N people roll a die in turn, following a prearranged list. The first to get a 6 names the drink. The second to get a 6 drinks it. The third, pays! Then the next on the list rolls and so on till the closing time.
The above goes on for 120 minutes, averaging 5 rolls per minute (the time for naming, drinking, plus settling disputes and paying the bills was discounted).
How many times was it the same player that named the drink, consumed it but did not pay for it?
a. Provide your estimates for N=3, N=6 and N=12
b. Please explain the meaning of the results.
Rem: Verification of analytical results by simulation is welcome.
I am not surprised that Charlie's simulations are lower than predicted.
There are two issues
1) If we are throwing away any incomplete rounds, then the expected number of completed rounds in 120 minutes is, I think, 100/3 - 1/2 = 32 5/6, not 33 1/3
2) Less obviously, there is an issue that comes in the closing minutes that the bar is open.
Let's say, for instance, that N = 6, and that a player rolls a 6 and names a drink at one minute before closing.
There is only time for 5 more rolls, and it is impossible to complete a cycle in which he is the drinker. It is, however, possible to finish a cycle in which somebody else is the drinker. So the last cycle is less likely than expected to produce a case where the same person chooses and drinks. The simulation reflects this, but the theory does not.
If the rule was a little different, and the players could keep rolling after 120 minutes to finish the round that was in progress at 120 minutes, then the bias against the last round chooser being the last round drinker disappears, and the simulation would more closely match the prediction, after the prediction was adjusted to reflect the new number of expected rounds, which is now, I think, 100/3 + 1/2 = 33 + 5/6
Edited on February 24, 2016, 6:23 am
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