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Math Question for computer Probability - Trainer ID in a videogame
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I think the best way to cope with this is to divide it into two cases, one where the first digit is 5 and one where the first digit is 6.

If the first digit is 5, the other four digits can be chosen from {5, 6, 7, 8, 9}, so clearly there are 5^(4) = 625 examples in this case.

I think there are, in fact, *no* examples with the first digit 6, which would make the answer to your question 625/65536, which is a smidgeon less than 1%. (That is, trying to grab such an ID will require you to try one or two hundred times.)
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Your minimum and maximum serve as a great jumping off point for the question. We can put an instant upper bound on the answer using those two figures - 4,444 possible options, narrowed down just from that list.

Let's take the minimum and work from there:

55,555  
5|5|5|5|5 (easier for me, at least, to see it with some pips between the digits)  
We know we can't decrement either the first or the second number, but we CAN increase the second number, so lets think of it like that. We don't actually care about WHAT the number is so we can abstract that away:

A (5) | B (5,6,7,8,9) | C (5,6,7,8,9) | D (5,6,7,8,9) | E (5,6,7,8,9)

Now we have our list of permutations. Let's see if we can find a pattern to work with so we don't have to just sit and count all that much. The first thing I notice is that this isn't so bad if we do "matching", like this: (I'll show what digits I'm "counting" in bold)

A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (**5**,6,7,8,9) | E (**5**,6,7,8,9)  
A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (**5**,6,7,8,9) | E (5,**6**,7,8,9)  
A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (**5**,6,7,8,9) | E (5,6,**7**,8,9)  
A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (**5**,6,7,8,9) | E (5,6,7,**8**,9)  
A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (**5**,6,7,8,9) | E (5,6,7,8,**9**)  
A (**5**) | B (**5**,6,7,8,9) | C (**5**,6,7,8,9) | D (5,**6**,7,8,9) | E (**5**,6,7,8,9)  
...

What jumps out at me here - to increment one in "D", we must iterate through 5 possibilities in "E". So, if it's just two "variables" (D & E), we'll have 25 different combinations, 5 combinations for each possible value we can take in D. Well, wouldn't a similar principle apply when considering C? Let's draw a smaller diagram with three variables to see what's at play here, something like:

A (0) | B (0,1,2) | C (0,1,2)  
\-------  
A (**0**) | B (**0**,1,2) | C (**0**,1,2)  
A (**0**) | B (**0**,1,2) | C (0,**1**,2)  
A (**0**) | B (**0**,1,2) | C (0,1,**2**)  
A (**0**) | B (0,**1**,2) | C (**0**,1,2)  
A (**0**) | B (0,**1**,2) | C (0,**1**,2)  
A (**0**) | B (0,**1**,2) | C (0,1,**2**)  
A (**0**) | B (0,1,**2**) | C (**0**,1,2)  
A (**0**) | B (0,1,**2**) | C (0,**1**,2)  
A (**0**) | B (0,1,**2**) | C (0,1,**2**)  
This approximates your question, just is a little smaller so we can see it all spelled out. I kept the A = 1 option factor, but as you might guess by this point, it's pretty irrelevant. So, we get 9 total permutations that work for us, and through the bolding, I think it's pretty clear to see we arrive at 9 because B has 3 options, which themselves are each mapped to 3 options, so 3\*3 = 9. And indeed, in our original question, 5\*5 (when considering only D & E) gives us 25 options which conclude the totality of those two columns. So let's apply this principle again to our bigger question:

A (5) | B (5,6,7,8,9) | C (5,6,7,8,9) | D (5,6,7,8,9) | E (5,6,7,8,9)  
1 \* 5 \* 5 \* 5 \* 5   
625

Now, to be a little more precise, your question was what is the probability, so it'd be 625/65,535, or 125/13,107
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Off-topic reply (if this isn't allowed mods I'm sorry) but I'm just curious: what is this for? I'm assuming this is for Pokémon, but what properties of the game check this specific trait of a trainer ID? E.g. is there something with RNG for say the Lottery corner?
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Assuming all numbers 00000 to 65535 are equally likely:

Do it a digit at a time. From the left, first digit can be 5 or 6. There are 10000 numbers that start with 5. Of those 10000, 5000, have a second digit that is 5-9, 2500 have a second and third that are 5-9 and so on. There’s a little trick for numbers that begin with 6.

Once you have the numbers that fulfil the criteria, all you gave to do to get the probability is divide that by the total of possible numbers.

Does that help any?

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