猜测（匹配）Guid 的概率是多少？
只是好奇，但匹配 Guid 的概率是多少？
说出 SQL Server 中的 Guid：5AC7E650-CFC3-4534-803C-E7E5BBE29B3D
它是阶乘吗？：(36*32)！ =（1152）！
讨论 =D
Just curious but what is the probability of matching a Guid?
Say a Guid from SQL server: 5AC7E650-CFC3-4534-803C-E7E5BBE29B3D
is it a factorial?: (36*32)! = (1152)!
discuss
=D
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目前尚不清楚你在问什么。我看到有两种方法可以解释你的问题。
给定一个 GUID
g，有人猜到它的概率是多少？为了简单起见，我们假设 GUID 的所有 128 位都可用。那么猜测g的概率是2^-128。那很小。让我们对此有一些直觉。假设我们的攻击者每秒可以生成 10 亿个 GUID。为了有 50% 的机会猜测g，我们的攻击者必须生成 2^127 个 GUID。以每秒 10 亿的速度计算，需要 5391448762278159040348 年才能生成 2^127 个 GUID。我们正在生成一个指南集合。发生碰撞的可能性有多大？也就是说，我们生成两个具有相同值的指导的可能性有多大？这就是生日悖论。如果随机生成一系列 n 个 GUID，则至少发生一次冲突的概率大约为
p(n) = 1 - exp(-n^2 / 2 * 2^128)（这是生日问题（可能的生日数量为 2^128）。np（n）
2^30 1.69e-21
2^40 1.77e-15
2^50 1.86e-10
2^60 1.95e-03
因此，即使生成 2^60 个 GUID，发生冲突的可能性也非常小。如果每秒可以生成 10 亿个 GUID，则仍然需要 36 年才会有 1.95e-03 的碰撞机会。
It's not clear what you're asking. I see two ways to interpret your question.
Given a GUID
g, what is the probability of someone guessing it? Let's assume for simplicity that all 128 bits of a GUID are available. Then the probability of guessinggis2^-128. That's small. Let's get some intuition around that. Let's assume that our attacker can generate one billion GUIDs per second. To have a 50% chance of guessingg, our attacker would have to generate 2^127 GUIDs. At a rate of one billion per second, it would take 5391448762278159040348 years to generate 2^127 GUIDs.We are generating a collection of guids. What is the likelihood of a collision? That is, what is the likelihood that we generate two guids with the same value? This is the birthday paradox. If you generate a sequence of n GUIDs randomly, then the probability of at least one collision is approximately
p(n) = 1 - exp(-n^2 / 2 * 2^128)(this is the birthday problem with the number of possible birthdays being 2^128).n p(n)2^30 1.69e-21
2^40 1.77e-15
2^50 1.86e-10
2^60 1.95e-03
So, even if you generate 2^60 GUIDs, the odds of a collision are extremely small. If you can generate one billion GUIDs per second, it would still take 36 years to have a 1.95e-03 chance of a collision.
可能的 GUID 数量（128 位值）为 2^128 或 3.4×10^38 — 地球整个体积的每立方毫米大约为 2 万亿个。
换句话说，有点低。
（地球卷参考来源：维基百科）
The number of possible GUIDs (128-bit value) is 2^128 or 3.4×10^38 — roughly 2 trillion per cubic millimeter of the entire volume of the Earth.
In other words, kind of low.
(Source for Earth volume reference: WikiPedia)
取决于 GUID 生成算法的类型。当前算法使用 124 个随机位，因此概率为 2^124 中的 1。
对于较旧的算法（使用时间和 MAC 地址），概率要高得多。
Depends on the type of GUID generation algorithm. Current algorithms use 124 random bits so the probability is 1 in 2^124.
With older algorithms (that use time and MAC address) the probability is much higher.
您的计算有很多错误。
首先，36*32 意味着任何字母数字字符都可以成为 GUID 的一部分。事实并非如此；只允许使用十六进制字符。
其次，唯一 GUID 数量的计算为
有效字符数 (16: 0-9, AF) ^ GUID 长度 (32 个字符)
所以我们有 16^32 ==> 2^(4^32)==> 2^128
猜测任何一个 GUID 的几率是 1 / 2^128。
There are a number of things wrong with your calculations.
First off, 36*32 implies that any alphanumeric character can be part of the GUID. This is not the case; only HEX characters are allowed.
Secondly, the calculation for the number of unique GUIDs is
Number of Valid Characters (16: 0-9, A-F) ^ length of GUID (32 characters )
So we have 16^32 ==> 2^(4^32) ==> 2^128
The odds of guessing any one GUID is 1 / 2^128.
这取决于生成的 GUID 数量。这类似于统计学中的生日问题。请参阅维基百科和 http://betterexplained.com/articles/understand-the-birthday-paradox / （这个具体有一个 GUID 示例）
一般来说，在 N 个可能的 guid 上生成 M 个 guid 的冲突概率为
1 - (1- (1/N))^C(M, 2)其中C(M,2)是“M 选择 2”It depends on how many GUIDs are generated. This is similar to the birthday problem in statistics. See Wikipedia and http://betterexplained.com/articles/understanding-the-birthday-paradox/ (this one specifically has a GUID example)
In general, the probability of a collision for generating M guids over N possible guids is
1 - (1- (1/N))^C(M,2)whereC(M,2)is 'M choose 2'它是 1 /（给定 UID 长度可能的唯一数字的数量）。在上面的示例中，我们看到 16 个字节，即 128 位。 2^128，因此匹配的概率为 1 / 2^128。
It is 1 / (number of unique numbers possible with the given UID length). In the above example we see 16 bytes, or 128 bits. 2^128, so the probability of a match is 1 / 2^128.
正确的计算方法是 2^128，因为我们处理的是 128 位。
但如果你想使用十六进制编号来计算它，那么你可以将其写为 16^32。
第一个数字有 16 种可能性 * 第二个数字有 16 种 * 第三个数字有 16 种……
在计算概率时，阶乘仅适用于某些组合相等的情况，例如在彩票数字中，156 相当于 561。
希望这会有所帮助。
The proper way to calculate this is 2^128 since we are dealing with 128 bits.
But if you want to calculate it using the hexadecimal numbering then you can write this as 16^32.
16 possibilities for the first number * 16 for the second * 16 for the third etc...
When calculating probabilities, factorials only apply if some combinations are equivalent such as in lottery numbers where 156 is equivalent to 561.
Hope this helps.