He should offer nothing, as it is clearly gambling and that just is not allowed here now is it!

Tut tut.

4s/3d

That's sexy

Really? Let me try to explain then, now it's evening!

The first go, B wins whatever is on the dice unless it's 6, in which case nada and game over. Hence the expected (average) win for one round is:

1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 0*1/6 = 15/6. Green being the colour of money.

There's then a 5/6 chance to play round two, a 5/6 chance from there to go on to round three, etc, so the winnings expected would be:

as from each round, there is a 5/6 chance you'll go to the next round and win on average 15/6 (and 1/6 chance you'll stop playing).

This is 15/6 + 5/6*15/6 + 5/6*5/6*15/6 + 5/6*5/6*5/6*15/6 + ... and on and on (and Ariston).

But this is 15/6 * ∑n=0,…,∞ (5/6)^n

To calculate the series, note that it looks like this: 1 + 5/6 + 25/36 + ...
Call this s = 1 + 5/6 + 25/36 + ...
But if we multiply each term by 5/6, we get the same progression but without the first term (of 1):
s*5/6 = 5/6 + 25/36 + ...

Now through the wonder of infinite series, we can substitute the second equation into the first:

s = 1 + s*5/6
s/6 = 1
s = 6

Therefore the expected value of B's winnings is 6*15/6 = 15

B should pay at most 15 to play.

Now you're talking dirty....

He's a maths slut!

Well, a degree in mathematics has to be good for something - might as well be that...

Having a degree in maths is akin to cheating............

Typical, the wording in the problem ("....willing to pay....") is significant. If I understand your answer correctly, 15 is the fee at which B would be 50% likely to lose (i.e. he wins 14 or less) and 50% likely to win (i.e. he wins 16 or more). The answer then should surely be 14, not 15? He should only be willing to pay a fee that gives him a probability, albeit small, of winnning, surely?

It's not a good idea to think of expected value in that way. For a start, it ignores the probability of actually getting 15 - but also it can be incorrect.

(E.g. imagine a very simple game where you throw two coins: HH wins 200 but any other outcome wins 100. What should you pay to play?
The expected outcome is:
HH: 200 * 1/4 +
HT: 100 * 1/4 +
TH: 100 * 1/4 +
TT: 100 * 1/4
= 125
So you should pay at most 125. But note that you're not equally likely to win more than this and less than this in any single game!)

However, if you played the game an infinite number of times, and calculated your average winnings from all those games, you would find the average to be 15.

Now, as a mathematician I say you should pay 15 at the most, since at 15 you would be (in probability terms) ambivalent about playing. Your expected net win is 0, but that's still ok - you won't care between playing or not playing so there's no reason not to play.

As an economist, it comes down to attitudes towards risk. Some people will pay much more than the expected outcome if the probability density function means there is a chance of a massive upside and limited downside (as there is in this game). See: any lottery that doesn't pay out all its revenue as prize money.

But in your game, the two coins are tossed, and the game is then definitely over - not the same thing. Your 125 reflects the total of ALL possible outcomes divided by the sample size - not the same thing.

You've given me some food for thought though........ my (non-mathematical) gut feeling told me to pay a lot less than 15...........

Ah but it IS (effectively) the same thing. That's the beauty.

Instead of looking at it as you write above, look at it as "the sum of all possible outcomes multiplied by the chance of each outcome occuring". This is the same thing as you suggest, but writing it this way helps understand why the massive outcomes with tiny probabilities in the infinite game case don't change the overall answer that much.

In my second game above, the sum of all possible outcomes multiplied by the chance of each outcome occuring is 125.
In the original game, the sum of all possible outcomes multipled by the chance of each outcome occuring is 15.

The expected value of the number of rounds played is 6. That might help you think about why the number may be higher than expected.

(I had something in here about how to use this to calculate expected winnings but haven't proved it so decided best not to risk saying it )

Also, if you really want to rake your noodle:
There is a higher probability of rolling a six and the game ending on the first go, than on the second go.

A simplistic way of looking at it.................

Probability of the game reaching a second roll = 0.8333
Ditto..... third roll = 0.6944
Ditto..... fourth roll = 0.5787
Ditto..... fifth roll = 0.4822

B should therefore assume that is unlikely (i.e. less than 50:50) that there will be more than four rolls of the dice.....

Four rolls of 5 = max 20 pounds winnings
Four rolls of 1 = min 4 pounds winnings
... etc......

Therefore, B would pay a MAXIMUM of 12 pounds.

Put it this way, as A I would not charge LESS than 15 quid to play...

No game then.............

Right, I'm off to the pub.............

(the above was my daughter's answer by the way - she's a Law student, not a mathematician - not bad I thought)