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
July 30th, 2021 at 9:29:19 AM permalink
Quote: ThatDonGuy

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?
.

Itís just Poisson.

e^(-5x/36) * (5x/36)^5/120 is the chance of rolling five 6s in time x, e^(-5x/36) * (5x/36)^4/24 is the chance of rolling four 6s in time x, etc. Sum these for values 0 - 5 and thatís the probability of rolling less than six 6ís.
Itís all about making that GTA
Ace2
Ace2 
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1041
August 3rd, 2021 at 5:39:52 PM permalink
When someone wins the Fire Bet, how many total pass bets will they have won, on average?
Itís all about making that GTA
Ace2
Ace2 
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1041
August 4th, 2021 at 9:46:20 AM permalink
Quote: Ace2

When someone wins the Fire Bet, how many total pass bets will they have won, on average?

Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?
Last edited by: Ace2 on Aug 4, 2021
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
August 4th, 2021 at 2:06:41 PM permalink
Quote: Ace2

Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?


Tough one!


I tried a Markov chain, where the probability of making a particular point was in proportion to establishing that point, but I got something like 17 - while a simulation returns around 7.4.

Presumably, it's low because the higher the number of points that are established, the more likely it is that you will seven out.

Wizard
Administrator
Wizard 
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
August 7th, 2021 at 2:24:26 PM permalink
Two ordinary six-sided dice are rolled until one of the following events happen:

A total of seven is rolled
A total of three is rolled 3x
A total of five is rolled 5x
A total of nine is rolled 5x
A total of 11 is rolled 3x

What is the probability the ending event is any of the last four listed?
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
August 7th, 2021 at 3:43:42 PM permalink
1 - 30026163533 / 31381059609 =~ 4.32%
Last edited by: Ace2 on Aug 7, 2021
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
August 8th, 2021 at 3:04:06 PM permalink
Quote: Ace2

1 - 30026163533 / 31381059609 =~ 4.32%


How about that - I actually managed to figure out the Poisson-based method for this one:

Only rolls of 3, 5, 7, 9, and 11 are concerned here, so they have probabilities of 1/9, 2/9, 1/3, 2/9, and 1/9 respectively

Note that all of these are derivatives from 0 to positive infinity, dx
In time x:
P(< 3 3s) = P(< 3 11s) = e^(-x/9) * (1 + x/9 + (x/9)^2 / 2)
P(< 5 5s) = P(< 5 9s) = e^(-2x/9) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6) + (2x/9)^4 / 24)
P(no 7s) = e^(-x/3)
P(all five) = e^(-x/3) * (e^(-x/9) * (1 + x/9 + (x/9)^2 / 2) * e^(-2x/9) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6) + (2x/9)^4 / 24))^2
= e^(-x) * ((1 + x/9 + (x/9)^2 / 2) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6 + (2x/9)^4 / 24))^2
P(rolling a 7 first) = P(all five, and then a 7) = e^(-x) * ((1 + x/9 + (x/9)^2 / 2) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6 + (2x/9)^4 / 24))^2

Wizard
Administrator
Wizard 
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23279
August 9th, 2021 at 5:28:07 AM permalink
Quote: Ace2

1 - 30026163533 / 31381059609 =~ 4.32%



I agree! I show a brief solution of all the difficult Repeater Bet Plus bets here.
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
August 16th, 2021 at 5:39:22 PM permalink
Quote: Ace2

Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255 /

6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

=~7.39415862291138014289 average total points won

Would any of our calculus ninjas like more time before I post the solution?
Last edited by: Ace2 on Aug 16, 2021
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
August 16th, 2021 at 7:12:29 PM permalink
Quote: Ace2

The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255 /
6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

=~7.39415862291138014289 average total points won

Would any of our calculus ninjas like more time before I post the solution?


Don't look at me; I'm surprised there is a rational number solution - I thought it would end up being an infinite series (calculate the probabilities of winning with 6, 7, 8, 9, ... points, then add them together).

  • Jump to: