Quote:
Originally Posted by Kender
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my table was wrong. had 1 too many .65 per line. but if you changed the second fd change to 50% the actual probability to be forgotten changes
1= 0.35 (1-.65)
2= 0.67 (1-.65*.5)
so you have a 67% chance to effect a mem blur if you fd twice
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Your first table was correct result wise. The total chance of being remembered would be this, based on number of flops.
1 = 65
2 = 42.3
3 = 27.5
4 = 17.9
5 = 11.6
6 = 7.5
7 = 4.9
8 = 3.2
9 = 2.1
That deviates already at 2 attempts, where they said it would have a 50% chance.
So I went with increasing rate of success, following repeat failures. Overall it works out to be pretty similar.
H