[機統] 基本機率事件 - 交大電子所乙組消失
Two gamblers play the game of "head or tails" . Each time a coin lands
heads up , then player A wins $1 from B ; otherwise (i.e. a coin lands
tails up) , player B wins $1 from A . Suppose that player A initially has
a dollars and player B has b dollars.
(b)If the game is not fair and on each play , player A wins $1 from B
with probability p , p is not 1/2 , 0 < p < 1 . What is the probability that
player A will be ruined?
解答 : A贏1$的機率為p，B贏1$的機率為q = 1-p，則
A輸光的機率 = Pa = [1 - (q/p)^a] / [1 - (q/p)^a+b)]
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(a)小題問A輸光的機率，答案是b/a+b，因為可以理解所以沒問這題
(b)小題中 想知道為何有(q/p)^a 以及(q/p)^a+b
道底是套甚麼式子?? 謝謝!!
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