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


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


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

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


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?

  • $\begingroup$ 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 ...) $\endgroup$ – Matti P. Feb 4 at 11:37
  • $\begingroup$ Also, the third approach assumes a different thing from the others. Is it possible that the same number appears several times on the board? $\endgroup$ – Matti P. Feb 4 at 11:38
  • $\begingroup$ 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. $\endgroup$ – robjohn Feb 4 at 11:40
  • $\begingroup$ I added explanation for second example $\endgroup$ – Zanko Feb 4 at 12:31
  • $\begingroup$ @MattiP. sorry for being unclear, there are 10 numbers for 5th prize. Yes same number can appear multiple time on the board $\endgroup$ – Zanko Feb 4 at 12:32

Your third calculation assumes that the numbers are drawn without replacement, that no number can be selected twice. That is the usual way lotteries are done.

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 $\frac {23}{10000}$.

You don't explain the second calculation. It appears the numbers are drawn with replacement. You can't just add the probabilities because the events are not disjoint. If you draw with replacement you could win both first and second. It sounds like you are computing the chance of winning at least one prize, in which case you must use inclusion/exclusion for this route.

  • $\begingroup$ I added explanation for second example $\endgroup$ – Zanko Feb 4 at 12:31
  • $\begingroup$ "Therefore the chance we draw it at least once must be less", shouldn't it be more? As in we have more chance since we can draw it multiple time? $\endgroup$ – Zanko Feb 4 at 12:34
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Feb 4 at 12:46
  • $\begingroup$ So according to this probability, there are more chances that 1234 will appear for the game without replacement than the game that has replacement? Even though for the game with replacement you stand to win multiple prizes at once. $\endgroup$ – Zanko Feb 4 at 12:52
  • $\begingroup$ There are chances that 1234 will appear more than once if you draw with replacement. You need to state things carefully and I don't understand "more chances that 1234 will appear". The chance that 1234 will appear at least once is less if you draw with replacement than if you do not. $\endgroup$ – Ross Millikan Feb 4 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.