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Many will find this ludicrous... - another example

Posted By: Daniel Murphy
Date: Tuesday, 11 September 2012, at 2:12 a.m.

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

It's surely possible, but I'm not sure you've found a mistake, Mike. Someone might tell us exactly how SHG adjusts for doublets, but if, as you suggest, the chance of rolling a particular doublet were 1.38% of 91.67, not 1.38% of 100, then the chance of rolling any doublet wouldn't be 8.33% (half of the normal chance), it would be .0138888/.9166666 = 9.09090%. It can't be both 8.33% and 9.09%, and I chose to hold the percentage at 8.33%, since that was what some comments here suggested (or that was one way of interpreting) what was happening in the red room. So, I think my arithmetic holds up, if the overall chance of rolling a doublet is to be half the normal chance, that is, 8.33%.

Which is not to say your method is wrong, or that SHG is rolling doublets at an 8.33% clips, but just that your method forces an overall chance of rolling a doublet of 9.09%, not 8.33%. SHG's RNG apparently uses yet another method since (as Michael's study of the dice in 2009 showed) the percentage of doublets on SHG was neither of these, but 9.22% (and hasn't changed?), which is why I added data for that overall chance of rolling doublets also.

So,

if P(doublet) is .0833, then P(nondoublet) is .9167.

If P(doublet) is .0909, then P(nondoublet) is .9091.

If P(doublet) is .0922, then P(nondoublet) is .9078. And

if P(doublet) is .1666, then P(nondoublet) is .8333. And the chance of rolling a particular doublet or nondoublet follows from these numbers.

If I'm still missing something, please let me know.

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