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 Pass the beer (Posted on 2013-04-09)
Suppose you and a bunch of friends are sitting around a table.
There are N of you.
You have a jug of beer in front of you, which no one has yet tasted.
So you take a swig of it, and then pass it to your left or right with probability 1/2.
Now suppose your neighbor does the same---he/she takes a swig of it and passes it to his/her left or right with probablity 1/2.
Each player continues in this fashion.
Because the beer is moving back and forth randomly around the table, it may be a while before some people get to taste the beer for the first time.

Which person around the table is most likely to be the last one to try the beer?
Is it a person near you or far from you?
(Assume that the jug is bottomless, and never runs out.)

Source: Math fun facts

 No Solution Yet Submitted by Ady TZIDON No Rating

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 re: computer solution | Comment 2 of 8 |
(In reply to computer solution by Charlie)

The true  answer is that ALL participants (except the first) are EQUALLY LIKELY (probability 1/(N-1)) to be last!

It can be easily verified manually for n=3 and n=4 .

Your results display about 1.3% difference between the lowest and the highest probabilities values and do not strongly support the theoretical conclusion.

Although the sgn(2*(rnd)-1) is supposed to equally distribute +1 and -1, -maybe it does not.

I suggest checking in parallel the deviation of "delta"- is it biased toward left or right: you may run 2 in parallel ,based on delta on one and minus delta or another etc

Changing the number of drinkers, quantity of experimental sessions is also an option.

Last, but not the least - why should any particular place be different?

 Posted by Ady TZIDON on 2013-04-10 09:45:14

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