Timeline for Men in a bar - stoch. processes
Current License: CC BY-SA 3.0
2 events
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| Jan 19, 2012 at 15:24 | comment | added | Barry Cipra | Johan, very nice! There ought to be a slick argument along the lines of, If the process sends one man home per round on average and hence takes, on average, $m$ rounds to empty the bar, then the average tippler survives (to drink again) through $m/2$ rounds, for a total, on average, of $m \times m/2$ drinks. But that just feels like it's playing too fast and loose with the law(s) of averages. | |
| Jan 19, 2012 at 9:31 | history | answered | Johan Wästlund | CC BY-SA 3.0 |