Enigmatic Code

Programming Enigma Puzzles

Enigma 1165: Bull’s-eye

From New Scientist #2321, 15th December 2001 [link]

The darts players at George’s local have devised a seeding system for themselves. Each has determined the number of darts he needs to throw, on average, in order to hit the Bull once. This is his rating — the lower the better. Thanks to remarkable foresight in their parents’ choices of names, the seedings proceeded alphabetically from Alan at No. 1 to Zak at No. 26, with no duplication of initial.

Fred and George met in the pub recently and decided to throw alternately, one dart at a time, until one hit the Bull. The other would then pay for the drinks. Although Fred is the more accurate darts player, they calculated from the ratings that if George threw first they would each have an even chance of winning the sudden-death game.

Duly refreshed, they then calculated from the ratings that this would be an even-money game for any pair of players who are adjacent in the seedings, provided the lower seed throws first. But if Zak played the game against Alan and threw first, his chance of winning would be exactly 10 per cent.

What is George’s rating?

[enigma1165]

One response to “Enigma 1165: Bull’s-eye”

1. Jim Randell 30 May 2016 at 7:14 am

I solved this one analytically:

If f and g are the probabilities of hitting the bull for players F and G.

F is the stronger player, so f > g.

Writing P(F, G, F) for “the probability that F gets a bull first, playing against G, F to throw”

P(F, G, G) = (g × 0) + (1 − g) × P(F, G, F)
P(F, G, F) = (f × 1) + (1 − f) × P(F, G, G)

P(F, G, G) = f (1 − g) / (f + g − fg)

but this last value is 1/2, so:

g = f / (f + 1)

So, starting with a, (a > b > c > … > z):

P(A, B, B) = 1/2

b = a / (a + 1)
c = b / (b + 1) = a / (2a + 1)
d = c / (c + 1) = a / (3a + 1)
e = a / (4a + 1)
f = a / (5a + 1)
g = a / (6a + 1)
h = a / (7a + 1)

z = a / (25a + 1)

also:

P(A, Z, Z) = a (1 − z) / (a + z − az) = 9/10

Solving the 2 equations in a and z gives:

a = 1/3, z = 1/28

and:

b = 1/4, c = 1/5, d = 1/6, e = 1/7, f = 1/8, g = 1/9, h = 1/10, …, y = 1/27

So:

A has a 1 in 3 chance of hitting the bull, so a rating of 3;
B has a 1 in 4 chance of hitting the bull, so a rating of 4;
C has a 1 in 5 chance of hitting the bull, so a rating of 5;

Y has a 1 in 27 chance of hitting the bull, so a rating of 27;
Z has a 1 in 28 chance of hitting the bull, so a rating of 28.

So:

Solution: George’s rating is 9.