November 13th, 2019 at 10:48:48 AM
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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 aren’t comeout rolls. For each of these rolls there’s a 5 in 6, or 1 in 1.2, chance a seven won’t be rolled.

On average there is a comeout roll every 5.06 rolls, so 4.06/5.06 = 0.802 rolls aren’t comeout rolls. For each of these rolls there’s a 5 in 6, or 1 in 1.2, chance a seven won’t be rolled.

It’s all about making that GTA

November 13th, 2019 at 1:57:10 PM
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So am I supposed to add one PB to the table every 6 rolls? #Damn7outs

November 14th, 2019 at 6:12:16 AM
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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

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

It’s all about making that GTA