模拟一个概率问题:3个独立的骰子
模拟一个概率问题:3个独立的骰子
提问于 2022-02-06 13:34:59
我决定模拟教科书中的概率问题：
三个公平的骰子是独立滚动的，一个骰子显示6，另两个显示两个不相等数的概率是多少(两者都不等于6)？
假设骰子是公平的，所以“理论”答案将是$\frac{2}{2}{2}{3}=0.5$；我决定在Julia中模拟这一点，下面是我为此编写的函数：
function simulation(n_experiments::Int = 100)
dice = [DiscreteUniform(1, 6) for i in 1:3] # 3 independent fair dice
successes = 0
for trial in 1:n_experiments
experiment = rand.(dice) # roll
one_six = sum(r == 6 for r in experiment) == 1 # check if there is only one "6"
others = filter(num -> num != 6, experiment)
no_duplicates = length(Set(others)) == 2 # check if other two are distinct
if no_duplicates
if one_six
successes += 1 # count "success"
end
end
end
return successes / n_experiments
end我预计这会返回0.5左右的内容，但实际上它在大约10^5次迭代(n_experiments)后返回~0.27。有人能帮我找出密码中的漏洞吗？
回答 1
Stack Overflow用户
回答已采纳
发布于 2022-02-06 14:08:22
我认为你的代码是对的，但你的数学错了：
julia> function simulation(n_experiments=100)
chains = rand.(fill(DiscreteUniform(1, 6), 3), n_experiments)
return (1/n_experiments) * mapreduce(+, chains...) do d1, d2, d3
(d1 == 6 && d2 != 6 && d3 != 6 && d2 != d3) || (d1 != 6 && d2 == 6 && d3 != 6 && d1 != d3) || (d1 != 6 && d2 != 6 && d3 == 6 && d1 != d2)
end
end
simulation (generic function with 1 method)
julia> simulation(10_000)
0.279
julia> count(Iterators.product(1:6, 1:6, 1:6)) do (d1, d2, d3)
(d1 == 6 && d2 != 6 && d3 != 6 && d2 != d3) || (d1 != 6 && d2 == 6 && d3 != 6 && d1 != d3) || (d1 != 6 && d2 != 6 && d3 == 6 && d1 != d2)
end
60
julia> 60 / 6^3
0.2777777777777778我希望能找到一种简洁的方式来对情况进行编码。因为分配，我避免了你的变体，我的只是详细的确保它是正确的--但是太长了.
附录
下面是具有生成条件的优化变量。与问题无关，而是因为我觉得它很酷。条件公式的功劳属于Bogumił。
julia> @generated function condition(xs::NTuple{N,<:Any}) where {N}
offdiagonals = ((i, j) for i = 1:N for j = (i+1):N)
names = Symbol.(:xs, 1:N)
assignments = [:($(names[i]) = xs[$i]) for i = 1:N]
comparisons = [:($(names[i]) != $(names[j])) for (i, j) in offdiagonals]
condition = Expr(:&&, :(maximum(xs) == 6), comparisons...)
return quote
$(assignments...)
$condition
end
end
condition (generic function with 1 method)
julia> @code_warntype condition((1,2,3))
MethodInstance for condition(::Tuple{Int64, Int64, Int64})
from condition(xs::Tuple{Vararg{var"#s17", N}} where var"#s17") where N in Main at REPL[71]:1
Static Parameters
N = 3
Arguments
#self#::Core.Const(condition)
xs::Tuple{Int64, Int64, Int64}
Locals
xs3::Int64
xs2::Int64
xs1::Int64
Body::Bool
1 ─ (xs1 = Base.getindex(xs, 1))
│ (xs2 = Base.getindex(xs, 2))
│ (xs3 = Base.getindex(xs, 3))
│ %4 = Main.maximum(xs)::Int64
│ %5 = (%4 == 6)::Bool
└── goto #7 if not %5
2 ─ %7 = (xs1 != xs2)::Bool
└── goto #6 if not %7
3 ─ %9 = (xs1 != xs3)::Bool
└── goto #5 if not %9
4 ─ %11 = (xs2 != xs3)::Bool
└── return %11
5 ─ return false
6 ─ return false
7 ─ return false
julia> function simulation(n_experiments=100)
(1/n_experiments) * sum(1:n_experiments) do _
dice = (rand(1:6), rand(1:6), rand(1:6))
condition(dice)
end
end
simulation (generic function with 2 methods)
julia> @time simulation(10_000_000)
0.157303 seconds
0.2777498这根本没有拨款。sum就像手动循环一样快！
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原文链接：
https://stackoverflow.com/questions/71007705
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