[BruceZ]

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It's true that the chance four randomly picked letters will be some permutation of "hues" is (4*3*2*1)/(26^4). But it's not true that the chance it will be a permutation of either "hues" or "hews" will be twice that.

[/ QUOTE ]

It is precisely twice that because these are mutually exclusive.


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If that were true, then C(26,4)*(4*3*2*1)/(26^4) would have to equal 1. In fact it equals 3.93. The problem is that some of the random 4 letter combinations have some letters the same.

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This is the sum of the probabilities of the C(26,4) letter combinations with 4 distinct letters. It is not 1 because this is less than the total number 26^4 of letter combinations, some of which have repeat letters.


I agree that Everett's method does not work though. From Everett's post:

[ QUOTE ]
There are (36 choose 4) ways to pick 4 letters in a line. The chance that a given set of 4 letters is some scrambled form of "hues" is (4*3*2*1)/(26^4). Same for "hews" so the chance that a given set of 4 letters matches either is 2*(4*3*2*1)/(26^4) (let's call this number P for simplicity). So the chance that a given set doesn't match either is 1-P. The chance that a line contains no instances of these words is (1-P)^(36 choose 4). (This is based on the assumption that knowing that one 4 letter combination does not contain "hues" or "hews" does not affect the chances of another combination containing them).

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This last assumption is the problem. The C(36,4) sets of 4 letters are certainly not independent of each other with respect to containing "hues" or "hews". This would be the case if they did not overlap. Since they share the same letters, the probability of one set containing these letters is strongly dependent on other sets containing these letters. Aaron has given an example of two sets which share 3 of the 4 letters. If one is "hues" and the other is "hue_", then the probability that the second is also "hues" has jumped from (1/26)^4 to just 1/26.


[ QUOTE ]
And the chance that one in 154 monkeys succeeds is:
1 - (1 - (1 - (1-P)^(36 choose 4))^14)^154)
Note that the bold section here is (hopefully) equal to the P that BruceZ used, if that makes the formula easier to read.

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I compute your value in bold to be [1 - 2*(4*3*2*1)/(26^4)]^C(36,4) =~ 0.2%, while my P is about 60.84%.