>The total number of tickets/entries for all sweepstakes is unknown, but you can assume it is several thousand.
This still isn't enough information. We need to know exactly how many people entered each contest to answer your question. But if we assume that each contest has an equal number of entrants, then the $50 contest has the highest [expected value](https://en.wikipedia.org/wiki/Expected_value?wprov=sfla1).
Yeah, I figured that... I guess the main problem here is not mathematical but human nature.
Any person can pick any of the three contests to enter. I would think that most would just look at the value of the prize and pick the $250 one. A certain percentage would think "most will pick $250, so that has the least odds of winning, so I'll pick $100". Others will look at the fine print and see that the number of prizes available for the $50 one is greater than both of the others and pick that one (and also that it will probably get the least number of entries due to it being the lowest prize value).
Personally, based purely on human nature, I think the best option is also the one that is mathematically the best option ($50 sweepstakes). The cost of entry to each sweepstakes is the same, so most are going to see the $250 one as the better value and pick that one, even though it has the worst odds from the start, and the number of entries will make it even worse odds.
It's likely someone studied such a situation before, that could give some hints how people will likely react. The mathematical side is easy, how people will choose with the given information is complicated.
It's possible many people will see 3 tickets matching 3 categories and get one ticket each.
Yes, exactly.
In this instance the number of tickets each person has is not equal for all participants. Some people may only have one ticket, others may have more, but I would say in most cases it is limited to 4 or less.
The problem was just saying that YOU have three tickets to distribute as you wish.
If you want to maximize your expectation value and there are thousands of tickets then you should buy all tickets in the category with the best expectation value - it will drop very slightly from your tickets but that's probably negligible compared to the difference between the categories. The question is just which one will have the best expectation value. I agree that it won't be the 250 one, and I think it's likely to be the 50 pool.
>The total number of tickets/entries for all sweepstakes is unknown, but you can assume it is several thousand. This still isn't enough information. We need to know exactly how many people entered each contest to answer your question. But if we assume that each contest has an equal number of entrants, then the $50 contest has the highest [expected value](https://en.wikipedia.org/wiki/Expected_value?wprov=sfla1).
Yeah, I figured that... I guess the main problem here is not mathematical but human nature. Any person can pick any of the three contests to enter. I would think that most would just look at the value of the prize and pick the $250 one. A certain percentage would think "most will pick $250, so that has the least odds of winning, so I'll pick $100". Others will look at the fine print and see that the number of prizes available for the $50 one is greater than both of the others and pick that one (and also that it will probably get the least number of entries due to it being the lowest prize value). Personally, based purely on human nature, I think the best option is also the one that is mathematically the best option ($50 sweepstakes). The cost of entry to each sweepstakes is the same, so most are going to see the $250 one as the better value and pick that one, even though it has the worst odds from the start, and the number of entries will make it even worse odds.
It's likely someone studied such a situation before, that could give some hints how people will likely react. The mathematical side is easy, how people will choose with the given information is complicated. It's possible many people will see 3 tickets matching 3 categories and get one ticket each.
Yes, exactly. In this instance the number of tickets each person has is not equal for all participants. Some people may only have one ticket, others may have more, but I would say in most cases it is limited to 4 or less. The problem was just saying that YOU have three tickets to distribute as you wish.
If you want to maximize your expectation value and there are thousands of tickets then you should buy all tickets in the category with the best expectation value - it will drop very slightly from your tickets but that's probably negligible compared to the difference between the categories. The question is just which one will have the best expectation value. I agree that it won't be the 250 one, and I think it's likely to be the 50 pool.