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

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

Guest May 26, 2014

#6
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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!.

Bertie  May 27, 2014
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#1
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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:

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

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

Alan  May 26, 2014
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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.

Bertie  May 27, 2014
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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 )

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

Melody  May 27, 2014
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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.

Bertie  May 27, 2014
#5
+91435
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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.

Melody  May 27, 2014
#6
+889
+5

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!.

Bertie  May 27, 2014

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