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The puzzle states that he has to be 100% sure that everyone had been to the room. "fairly confidently" is not good enough here. And, it won't matter if you incr

by barns 19y ago

The puzzle states that he has to be 100% sure that everyone had been to the room. "fairly confidently" is not good enough here. And, it won't matter if you increase the number of visits in a row to 5, 20, or more, there is still no guarantee.

mxh 19y ago
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.
- run4yourlives 19y ago
  I'm with you. Hell, even 1000 days is a long time, I'd give 'er after 800. If I'm wrong, such is life. If I'm right, I saved myself 22 years in solitary confinement.
  - lotu 19y ago
    You will also save your self 22 years in solitary confinement if your wrong. So win win.

by barns 19y ago

mxh 19y ago

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.

run4yourlives 19y ago
I'm with you. Hell, even 1000 days is a long time, I'd give 'er after 800. If I'm wrong, such is life. If I'm right, I saved myself 22 years in solitary confinement.
- lotu 19y ago
  You will also save your self 22 years in solitary confinement if your wrong. So win win.