先手还是后手？
To Begin or Not to begin?
题目详情
A、B 两人轮流从袋子里不放回摸球。袋中有 个黑球和 1 个红球。
摸到红球的人获胜。
问：先手更占优还是后手更占优？
提示：先看 。
A & B are alternately picking balls from a bag without replacement. The bag has k black balls and 1 red ball. Winner is the one who picks the red ball. Who is more likely to win, the on who starts first, or second?
Hint
Look at k=0,1,2,3... Write down probability of starter being a winner.
解析
结论：
- 当 为奇数时，先手胜率为 ，先后手一样。
- 当 为偶数时，先手胜率大于 ，先手更占优。
直观解释：摸球等价于把 个球随机排成一列，红球落在奇数位则先手赢、偶数位则后手赢。若总数为偶数（ 奇），奇偶位数量相同；若总数为奇数（ 偶），奇数位多 1 个，因此先手更大概率拿到红球。
Original Explanation
Begin!
Solution
A & B keep on picking the balls without looking at their color. After all balls have been picked, the one who starts the game will have more balls (if k=even, total balls=odd) and hence higher probability of winning.
(prob of A winning in first chance = 1/(k+1)) + (prob of A winning in 3rd chance = (1-(1/k+1)) * (1-1/k) * 1/(k-1) = 1/(k+1)) + ... + (prob of A winning in 2r+1-th chance = 1/(k+1)) + ... . When k=2n+1, there are n+1 such terms, giving the prob as (n+1) * 1/(k+1)=1/2. When k=2n, there are n+1 such terms, giving prob as (n+1) * 1/(k+1)=(n+1)/(2n+1)>1/2. Hence, doesnt matter who starts first when k is odd. The first player has higher chance of winning when k is even