Post-hoc t-tests for ANOVA My sociology professor said that when performing multiple t-tests between two groups after the ANOVA f-test, the likelihood to make at least one type-1 error adds up by 5% with each t-test. So for one test it's 5% (assuming a threshold of p<0.05), for two tests it's 10% and so on.
I commented that it cannot be the case (like that you could surpass 100% chance after 20 tests). The likelihood to make at least one error would be 1-(95%*Number_of_t-tests). He said he would check this with an expert. Now he got back to me saying that my argument is not correct because the post-hoc t-tests are not independent of one another (he mentioned something about a binomial distribution).
He wrote on the board X=m*p, where X is now the expected value of the error, 'm' is the number of t-tests, and p is the likelihood to make at least one error, which according to him still adds up as I mentioned. So for 30 t-tests, we get X=30*1.5. When performing 30 t-tests, we expect to have 45 errors and the likelihood to make at least one error is 150% (huh?)
Am I missing something?
 A: You are not missing something. You have a poorly informed professor. The truth is, unless we believe all of the post hoc test are independent, then there is no way to precisely calculate the probability of a type-I error. 
If we do assume they're independent, you could model them with a Poisson distribution. So if you run 45-post hoc tests (with a alpha of .05), the expected number of false positives is 2.25.
Again, if we assume independence, we can use the probability mass function associated with the Poison distribution to calculate the probability of getting one or more errors in a certain time period.
The probability mass function is: $f(k; \lambda) = Pr(X=k) = ( \lambda^k e^ {-\lambda} )/(k!)$
In this case lambda is the number of tests we're running. Let's say 45.
So, 
$ Pr(k>1) = \sum_{ k=1}^{\infty}  ( 2.5^k e^ {-2.5} )/(k!)$
$ = 1 - Pr(k=0) =1 - ( 2.5^0 e^ {-2.5} )/(0!)  = 1- e^{-2.5} = 0.9179 $
So we can see using this calculation our probability approaches 100%, but we don't do absurd things like your professor. Again, this assumes independence, though, which is not appropriate, which is why there is not single method that is agreed on for p-value adjustments for mutliple adjustments. Your professor would probably favor the Bonferrnoi test, which simply divides your alpha value by the number of tests, but there are many, many different ways to adjust for type-I error.
