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I'm having difficulty with the following problem:

You choose a letter at random from the word Mississippi eleven times without replacement. What is the probability that you can form the word Mississippi with the eleven chosen letters? Hint: it may be helpful to number the eleven letters as $1, 2, . . . , 11$.

This is how I approached it: the number of possible outcomes is $11!$, since we're taking one word at a time without replacement, so there are $11$ choices for the first letter, $10$ choices for the second letter and so on. Now, as to the number of 'success' outcomes, I noticed that there are $4$ s's to choose from, $4$ i's, $1$ M and $2$ p's. So, in order to form the word Mississippi, we have for the first letter $1$ option, $4$ for the second and third letters, $3$ for the fourth (since we've already used one "s") and so on, which amounts to a total of $4^2*3^2*2^3=1152$ different ways of doing so.

However, my answer does not match the one provided in my book (Henk Tijn's Understanding Probability 3rd edition). What am I doing wrong? Thanks very much in advance.

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    $\begingroup$ If you choose eleven times without replacement then you will have all the letters necessary to form the word Mississippi. Or: the order is important? $\endgroup$
    – zoli
    Commented Dec 10, 2017 at 22:42

2 Answers 2

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There are $11!$ permutations of 11 letters but the order of the 4 s's, 4 i's and 2 p's doesn't matter. This means there are $2! 4! 4!$ indistinguisable permutations for any permutation of the 11 letters. Therefore there are \begin{equation} \frac{11!}{2! 4! 4!} = 34650 \end{equation} ways of arranging the letters of Mississippi, making the probability $1/34650$ that a random permuatation spells Mississippi.

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Here is the word $$\begin{matrix}M&I&S&S&I&S&S&I&P&P&I\\1&2&3&4&5&6&7&8&9&10&11\end{matrix}$$with a numbering of the drawings.

1 The probability that you choose letter $M$ the first time is $\frac{1}{11}.$

2 Then, given 1, the probability that you choose letter $I$ the second time is $\frac4{10}$.

From this point on the phrase: "given the result of the previous drawings" will be omitted.

3 The probability that you choose letter $S$ the third time is $\frac49$.

4 The probability that you choose letter $S$ the fourth time is $\frac38$.

5 The probability that you choose letter $I$ the fifth time is $\frac 37$.

6 The probability that you choose letter $S$ the sixth times is $\frac26$.

7 The probability that you choose letter $S$ the seventh times is $\frac15$.

8 The probability that you choose letter $I$ the eighth time is $\frac24$.

9 The probability that you choose letter $P$ the ninth time is $\frac23$.

10The probability that you choose letter $P$ the tenth time is $\frac12$.

11The probability that you choose letter $I$ the eleventh time is $1$.

So, the probability that you get the word again is

$$\frac{1}{11}\frac4{10}\frac49\frac38\frac 37\frac26\frac15\frac24\frac23\frac12=\frac{2(4!)^2}{11!}$$

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  • $\begingroup$ "The probability that you choose letter $I$ the second time" given that you selected an $M$ the first time. Without that added phrase, the probability of choosing an $I$ the second time would be $\frac{4}{11}$ instead of $\frac{4}{10}$. Similarly, the rest of the steps phrasings could be modified. $\endgroup$
    – JMoravitz
    Commented Dec 10, 2017 at 23:19
  • $\begingroup$ @JMoravitz: Yes, Thx, I'll edit. $\endgroup$
    – zoli
    Commented Dec 10, 2017 at 23:20
  • $\begingroup$ @zoli Your answer makes sense to me. However, according to my book, the correct probability is $\binom{11}{4}·\binom{7}{4}·\binom{3}{2}·4^4·4^2·2^2·1/11^{11}=0.0318$. Could you help me understand the reasoning behind this answer? Thanks. $\endgroup$
    – Sasaki
    Commented Dec 28, 2017 at 23:48
  • $\begingroup$ @Sasaki The equation isn't even valid, the left and right hand sides of the equation are not equal. $\endgroup$
    – jodag
    Commented Jan 8, 2018 at 5:17

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