问题描述
目前我使用均匀分布来生成获胜概率.
Currently I'm using a uniform distribution to generate a winning probability.
假设概率为 1% 我定义一个中奖号码 47 然后我做一个 mt_rand(1,100) 并且如果号码是 47 用户获胜.哪个没问题.
For a probability of let's say 1% I define a winning number 47 then I do a mt_rand(1,100) and if the number is 47 the user win. Which is fine.
这对于像 1/100'000 这样的小概率来说效果很好，但是当我想要一个概率为 40% 即 1/0.4 = 2.5
This work well for small probability like 1/100'000 but when I want a probability of let's say 40% which is 1/0.4 = 2.5
我不能做 mt_rand(1,2.5) 我必须做 mt_rand(1,2) 或 mt_rand(1,3)分别表示 50% 和 33%.
I cannot make mt_rand(1,2.5) I have to do mt_rand(1,2) or mt_rand(1,3) which mean respectively 50% and 33%.
我该怎么做才能获得 40% 的概率?
How should I do to get a 40% probability?
推荐答案
if (mt_rand(1, 100) / $probability <= 1) {
// success
}
例如如果你现在有 $probability = 10;您有十个(共一百个)案例，其中 1 到 100 之间的数字小于或等于 1.
E.g. if you have now $probability = 10; you have ten (of hundred) cases where the number between 1 and 100 is smaller or equal to one.
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