In your third year at Warwick you often have a drink with a friend who enjoys probability (yes, you make some strange friends here). On the Sunday evening before the start of Term 2 she challenges you to a curious game of chance. Hiding a fair six-sided die (numbered 0,1,2,3,4,5) from your observation, she throws it ten times and secretly notes each outcome on a convenient beer mat. Then she turns to you and announces:

"I shall now read you the running averages of the dice throws, but in reverse order.''

(You must have looked puzzled, because she breaks off to explain)

"Let *X*_{n} be the result of the *n*^{th} throw, so the *n*^{th} running average is *Y*_{n}=(*X*_{1}+...+*X*_{n})/*n*. I will read out the numbers *Y*_{10}, *Y*_{9}, ..., *Y*_{1} in that order.''

Clearly a flicker of comprehension must have somehow crossed your face, because she now continues:

"At any stage you can stop me reciting the sequence. If the last average I read out was *Y*_{n} then with probability *Y*_{n}/5 (arranged using that book of random numbers which I carry everywhere) I will buy you a pint and the game ends. If you don't stop me at all then clearly you get a pint with probability *Y*_{1}/5. Otherwise you buy me a pint.''

You calculate rapidly: **E**[*Y*_{1}/5]=**E**[*X*_{1}/5]=(1/6) (0+1+...+5)/5=1/2 even if you wait till the last number, and you get to choose when to stop. Must be to your advantage!

"Seems a reasonably good deal to me,'' you say.

"I'm glad you think so;'' she replies, "before we begin you can buy me a pint to compensate. Then we can play the game *K* times, where *K* is as large as you think it should be so that on average you come out ahead.''

How big should *K* be?

The very next day you hear about this course, ST318 *Probability Theory*; by the end of the course you'll have learned enough to be able to answer the above question instantly *without any calculation at all*. All this and 15 CATS credit too. Now that is what you call fair! How can you resist?