rounding errors occurred when add all fractions from the 1024 different states you can end up on.
ďMan BabesĒ #AxelFabulous
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
June 15th, 2021 at 3:14:53 PM permalink
Quote: Ace2

I would be interested in seeing everyone's method...I'm guessing the Wizard integrated a Poisson formula like I did.



Yes, of course. You taught me and won't let me forget the method.

Here is the integral in text form: exp(-7303x/13860)*(1-exp(-x/252))^2*(1-exp(-x/72))^2*(1-exp(-x/36))^2*(1-exp(-2x/45))^2*(1-exp(-25x/396))^2*(7303/13860)

Here it is in proper notation:



Here is the screen with the answer. One has to scroll way over to the right to see all the digits.



I hope this proves I at least had the right answer and just didn't report it properly.
Last edited by: Wizard on Jun 28, 2021
It's not whether you win or lose; it's whether or not you had a good bet.

  • Jump to: