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Many will find this ludicrous... - another example

Posted By: Taper_Mike
Date: Tuesday, 11 September 2012, at 1:14 a.m.

In Response To: Many will find this ludicrous... - another example (Daniel Murphy)

–

Daniel: There are 6/36 doublets and 30/36 nondoublets. With normal dice, doublets occur 16.66% of the time, so nondoublets must occur 83.33% of the time. But if you halve the chance of rolling any doublet, then that chance is only 8.33%. And the chance of rolling any nondoublet must be 91.67%. Which means that the chance of rolling a specific doublet with "Red Room" dice is only 1.38%, and the chance of rolling a specific nondoublet has to be 6.11%.

Your concept is good, and, in general, so are your conclusions. There is a small mistake, however, in the computations.

Chance of non-doublets = 30/36
Chance of doublets = 3/36
Chance of reroll = 3/36

Chance of specific doublet = (3/36) / 6 = 1/72 = 1.39%
Chance of specific non-doublet = (30/36) / 15 = 2/36 = 5.56%

One easy way to take care of the rerolls is just to reduce the size of the universe from 100% to 33/36 = 91.67%, and then scale the foregoing probabilities accordingly.

The odds of rolling a specific doublet are 1.39% out of 91.67% = 1.39/91.67 = 1.52%.
The odds of rolling a specific non-doublet are 5.56% out of 91.67% = 5.56/91.67 = 6.06%

Mike

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