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I have $100$ keys and $100$ safes. Each key opens only one safe, and each safe is opened only by one key. Every safe contains a random key. 98 of these safes are locked. What's the probability that I can open all the safes?

This question is confusing for me. Can you walk me through it step-by-step?

user72273
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1 Answers1

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In order to open the safes, there must be at most two 'cycles' of safes and keys, and one unlocked safe in each cycle. Let $A \to B$ denote that safe A unlocks safe B. For example, if 1 and 98 are unlocked,

$$1 \to 2 \to 3 \to 4 \cdots \to 72 \to 1 \textrm{ and } 98 \to 99 \to 100 \to 73 \to 74 \to 75 \cdots \to 97 \to 98$$

is unlockable, but not if 1 and 2 are unlocked.

How many such pairs of cycles can you make? Keep in mind that there must be an unlocked safe in each cycle.

This should be $\frac{99}{2} \cdot 100!$ Say we have a cycle of size $n$. There are $\binom{100}{n}$ safes we can put in that cycle, and $n!$ ways we can arrange them. There are $(100 - n)!$ ways to arrange the other cycle. So there are $\binom{100}{n}n!(100-n)! = 100!$ arrangements where a cycle is size $n$. Thus, for all cycle sizes, we have $\sum_{i=1}^{99} 100! = 99 \cdot 100!$. But since we double-counted everything (every time we count a cycle of size $n$, we count one of size $100-n$ too), we must divide by 2: $\frac{99}{2} \cdot 100!$

The other possibility is that they form one large cycle, like $1 \to 2 \to 3 \to 4 \cdots \to 100 \to 1$. Count those up too. If there is a large cycle, then all combinations of unlocked safes allow everything to be unlocked.

This is (oddly enough) also $\frac{99}{2} \cdot 100!$. Say one of our unlocked safes is at the beginning of the sequence. There are $100!$ such sequences. Each of those has a possible 99 choices for the other unlocked safe. Again, we double count: a sequence starting with 1, where 2 is chosen, is equivalent to starting with 2, with 1 chosen. So we have $\frac{99}{2} \cdot 100!$ This means we have a total of $99 \cdot 100!$ arrangements that are unlockable.

Then, we compute how many arrangements are possible, unlockable or not.

$4950 \cdot 100!$. There are 100 keys we can put in the first box, 99 in the second, etc. Then we choose our unlocked safes: $\binom{100}{2} = 4950$ possible combinations. So the chance your arrangement is unlockable is $\frac{99 \cdot 100!}{4950 \cdot 100!} = \frac{1}{50}$, which is higher than I expected.

I think this generalizes nicely, actually:

It's always $\frac{2}{n}$. Weird.

Henry Swanson
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  • Nice answer but this part seems wrong: "There are $\binom{100}{n}$ safes we can put in that cycle, and $n!$ ways we can arrange them." When you have $n$ safes not every arrangement is guaranteed to be just one loop... – timidpueo Apr 14 '13 at 13:44
  • I wasn't super clear about it, but I meant "there are $\binom{100}{n}$ cycles of this size. And note that I'm really picking where the keys go, so that a cycle forms; the safes themselves don't move. – Henry Swanson Apr 14 '13 at 14:45
  • Wow, thanks for explaining this so that a high school student could understand it. Someone I was talking to in real life said something about Markov Chains, which is way beyond me. – user72273 Apr 15 '13 at 10:07
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    With respect to your last comment: I showed in another answer here that for general $k$ and $n$, the answer is always $k/n$. – ShreevatsaR Apr 15 '13 at 17:30
  • @ShreevatsaR While I'd like to study your answer, it's too advanced for me I think. The answer by Mr. Swanson makes sense to me, and so is useful to me. Maybe this question should be revised to "explainable at a high school level". – user72273 Apr 15 '13 at 19:11
  • @user72273: Indeed, that answer is for the more general case ($k$ and $n$ instead of $2$ and $100$), but I believe it is at the high school level (and, once understood, actually simpler than this explicit counting). Could you tell me what part of it you found hard to understand, so that it will help me make the exposition clearer too? (Are you familiar with the notation: see this and this?) – ShreevatsaR Apr 16 '13 at 05:07
  • Oh wow that's beautiful. @user72273, definitely take a look at this, it's much more general than my answer. – Henry Swanson Apr 16 '13 at 14:31