Riddle -- Probability of Expectation

[BruceZ]

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If one of the letters is an H it really doesn't effect the chance of an E showing up because if it wasn't an H it would still be something other than E.

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But it does affect it. If the first letter is an "h," then there are only 35 chances to get an "e." If it is not an "h," then there is 1/25 instead of 1/26 that it is an "e."

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Doing a binomial distribution calculation or taking an events liklihood to the power of trials takes this into account.

In this example the binomial distribtion of an event with the liklihood of an event that has the probability of occuring equal to 1/26 not occuring in 36 trials is 0.24366872185316. This is the same as (1/26)^36.

I am sorry I don't know how to explain it any better.

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Of course Aaron and alThor are correct. This step is incorrect:

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The chance of all 3 letters occurring in the same 36-character line is

0.756331278146836^3 = 0.432649477099284

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Raising this probability to the 3rd power would only be correct if the events of each letter occuring in 36 trials were independent of each other, and they are not. The selection of each of the 36 letters are independent, but you are confusing that with the overall selection of each letter in 36 trials, and these do not form independent Bernoulli trials, so the binomial distribution does not apply. If one of these letters occurs, then the probability that one of the other letters occurs decreases. You want to multiply P(h occurs)*P(e occurs | h occurs)*P(s occurs | h and e occur). This is not the same as [P(h occurs)]^3 since these events are not independent.

You also cannot multiply the probability that h,e, and s occur by the probability that u or w occur, as you do at the end of the calculation.