131,115,985 to 1 [...] this amount of evidence would only be enough to give you an even chance of winning the lottery.
The number of false bleeps is distributed almost exactly Poisson with λ=1. The important figure is not the expected number of bleeps (Ex1, which is indeed 2). It’s the expected probability that a random bleep is the true one, E1x1. At the moment I can’t find an analytic solution (and a short search suggests none is known), but a computation shows the result is around 63.2%, much better than 50%.
Similarly, with 14 boxes (arguably “28 bits of evidence”), the chance of winning is about 79.1% on average, much better than 23.
The number of false bleeps is distributed almost exactly Poisson with λ=1. The important figure is not the expected number of bleeps (Ex1, which is indeed 2). It’s the expected probability that a random bleep is the true one, E1x1. At the moment I can’t find an analytic solution (and a short search suggests none is known), but a computation shows the result is around 63.2%, much better than 50%. Similarly, with 14 boxes (arguably “28 bits of evidence”), the chance of winning is about 79.1% on average, much better than 23.