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# Who buys coffee?

### #1

Posted 16 July 2013 - 01:47 PM

You're having coffee with friends sitting around a table on a work break. You paid for everyone's coffee the last time, and so you propose a method for randomly assigning that task to another person this time. Your proposal involves flipping a fair coin until a particular event happens. Here are the details:

You flip first. If it's heads, you pass the coin to your left; tails, you pass the coin to your right.

The person receiving the coin flips it and passes it left or right according to the same algorithm.

And so on.

As the process continues, more and more of the persons will receive the coin.

At some point, there will be just one person who has not received the coin.

That person must pay the bill.

Is this a fair method?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 16 July 2013 - 02:17 PM

### #3

Posted 16 July 2013 - 03:53 PM Best Answer

### #4

Posted 16 July 2013 - 05:00 PM

**Edited by Anza Power, 16 July 2013 - 05:01 PM.**

### #5

Posted 16 July 2013 - 08:53 PM

Got it...

### #6

Posted 17 July 2013 - 08:20 AM

To decide, and get a clearer view of the processes,

conside a simpler case.

The case of three people (two others) is trivial.

Four people is rich enough to be instructive, and still tractable by hand.

It's an interesting result.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #7

Posted 17 July 2013 - 09:25 AM

Four people is rich enough to be instructive, and still tractable by hand.

In case of 4 people, the immediate neighbours of the starter each have a starting probability of 1/2 of not being the last.

For the one opposite, it does not matter whether the first toss is heads or tails. After the first toss, he also has a probability of 1/2 of not being the last. So his effective starting probability is also 1x1/2 = 1/2

All have equal probability of getting the coin before the last person.

### #8

Posted 17 July 2013 - 02:18 PM

Someone has the right answer, and someone is close.

To decide, and get a clearer view of the processes,

conside a simpler case.

The case of three people (two others) is trivial.

Four people is rich enough to be instructive, and still tractable by hand.

It's an interesting result.

The cases of 3 and 4 both match my answer above, in case of four people you and the person right in front of you have 1/3 chance each and the two others who are beside you have 1/6 chance each...

**Edited by Anza Power, 17 July 2013 - 02:18 PM.**

### #9

Posted 17 July 2013 - 11:13 PM

The cases of 3 and 4 both match my answer above, in case of four people you and the person right in front of you have 1/3 chance each and the two others who are beside you have 1/6 chance each...Someone has the right answer, and someone is close.

To decide, and get a clearer view of the processes,

conside a simpler case.

The case of three people (two others) is trivial.

Four people is rich enough to be instructive, and still tractable by hand.

It's an interesting result.

I didn't read your answer carefully enough to see that you included the initial flipper.

So that degrades the chances of the adjacent persons.

Do you get the expected result now if you exclude him?

The process is supposed to find "another" person. I.e. a person **other than** the initial flipper.

The one who buys is the one who never touches the coin.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #10

Posted 18 July 2013 - 08:21 PM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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