SOLUTION: A box contains 11 black balls and one red ball. Ramon and Paola, without looking, draw balls without replacement from this box until one of the two removes the red ball. The game i

Question 887119: A box contains 11 black balls and one red ball. Ramon and Paola, without looking, draw balls without replacement from this box until one of the two removes the red ball. The game is won by the player who removes the red ball. Ramon is a gentleman and allows Paola decide if she wants to start or not. Paola has a feeling that she might have a better chance of winning if she starts; after all, she could win on the first try. On the other hand, if in her first attempt she loots a black ball, then the probability that Ramon loots red ball on his first attempt increases, because then there is one less black ball in the box. Should Paola start the game to maximize her chances of winning?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the probability of getting red on the first draw is 1/12

the probability of getting red on the second draw is 11/12 * 1/11 = 1/12

this is because the 11 in the numerator and the denominator cancel out and you are left with 1/12.

the probability of getting red on the third draw is 11/12 * 10/11 * 1/10 = 1/12

this is because the 11 and 10 in the numerator and the denominator cancel each other out and you are left with 1/12.

the probability of getting red on the fourth draw is 11/12 * 10/11 * 9/10 * 1/9 = 1/12

this is because the 11 and 10 and 9 in the numerator and the denominator cancel each other out and you are left with 1/12.

as strange as it may seem, the probability of getting red on any draw is always 1/12.

while it is true that the probability of getting red on the second draw given that black was drawn on the first draw is 1/11, you don't know that black would be drawn on the first draw so it is not given.

in my mind, you should go first.

here's an answer from the web that supports the fact that the odds are the same, although i would still recommend going first because the probabilities are the same so why wait and give the other person first shot at it?

http://math.stackexchange.com/questions/866243/starting-first-or-second-would-have-better-chance-in-this-game

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