To get even chances in the game, Ben must be able to get a bull's eye on his first throw such that he has a probability the same as Bill's on his first go, which is a probability of 10% after Bill has thrown and missed. Bill misses on his first throw with probability 90%. So Ben's probability multiplied by 90% must equal 10%; in other words, Ben must be able to get a Bull's eye once every 9 throws. That is the answer.

If there were three or more players where each player was to have an equal chance of winning, subsequence players would have to be able to hit a bull's eye once every 11-n throws where n is the number of players, for n < 11. Where n = 10, therefore, the last player must get a bull's every time without fail!

No one got the right answer. John Stafford did an excellent mathematical analysis which added up the sequence of probabilities to infinity to prove the result but mixed up who went first so got the wrong answer of 1 in 11. John Bownas did a Monte-Carlo simulation (using a random number generator and applying statistical analysis on the results) but as a consequence only got the approximate answer of 11 in 100 which was almost right.

John Stafford proved that the average length of a game was 5 throws each.

Many thanks for your efforts.

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