if a combination has four letters and four numbers what is the probability of someone guessing the combination?

Oops !

My turn to be embarrassed - hope that makes you feel better. I shall blame that cheese on toast that I had for lunch, it always makes me feel burpy.

Yes, the 26^4*10^4 allows for each letter and digit to be repeated, so how can they all be different ?

Rather than 26^4*10^4*8! the number of possible permutations is 26C4 * 10C4 * 8!. 

If there are eight "wheels", four of which have the letters A to Z and four of which have the digits 0 to 9, then there are 26⁴*10⁴ possible combinations, so the probability that a random guess will get the right combination is:

$${\mathtt{p}} = {\frac{{\mathtt{1}}}{\left({{\mathtt{26}}}^{{\mathtt{4}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{4}}}\right)}} \Rightarrow {\mathtt{p}} = {\mathtt{0.000\: \!000\: \!000\: \!218\: \!829\: \!9}}$$

or about 2*10^-10

Of course if a different alphabet is used that has a different number of letters, the probability will be different!

Combination locks should really be called Permutation locks shouldn't they ?

Afterall, if I told you that  the three numbers on a simple lock were 2, 3 and 7 you wouldn't necessarily expect them to be in that order. You would have 3! = 6 different orders to try.

On that basis I think that the previous answer needs to be divided by 8! = 40320.

Okay Bertie, it is a permutation lock.

 Except, the letters are in a set postion and the numbers are in a set postition.

So why would you want to divide by 8!  I don't get it.  

(Okay, I just got it,you are just being padentic about the lay persons word combination )

In your example

Afterall, if I told you that  the three numbers on a simple lock were 2, 3 and 7 you wouldn't necessarily expect them to be in that order. You would have 3! = 6 different orders to try.  Yes that is true

It is a 3 digit tumbler - order obviously matters - the number of possiblilities is 10*10*10=1000

The right combination (lay speak)  is obviously a permution - order matters so the probablilty of getting the right combination is 1/1000

The probablity of getting 2,3 and 7 in any oder is 6/1000

Don't worry - I am really just talking to myself here - I do that.   

If order does not matter than the probability would be greater.  I'd multiply by 8! (not divide) but this would not be right I do not think.  It would be ok if every number/letter were different but what if all the numbers were 7 and all the letters were A then it would not work!  Plus, in Alan's implied scenario the letters occupy 4 specific spots and he numbers occupy a different 4 specific spots.

Ummm  

Yes, in Alan's scenario I was making the (unwarrented) assumption that the four letters and the four digits were all different.

The number of four letter combinations together with four digit combinations is 26^4*10^4, and if it's just the case of guessing what the letters were and what the numbers were, without any consideration of the order, the probability of a correct guess would be 1/(26^4*10^4).

If all of the letters and numbers in each combination are different, then each can be arranged in 8! different orders, giving rise to 26^4 times 10^4 times 8! different permutations, and the probability of someone guessing the correct one will be the reciprocal of this.

Let me think about this.

If all the letters and all the numbers were different then the number of permutations would be:

(This is assuming that number and letter positions are pre-allocated)

26*25*24*23*10*9*8*7=

$${\mathtt{26}}{\mathtt{\,\times\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{23}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{1\,808\,352\,000}}$$

or

26C4*10C4*8P8

$${\left({\frac{{\mathtt{26}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{10}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!} = {\mathtt{1\,808\,352\,000}}$$

Now, if they all have to be different but the number and letter positions are NOT pre-allocated then

$${\left({\frac{{\mathtt{26}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{10}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{8}}{!} = {\mathtt{126\,584\,640\,000}}$$

That is what I think anyway.