Balls to that. I'm armed with a bad attitude and worse math. By my reckoning: On any given day, a given prisoner has a .99 probability of NOT visiting the ro

Balls to that. I'm armed with a bad attitude and worse math. By my reckoning: On any given day, a given prisoner has a .99 probability of NOT visiting the room. Over a period of N days, a given prisoner has a pow(.99, N) probability of having _NEVER_ visited the room. Therefore, the probability a given prisoner HAS visited the room at least once over a period of N days is (1-pow(.99, N)). Therefore, the probability that X prisoners have each visited the room at least once over a period of N days is pow((1-pow(.99, N)), X)[1]. If I'm in the room, the question I need to answer is: How likely is it that each of the 99 other prisoners have visited the room at least once in the preceeding N days? Let's visit Mr. Python: N P(safe) Cost[2] ----- -------- ---------- 100: 0.000000 (18250.00) 200: 0.000001 (18249.99) 300: 0.006887 (18124.31) 400: 0.166419 (15212.86) 500: 0.520678 ( 8747.63) 600: 0.787901 ( 3870.80) 700: 0.916504 ( 1523.81) 800: 0.968598 ( 573.08) 900: 0.988391 ( 211.87) 1000: 0.995735 ( 77.83) 1100: 0.998437 ( 28.53) 1200: 0.999428 ( 10.45) 1300: 0.999790 ( 3.82) 1400: 0.999923 ( 1.40) 1500: 0.999972 ( 0.51) 1600: 0.999990 ( 0.19) 1700: 0.999996 ( 0.07) 1800: 0.999999 ( 0.03) 1900: 0.999999 ( 0.01) 2000: 1.000000 ( 0.00) 2100: 1.000000 ( 0.00) 2200: 1.000000 ( 0.00) 2300: 1.000000 ( 0.00) 2400: 1.000000 ( 0.00) 2500: 1.000000 ( 0.00) 2600: 1.000000 ( 0.00) 2700: 1.000000 ( 0.00) 2800: 1.000000 ( 0.00) 2900: 1.000000 ( 0.00) 3000: 1.000000 ( 0.00) 3100: 1.000000 ( 0.00) 3200: 1.000000 ( 0.00) 3300: 1.000000 ( 0.00) 3400: 1.000000 ( 0.00) 3500: 1.000000 ( 0.00) After 1000 days, I'm guessing everyone's visited. Screw you guys, I'm going home.[3] [1]I know this can't be exactly right, because prisoner visits aren't independent events, but I figure it can't be that wrong, and I want to go home. [2]Cost is calculated in expected days of life forgone, multiplying (1-P(safe)) by 50 years by 365. Arbitrary, but it gives some idea of how much expected life you gain by waiting longer to roll the dice. [3]I know this is ducking the intention of the question, but I think the point that death isn't that bad a risk if the probability is low is a legitimate one.