Page:Logic of Chance (1888).djvu/52

viz.$(n+1)2^{n-2}$.

This being spread over $2^{n-1}$ different occasions of gain his average gain will be ${\tfrac {1}{2}}(n+1)$.

Now when we are referring to averages it must be remembered that the minimum number of different occurrences necessary in order to justify the average is that which enables each of them to present itself once. A man proposes to stop short at a succession of ten heads. Well and good. We tell him that his average gain will be £5. 10s. 0d.: but we also impress upon him that in order to justify this statement he must commence to toss at least 1024 times, for in no less number can all the contingencies of gain and loss be exhibited and balanced. If he proposes to reach an average gain of £20, he will require to be prepared to go up to 39 throws. To justify this payment he must commence to throw $2^{39}$ times, i.e. about a million million times. Not before he has accomplished this will he be in a position to prove to any sceptic that this is the true average value of a ‘turn’ extending to 39 successive tosses.

Of course if he elects to toss to all eternity we must adopt the line of explanation which alone is possible where questions of infinity in respect of number and magnitude are involved. We cannot tell him to pay down ‘an infinite sum,’ for this has no strict meaning. But we tell him that, however much he may consent to pay each time runs of heads occur, he will attain at last a stage in which he will have won back his total payments by his total receipts. However large $n$ may be, if he perseveres in trying $2^{n}$ times he may have a true average receipt of ${\tfrac {1}{2}}(n+1)$ pounds, and if he continues long enough onwards he will have it.

The problem will recur for consideration in a future chapter.