Let's say we had the word STATISTICS, and it was hanging on a wall, then two letters in the word blew off, and somebody found them. What are the chances they put them in the correct order?

here is what I have.

A = letters are put in order I = letters are identical

P(A given I) = 1 P(A given I compliment) = 1/2

I want to use the total probability theorem, but I have a hard time grasping what P(I) could be, I'm sure it would be 1 / 10 NCR something, but I'm not sure what to put.

Any ideas?

Here is a solved version of this problem.

Word = CANAL

A = words are in the right order I = words are identical

P(A given I ) = 1 P(A given I^c) = 1/2 P(I) = 1/5 choose 2

P(A) = 1*1/10 + 1/2*9/10

by the total probability theorem.

• Are you assuming the remaining letters stay in place? Note that under this assumption if both lost are the same letter you automatically get the right order. – Michael R. Chernick Dec 9 '16 at 19:53
• So this applies to T, S, I. 3 letters are S, 3 are T and 2 are I. – Michael R. Chernick Dec 9 '16 at 19:56
• Lets look at the CANAL word as it is simpler. Now only the letter A appears twice. So if both letters are A you get it correct. Everything else could lead to a mistake. Suppose you get C and L then the probability of getting it write is 1/2. Two possible spellings CANAL and LANAC. – Michael R. Chernick Dec 9 '16 at 20:40
• If you get C and N the probability is 1/2 with NACAL and CANAL the two possibilities. The logic is the same for N and L. But if you have A and C or A and N or A and L the story is different. These three cases work the same way so lets take A and C, The possible spellings are ACNAL, AANCL and CANAL. The probability here is 1/3. Now these are conditional probabilities and must be multiplied by the probability of the choice of letters. You (the OP) got 11/20. Are you sure that you took all of this into account? – Michael R. Chernick Dec 9 '16 at 20:54
• What are the chances conditional on? The letters that blew off or the mere fact that two blew off? If it's the former, the answer depends on the letters. If it's the latter, the answer depends on the probability distribution you assume concerning which letters blow off. – whuber Dec 9 '16 at 21:01

So here goes another answer using a different interpretation of the question: two letters fall off. They will be replaced randomly. A correct replacement is such that the spelling of the word is unchanged.

$P(A) = P(same\,letters\,fell\,off) + P(different\,letters\,fell\,and\,replaced\,correctly)$

We have 4 bins (S, T, A, I, C) with (3, 3, 1, 2, 1) letters each.

$P(same\,letters\,fell\,off) = P(2\,S\,fell) + P(2\,T\,fell) + P(2\,A\,fell) + P(2\,I\,fell) + P(2\,C\,fell)$ $= 3 / 10 * 2 / 9 + 3 / 10 * 2 / 9 + 0 + 2 / 10 * 1 / 9 + 0 = 14 / 90$

$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 76 / 90$

$P(different\,letters\,fell\,and\,replaced\,correctly) = P(different\,letters\,fell\,off) * P(replaced\,correctly) = 76 / 90 * 1 / 2 = 76 / 180$

$P(A) = 14 / 90 + 76 / 180 = 104 / 180$

Using the word CANAL, buckets are (C, A, N, L) with sizes (1, 2, 1, 1).

$P(same\,letters\,fell\,off) = P(2\,C\,fell) + P(2\,A\,fell) + P(2\,N\,fell) + P(2\,L\,fell)$ $= 0 + 2 / 5 * 1 / 4 + 0 + 0 = 2 / 20$

$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 18 / 20$

$P(different\,letters\,fell\,and\,replaced\,correctly) = P(different\,letters\,fell\,off) * P(replaced\,correctly) = 18 / 20 * 1 / 2 = 18 / 40$

$P(A) = 2 / 20 + 18 / 40 = 22 / 40 = .55$

This was my first (incorrect) interpretation of the OP's question (I've added another answer which I believe has the correct interpretation).

Two letters fall off. Assuming the person replacing them knows how to spell, if different letters fall they will certainly be replaced correctly. If the same letters fall off, they must be replaced in the same order in which they were originally (correct spelling AND correct original position). Under these assumptions, the answer is:

$P(A) = P(different\,letters\,fell\,off) + P(same\,letters\,fell\,and\,replaced\,correctly)$

We have 4 bins (S, T, A, I, C) with (3, 3, 1, 2, 1) letters each.

$P(same\,letters\,fell\,off) = P(2\,S\,fell) + P(2\,T\,fell) + P(2\,A\,fell) + P(2\,I\,fell) + P(2\,C\,fell)$ $= 3 / 10 * 2 / 9 + 3 / 10 * 2 / 9 + 0 + 2 / 10 * 1 / 9 + 0 = 14 / 90$

$P(different\,letters\,fell\,off) = 1 - P(same\,letters\,fell\,off) = 76 / 90$

$P(same\,letters\,fell\,and\,replaced\,correctly) = P(same\,letters\,fell\,off) * P(replaced\,correctly) = 14 / 90 * 1 / 2 = 14 / 180$

$P(A) = 76 / 90 + 14 / 180 = 166 / 180$

• I do not think this is correct, for example the answer to this question if the word was CANAL would be 55% I know this is true, and when I followed your steps it ended up being in the 90's – Itachi San Dec 9 '16 at 20:05
• Then I have misinterpreted your question. Can you please clarify what constitutes a correct substitution? You can use CANAL as an example – Alexandre Dec 9 '16 at 20:18
• Here is how it is solved in the book. – Itachi San Dec 9 '16 at 20:26
• Look at original post I will edit it in. – Itachi San Dec 9 '16 at 20:27
• If the assumption is that the other letters stay in place the person replacing doesn't have to know how to spell. – Michael R. Chernick Dec 9 '16 at 20:29