It seems some information is missing, so I’ll try reformulating the problem my way:
A jar contains 1 fair coin (equal odds for heads and tails), and 99 unfair coins (99.9% chances of tails, 0.1% of heads). You pull out a coin, flip it, and it comes up tails. What are your expectations for the second flip?
P(fair & heads) = 0.01 * 0.5 = 0.005
P(unfair & heads) = 0.99 * 0.001 ~= 0.001
so P(heads on first throw) = 0.006
so P(fair | heads on first throw) = 5⁄6
so P(heads on second throw | heads on first throw) = 5⁄6 0.05 + 1⁄6 0.001 ~= 0.42
… at least, that’s one way of interpreting “you are 99% confident in your results”, I consider that the remaining 1% is “your analysis was completely wrong and the coin is just as likely to land on heads or tails”. A more realistic situation would be one where your confidence would be distributed among possible coins in coinspace, something like “90% confident of less than 0.01% odds for heads, 99% confident in less than 1% odds for heads, 99.9% confident in less than 10% odds for heads, etc.”.
I like the reinterpretation of the problem, but is
P(unfair & heads) = 0.99 * 0.0001 ~= 0.001
a typo? Just running the numbers through SpeedCrunch gives 0.99 * 0.0001 = 0.000099, and 0.000099 ~= 0.0001, which seems intuitively right as 0.99 is “almost” 1.
It seems some information is missing, so I’ll try reformulating the problem my way:
P(fair & heads) = 0.01 * 0.5 = 0.005
P(unfair & heads) = 0.99 * 0.001 ~= 0.001
so P(heads on first throw) = 0.006
so P(fair | heads on first throw) = 5⁄6
so P(heads on second throw | heads on first throw) = 5⁄6 0.05 + 1⁄6 0.001 ~= 0.42
… at least, that’s one way of interpreting “you are 99% confident in your results”, I consider that the remaining 1% is “your analysis was completely wrong and the coin is just as likely to land on heads or tails”. A more realistic situation would be one where your confidence would be distributed among possible coins in coinspace, something like “90% confident of less than 0.01% odds for heads, 99% confident in less than 1% odds for heads, 99.9% confident in less than 10% odds for heads, etc.”.
I like the reinterpretation of the problem, but is
a typo? Just running the numbers through SpeedCrunch gives 0.99 * 0.0001 = 0.000099, and 0.000099 ~= 0.0001, which seems intuitively right as 0.99 is “almost” 1.
Augh! Serves me right for modifying the values while writing the comment and not thoroughly rechecking the calculations, thanks!