我想用wolframalpha找出一条线y = ax + b通过point [2,8]的概率，当a和b由公平的掷骰子决定时。
这就是我想要的：
Count[Flatten[Table[a 2 + b, {a,6},{b,6}]],8]/ Length[Flatten[Table[a 2 + b, {a,6},{b,6}]]]，但我不喜欢重复。 我不完全确定为什么以下不会工作：
Count[x, 8]/Length[x] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]我能解决这个问题吗？发生了什么？
I want to use wolframalpha to find the probability of a line y = a x + b passes through the point [2,8], when a and b are determined by fair dice roll.
This does what i want:
Count[Flatten[Table[a 2 + b, {a,6},{b,6}]],8]/ Length[Flatten[Table[a 2 + b, {a,6},{b,6}]]], but I don't like the repetition. I'm not fully certain why following will not work:
Count[x, 8]/Length[x] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]Can i get around this and what is happening?
最满意答案
评估顺序不是你想要的：
Count[x, 8]/Length[x] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]/.的左侧/. 在替换前评估，因此变为： Indeterminate
你需要推迟评估。 正常的方法是使用“纯函数”。 请参阅功能＆和插槽＃ ：
Count[#, 8]/Length[#] & @ Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]可以强制ReplaceAll（简写形式）工作，但它是非标准的：
Unevaluated[ Count[x, 8]/Length[x] ] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]这里的评估Unevaluted让左侧过早评估。
The order of evaluation in this is not what you desire:
Count[x, 8]/Length[x] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]The left side of /. evaluates before replacement, and therefore becomes: Indeterminate
You need to delay evaluation. The normal method for this is to use a "pure function." See Function & and Slot #:
Count[#, 8]/Length[#] & @ Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]It is possible to force ReplaceAll (short form /.) to work, but it is nonstandard:
Unevaluated[ Count[x, 8]/Length[x] ] /. x -> Flatten[Table[a 2 + b, {a, 6}, {b, 6}]]Unevaluted here keeps the left-hand side from evaluating prematurely.
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