Originally posted by Billy Bobbins:
For the first case I can help out. For at least 1 success, there must be no non-successes. Non-success must mean that all 3 dice are 4,5, or 6. There is a 50% chance of rolling a 4,5, or 6 on any particular die. So the chance that all 3 roll 4, 5, or 6 is 50% * 50% * 50%, or 1/8. Therefor, the chance of at least 1 success is (1 - 1/8), or 7/8.
Originally posted by Corey McGlone:
But, in the first case, only 1 die fails on a 4, 5, or 6. The other 2 dice fail on 3, 4, 5, and 6. That means that the first die would fail 50% of the time while each of the other two would fail 66% of the time, right? Using the same formula, I would get this:
50% * 66% * 66% = 1/2 * 2/3 * 2/3 = 4/18 = 22.2%
Therefore, I would get at least one success 100% - 22.2% = 77.8% of the time.
Correct?
Now, if I add a fourth die that succeeds on a 1, 2, 3, or 4, I would have this:
33% * 50% * 66% * 66% = 1/3 * 1/2 * 2/3 * 2/3 = 4/54 = 7.4%
In such a case, I would get at least one success 100% - 7.4% = 92.6% of the time.
Okay, so that seems simple enough - I just multiply the individual probabilities together. But what about the second situation where I need more than just one success?
"I'm not back." - Bill Harding, Twister
Originally posted by Eugene Kononov:
From multiple runs, the variation didn't exceed 0.0003, so I am questioning your 59.3%.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
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