>During the birthdayparty of her daughter, a father takes a chololate bar
and
>>shows it to the ten girls that are standing in front of him.. He says, "I
>>will think of a number between 1 and 10. (1 and 10 included) Whoever
guesses
>>the number is the winner of this chocolate bar.. Jane.. You may start"
>>Jane - al excited - says "Four"
My friends and I struggled with this problem over many beers. I strongly
beleive that if you assume that there is no forgetting of wrong answers all
positions are of equal value.
Girl one has a 1/10 chance of guessing correctly. Girl two has a 9/10
chance of getting a turn and a 1/9 th chance of winning if she gets a turn.
The compround probability is 1/10. Girl threes chance is 9/10*8/9*1/8 =
1/10. The pattern persists.
My friends agreed with the math, but inisited that an intermediate position
was still preferable to on of the extremes.
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