Question 887119
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?


<a href = "http://math.stackexchange.com/questions/866243/starting-first-or-second-would-have-better-chance-in-this-game" target = "_blank">http://math.stackexchange.com/questions/866243/starting-first-or-second-would-have-better-chance-in-this-game</a>