Generate 15 random letters , what is the probability we can spell MISSISSIPPI

combinatoricsprobability

Suppose 15 characters are generated one by one, what is the probability we can rearrange the characters to spell MISSISSIPPI?

My answer was

$${15\choose 11}\times{11! \over 4!4!2!1!}\times{26^{-15}}.$$

Just wanted to clarify if this was correct. This is because we have the multinomial coefficient corresponding to the number of ordered sequences associated with the multi-set of MISSISSIPPI and 15C11 different ways of selecting where to place these letters.

Best Answer

I would approach this problem via inclusion-exclusion based on the events:

  • $M$: we have strictly fewer than $1$ M selected

  • $I$: we have strictly fewer than $4$ I's selected

  • $S$: we have strictly fewer than $4$ S's selected

  • $P$: we have strictly fewer than $2$ P's selected

The probability of failure then is $Pr(M\cup I\cup S\cup P) = Pr(M)+Pr(I)+Pr(S)+Pr(P)-Pr(M\cap I)-Pr(M\cap S)-Pr(M\cap P)-Pr(I\cap S)-\dots+\dots-Pr(M\cap I\cap S\cap P)$

You should be able to calculate each of $Pr(M),Pr(M\cap I),Pr(M\cap I\cap S),\dots$ and complete the calculations that way, though this will be rather tedious to do as you will have to potentially use case-work on the exact number of I's and S's appearing, etc...

For example, $Pr(M\cap I)$ is the probability no M's and at most $3$ I's were used. Breaking into cases based on the number of I's used we have $Pr(M\cap I) = \binom{15}{0}\left(\frac{24}{26}\right)^{15} + \binom{15}{1}\left(\frac{1}{26}\right)\left(\frac{24}{26}\right)^{14}+\dots+\binom{15}{3}\left(\frac{1}{26}\right)^{3}\left(\frac{24}{26}\right)^{12}$

Doing it like this though, you might be breaking $M\cap I\cap S\cap P$ into $32$ events, which is incredibly tedious.

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