如果6,8的顺序不固定的话,赌场赢的概率大 顾客赢的概率:18/43 赌场:25/43 ? |
好像不对呀? 怎么算的? |
Let P be the probability that you will win. Let P7,P6,P8 be the probability that you will win given the last round the dices add up to 7,6 and 8 respectively. Then we end up with the following system of linear equations: P = 6/36 P7+5/36 P6 + 5/36 P8 +20/36P P7 = 6/36+5/36P6+5/36P8+20/36 P P6 = 20/36 P+5/36 P6+6/36P7 P8 = 20/36 P+5/36 P8+6/36P7 |
(昨晚 我以为双方各扔自己的,看谁得到所要的数字概率大。)因为除了6,7,8 这三个和,其它对双方都没有意义,所以就只考虑这3个数字出现的情况。 设出现了7以后顾客赢的概率为P1, 出现了6 或 8以后顾客赢的概率为P2, 出现了7和6(或8)以后顾客赢的概率为P3 P3=3/8+5/16*P3, P3=6/11 P1=3/8+5/8*P3, P1=63/88 P2=3/8*P3+5/16*P2, P2=6/11*P3=36/121 则顾客赢的概率为 P=3/8*P1+5/8*P2=3/8*63/88+5/8*36/121=0.4545 不知对不对? |
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