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<blockquote data-quote="David Johnson2" data-source="post: 9443654" data-attributes="member: 6911800">Assuming that after each round, the dice pool is 5/6 as large as the previous round, the number of turns until no dice are left is given by t=log⅚(1/2n), where n is the number of dice. This is a huge approximation because it yields fractional dice numbers for each round, but they largely agree with previous posts.<table style='width: 100%'><tr><td>On average, a pool of 1 countdown dice will run out in 3 turns. 
On average, a pool of 2 countdown dice will run out in 7 turns. 
On average, a pool of 3 countdown dice will run out in 9 turns. 
On average, a pool of 4 countdown dice will run out in 11 turns. 
On average, a pool of 5 countdown dice will run out in 12 turns. 
On average, a pool of 6 countdown dice will run out in 13 turns. 
On average, a pool of 10 countdown dice will run out in 16 turns. 
On average, a pool of 15 countdown dice will run out in 18 turns. 
On average, a pool of 20 countdown dice will run out in 20 turns. 
On average, a pool of 30 countdown dice will run out in 22 turns. 
On average, a pool of 60 countdown dice will run out in 26 turns. 
On average, a pool of 100 countdown dice will run out in 29 turns.
</td></tr></table>Using this method, the number of dice needed in a pool that will average lasting n turns is 0.5*1.2^t.In these formulas, change 5/6 to the probability that a dice remains in the pool and 1.2 to the reciprocal of the probability for other dice types. For instance, for a d20 removed on 18-20, these are 17/20 and 20/17, respectively. 100d20 last 32 turns compared to 29 for 100d6 under this modification.</blockquote>

[QUOTE="David Johnson2, post: 9443654, member: 6911800"] Assuming that after each round, the dice pool is 5/6 as large as the previous round, the number of turns until no dice are left is given by t=log[SIZE=1]⅚[/SIZE](1/2n), where n is the number of dice. This is a huge approximation because it yields fractional dice numbers for each round, but they largely agree with previous posts. [TABLE] [TR] [TD][LEFT]On average, a pool of 1 countdown dice will run out in 3 turns. On average, a pool of 2 countdown dice will run out in 7 turns. On average, a pool of 3 countdown dice will run out in 9 turns. On average, a pool of 4 countdown dice will run out in 11 turns. On average, a pool of 5 countdown dice will run out in 12 turns. On average, a pool of 6 countdown dice will run out in 13 turns. On average, a pool of 10 countdown dice will run out in 16 turns. On average, a pool of 15 countdown dice will run out in 18 turns. On average, a pool of 20 countdown dice will run out in 20 turns. On average, a pool of 30 countdown dice will run out in 22 turns. On average, a pool of 60 countdown dice will run out in 26 turns. On average, a pool of 100 countdown dice will run out in 29 turns.[/LEFT][/TD] [/TR] [/TABLE] Using this method, the number of dice needed in a pool that will average lasting n turns is 0.5*1.2^t. In these formulas, change 5/6 to the probability that a dice remains in the pool and 1.2 to the reciprocal of the probability for other dice types. For instance, for a d20 removed on 18-20, these are 17/20 and 20/17, respectively. 100d20 last 32 turns compared to 29 for 100d6 under this modification. [/QUOTE]

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