Think of this as a mathematical intuition puzzle. You *could*
work it out analytically, or you *might* write a computer
program and beat it to death numerically (don't bother, since such a
program written in JavaScript is embedded in this page), but
the point is to first see if you can intuit the correct answer from
the mathematical properties of the components of the problem.
It's well worth a little head scratching, as the result is quite
beautiful and shows how seemingly unrelated things in mathematics
show a deeper unity when you look more closely at them.

The problem involves a game something like blackjack, but
played with random numbers between zero and one instead
of a deck of cards. To be precise, the random numbers are
*uniformly distributed*, which means that all numbers
in the range are equally probable, and greater than or equal
to zero and less than one. It doesn't matter whether the numbers
are true hardware-generated
random values or a pseudorandom sequence generated by a computer.

To play the game, start with a sum of zero. “Draw” a random number from the generator, and add it to the sum. Since the generator always returns numbers less than one (although they can be arbitrarily close to one), the sum after one draw will always be less than one. Tally one number drawn. Now draw another random number from the generator, add one to the tally of numbers drawn, and add it to the sum. If the sum is now greater than one, you've “gone bust”—record the tally of numbers drawn, reset the sum to zero, and continue with another “hand”. While the sum remains less than one, continue to draw numbers, increment the tally, and add them to the sum until it exceeds one, then record the tally at which you went bust.

Here's an example of several “hands” of the game.

Hand 1 | |||
---|---|---|---|

Tally | Draw | Sum | |

0 | 0.0000 | ||

1 | 0.4770 | 0.4770 | |

2 | 0.0516 | 0.5286 | |

3 | 0.1793 | 0.7079 | |

4 | 0.2250 | 0.9329 | |

5 | 0.0438 | 0.9767 | |

6 | 0.3006 | 1.2774 | Bust! |

In this first hand, it took six “draws” from the random number generator before the sum exceeded one. We record the final tally of 6 as the result of the hand and try another.

Hand 2 | |||
---|---|---|---|

Tally | Draw | Sum | |

0 | 0.0000 | ||

1 | 0.8474 | 0.8474 | |

2 | 0.3005 | 1.1479 | Bust! |

This time we went bust on only the second draw. Again, recording the tally of 2, we try again.

Hand 3 | |||
---|---|---|---|

Tally | Draw | Sum | |

0 | 0.0000 | ||

1 | 0.5535 | 0.5535 | |

2 | 0.2721 | 0.8257 | |

3 | 0.5390 | 1.3646 | Bust! |

This time we bust out on the third draw, so we note the final tally of 3. At this point the list of tallies from the three hands so far reads:

Hand | Tally |
---|---|

1 | 6 |

2 | 2 |

3 | 3 |

Now let us compute the average (to be precise, the *arithmetic mean*) of the
number of draws it took to go bust. This is just the sum of the tallies for all
the hands played so far divided by the number of hands. After three hands, we
compute this average (rounded to four decimal places) as:

(6 + 2 + 3) / 3 = 3.6667

Now we can pose the puzzle question:

As the number of hands increases, does the average number of draws per hand converge to a specific value, and if so, what?

- Derbyshire, John. Prime Obsession. New York: Plume, [2003] 2004. ISBN 0-452-28525-9. Pages 366–367.
- Weisstein, Eric W. “Uniform Sum Distribution.” From MathWorld—A Wolfram Web Resource.

January, 2006

*This document is in the public domain.*