Probability of At Least 2 Identical Letters in 4-Letter String

[Seda]

Homework Statement

What is the probability of having at least two identical letters in a randomly selected string of four letters.

Homework Equations

None, except maybe a basic idea of permutations/combinations

The Attempt at a Solution

Well, probability seems like an easy topic, but I'm having trouble on this one.

First off, I'm assuming that the letters are picked simultaneously.

Well, I can see that obviously, the total number of possibilities is 26^4.

ANd I know I need to account for having 2 letters the same, 3 letters the same, and 4 letters the same.Well, I know how to do this problem easy if I was given a specific letter to worry about. The result would be like this:Probability = [25/26 X 25/ 25 X 1/26 X 1/26] + [25/26 X (1/26)^3] + [(1/26)^4]

---------------^2 the same -------------------------^ 3 the same ------ ^ all four the sameBut this would only be the answer if I was given a specific letter to worry about being the same. How do I solve when it can be any letter?

 

[sutupidmath]

whad does a string of four letters consist? What is that?

 

[Seda]

4 letters chosen at random.


aghi

iopl

hujg

futy



etc...

 

[kamerling]

Well, I can see that obviously, the total number of possibilities is 26!

No. The total number of possibilities = (number of ways to choose 1st letter) * (number of ways to choose second letter) * etc.

ANd I know I need to account for having 2 letters the same, 3 letters the same, and 4 letters the same.

It might be easier to use 1 - (probability of no letters the same)

 

[Seda]

Sorry, I meant to say 26^4, i have no idea why I said factorial...ill edit that...

 

[sutupidmath]

4 letters chosen at random.


aghi

iopl

hujg

futy



etc...

Blahhhh, damn it, i confused it with the meaning of "letter" in my native language, because it means completely sth else, and it made no sens to me.

 

[Seda]

Hmm


How would I figure the probability of none of the letters being the same if the letters are chosen simultaneously?

If the letters were "rolled" in order, I could easily do (26*25*24*23)/(26^4)...but 1 - that = about 21% and that seems pretty high...

 

[Seda]

Im stumped

 

[Redbelly98]

It might be easier to use 1 - (probability of no letters the same)

I'd say that is exactly the simplest way to solve this one, so the math is really in finding the probability that all letters are different.

To get started, figure out the probability that 2 letters chosen at random are different. Then go on to 3 letters, and finally 4 letters.

 

[Tedjn]

Well, the easiest way I find to think about probability problems is to go back to definitions, which is to say,

probability of no letters the same = (# of strings with four letters different) / (# of total four letter strings)

 

[Redbelly98]

True Tedjn, it's easier to conceptualize the way you put it.