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Ace2
Ace2 
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 587
November 13th, 2019 at 10:48:48 AM permalink
A decent estimate for this probability, expressed as 1 in x, is 1.2^((n-1) * 0.802). So the estimate of lasting 154 rolls is 1.2^122.7 = 1 in 5.2 billion. Accurate within 7%

On average there is a comeout roll every 5.06 rolls, so 4.06/5.06 = 0.802 rolls arenít comeout rolls. For each of these rolls thereís a 5 in 6, or 1 in 1.2, chance a seven wonít be rolled.
Itís all about making that GTA
ChumpChange
ChumpChange
Joined: Jun 15, 2018
  • Threads: 24
  • Posts: 1052
November 13th, 2019 at 1:57:10 PM permalink
So am I supposed to add one PB to the table every 6 rolls? #Damn7outs
Ace2
Ace2 
Joined: Oct 2, 2017
  • Threads: 19
  • Posts: 587
November 14th, 2019 at 6:12:16 AM permalink
For more accuracy (within 1%), use:

3.131 * 1.159456^(n-10)

3.131 being the value for 10, 1.159456 being the sustained rate for each iteration after 10.

So the value for 154 is 3.131 * 1.159456^144 = 5.601 x 10^9
Itís all about making that GTA

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