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Posted

Hi guys,



Currently, trying to determine the percentage chance of winning an imaginary giveaway for 500 prizes. It uses a single entry per person.



Say there is 10000 entries and a total of 500 can win with each winner being unique (1 per person).



So far I figure that since you can only win once the odds would be:



1/10000 + 9999/10000 * 1/9999 + 9999/10000 * 9998/9999 * 1/9998 + 9999/10000 * 9998/9999 * 9997/9998 * 1/9997 + ... etc



How would I go about calculating that if one person was to have 1000 entries instead of just one?



For example, say one person cheats and gets 1000 tickets in a raffle instead of just one with the same prize pool and total entries.




Posted (edited)

Well, I am treating all prizes as being equal for the sake of simplicity. So all 500 prizes are effectively the same thing.

 

And for the sake of an overall chance of winning I have combined them.

Edited by DyslexicEngineer
Posted

Well, I am treating all prizes as being equal for the sake of simplicity. So all 500 prizes are effectively the same thing.

 

And for the sake of an overall chance of winning I have combined them.

So are you saying that all the prises are of equal value. So there is no benefit of being the first number drawn.

Are you drawing them one at a time, and not returning that number back to the pool?

Posted (edited)

So are you saying that all the prises are of equal value. So there is no benefit of being the first number drawn.

Are you drawing them one at a time, and not returning that number back to the pool?

 

Correct, there is no benefit to being drawn first.

 

Once, a winning entry is drawn then the prize and that number is gone from the pool. Additionally, nobody has the same number.

Edited by DyslexicEngineer
Posted

500 goes into 10,000 twenty times.

 

so this sounds too easy but would not the odds of winning then be 20-1?

 

LOL. I like my math easy. As well as minimal.

Posted

500 goes into 10,000 twenty times.

 

so this sounds too easy but would not the odds of winning then be 20-1?

 

LOL. I like my math easy. As well as minimal.

 

You get something close to that either way.

 

What I want to know is if one person has more than one entry what the effect on the odds would be.

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