(adsbygoogle = window.adsbygoogle || []).push({}); Suppose we have "n" envelopes and "n" letters and we put letters to envelopes at random.

Let I_i = I({i-th letter matches}) =1 if the i-th letter matches and 0 otherwise.

Then I_i ~ Bernoulli(1/n) for i=1,2,...,n.

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My questions:

1) So we put letters to envelopes at random one by one.

I can see that when we put the FIRST letter into an envelope(i.e. i=1), then I_1 = I({1-st letter matches})~ Bernoulli(1/n).

But I don't understand why I_i ~ Bernoulli(1/n) for i=2,3,...,n. When we put the second letter into an envelope, P({2-nd letter matches}) depends on wether the first letter matched or not, right? And P({3-rd letter matches}) depends on wether the first and second letter matched or not...etc. Then why would I_i ~ Bernoulli(1/n) still be true for i=2,3,...,n ? Could someone explain, please?

2) I assume the result I_i ~ Bernoulli(1/n) for i=1,2,...,n is true. Since p=1/n is the same for i=1,2,...n, it seems to me that the probability of a success stays constant from trail to trail at 1/n. Is it true that the "number of matches" ~ Binomial(n,1/n)? Why or why not?

Any help is greatly appreciated!

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# Envelope-letter matching problem

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