Probability of number appearing on a board of randomly generated numbers?

Let's say I have a board that generates 23 number randomly (0000 - 9999).

1st prize has 1 number

2nd prize has 1 number

3rd prize has 1 number

4th prize has 10 number

5th prize has 10 number

Let's say I pick number $$1234$$. The probability that my $$1234$$ will appear can be calculated in different way

1st

1 - $$P(1234 \text{ not appearing}) = 1 - (9999/10000)^{23} = 0.002297471770114836...$$

2nd

$$P(\text{appear in }1st) + P(\text{appear in }2nd) ... + P(\text{appear in }5th) = (1/10000)*3 + 2(1 - (9999/10000)^{10}) = 0.002299100239958...$$

Probability that 1234 appear in 1st, 2nd or 3rd are 1/10000 each. So 1/10000*3

Probability that 1234 appear in 4th are 1 - P(Not appear in 4th). Using binomial it is 1 - 0.9999^10

Probability that 1234 appear in 5th are 1 - P(Not appear in 5th). Using binomial it is 1 - 0.9999^10

Adding them together should yield probability of 1234 appearing on the board

3rd

I can simply say there are 23 numbers on the board so my number has 23/10000 chance = 0.0023

Why are these three methods giving slightly different value? Which method is the most accurate?

• In the second equation, what do you mean by "appear in 1st/2nd (...)" ? First what? Also, I don't understand the prixes. "5th price = 10 number"? I don't understand what that means in this context. (Also, the prizes are not regarded in the problem, but still ...) – Matti P. Feb 4 at 11:37
• Also, the third approach assumes a different thing from the others. Is it possible that the same number appears several times on the board? – Matti P. Feb 4 at 11:38
• The difference is that the first method uses replacement; the third method does not allow replacement. I am not sure what is going on in the second method. – robjohn Feb 4 at 11:40
• I added explanation for second example – Zanko Feb 4 at 12:31
• @MattiP. sorry for being unclear, there are 10 numbers for 5th prize. Yes same number can appear multiple time on the board – Zanko Feb 4 at 12:32

Your first approach assumes the numbers are drawn with replacement, that the first number is drawn and we check whether it is $$1234$$, then another number is drawn from all the numbers and we check that one, and so on. The probability is a little less than $$\frac {23}{10000}$$. The expected number of draws of $$1234$$ is $$\frac {23}{10000}$$ but we can draw it more than once. Therefore the chance we draw it at least once must be less than $$\frac {23}{10000}$$.
• No. If the average number of wins is $\frac {23}{10000}$ and sometimes you get $2$ wins, the fraction of the time you get at least one is less that $\frac {23}{10000}$ because the times you get $2$ are counted twice. – Ross Millikan Feb 4 at 12:46