What's the probability of a pangram in a crossword occuring by chance? There are several sources of the percentage occurence of each letter in the English language.
So is it possible using these to calculate the probability of a crossword containing each letter of the alphabet at least once. (A pangram)
I assumed a 15X15 crossword and that 160 of the 225 squares had letters in them and all entries are real words.
I tried the approach of calculating the chance of at least one of each letter occurring over 160 squares. 1 -((1 - percentage of letter) ** 160).
I then took the product of these 26 values which came to 0.000300021186131477.
This seems far too low! Wht am I doing wrong?
 A: For one, the model used may be wrong due to the many interdependencies in general (it is not just very unlikely, it is impossible that none of the letters occurs) and especially those inherent in a cross-word, but let's ignore this (as it is hard to account for it in any way that could be called right and we may assume the error is quite small). Also, crosswords are hand-crafted and may have other letter frequencies than plain English texts.
If all 26 letters had the same frequencies, you'd have $1-(\frac{25}{26})^{160}\approx 0.9981$ and this to the 26th power is $\approx 0.95\approx 1$. In other words: For a rough estmiate we may ignore the contribution of common letters, we have to concentrate on rare letters, which can spoil the high result.
Most letters have a frequency $>0.01$, but the rarest is Z with a frequency of $0.074\%=0.00074\approx 0.11$. Likewise, the next rarest letter Q contributes a factor $1-0.99905^{160}\approx0.14$, then X contributes $1-0.9985^{160}\approx 0.21$. Already the product of these numbers is merely $\approx0.003$. It is not surprising that the combined effort of the remaining (rare) letters can decrease this number by another factor of $0.1$ to the $0.0003$ you found.
