Probabilty of a specific 4 digit number I'm having an argument with a colleague about the probability of a 4 digit number.
The assumption is that the number is randomly and independently created.
Assume the probability of generating the number 4109.
Now, my colleague reckons that it's a simple P(4109) = 1/10 * 1/10 * 1/10 * 1/10. (a bit like simply tossing a coin).
But I'm not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc.
Obviously the probability of that specific sequence, 4109 is a probability found by calculating the permutation, not simple combination....
Please confirm my believe or point me to some proper material to support this?
 A: It's very simple.  In 4 decimal digits there are 10,000 (0000 to 9999) possible values.  The odds of any one of them coming up randomly is one in 10,000.
A: A specific "4 digit number" would have 1/9000 chance, since there are 9000 4 digit numbers (1000-9999). Another way to look at this would be to choose the first digit (9 options: 1-9), then the second digit (10 options: 0-9), etc so you get 9*10*10*10 = 9000 options, you can multiply the options like this in this scenario because ceach combination of the option is a different valid results and all valid result can be created by such combination.
Same argument if you want to consider "0421" a "4 digit number", except the answer will be 1/10000
A: Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:


*

*the chance that the first digit is 4

*the chance that the second digit is 1

*the chance that the third digit is 0

*the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it? 
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:


*

*the probability that the first digit is 0,1,4, or 9 = 4/10

*the probability that the second digit is one of the three not yet selected = 3/10

*the probability that the third digit is one of the two not yet selected = 2/10

*the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $\prod_{i=1}^{4} \frac{i}{10}$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
A: Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number). 
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that. 
