# How to run a fair drawing for door prizes?

#### msmith

##### New Member
Given that you're the organizer for an door prize drawing that has 10 prizes ranging in value from $100 to$1. Everyone in attendance for the drawing has submitted one ticket into the drawing box. You want the drawing to be fair and have everyone have a chance at the biggest prize. Once a ticket is drawn it's not returned to the drawing box.

Obviously if you draw for the biggest prize first then everyone has an equal chance.

But what if you draw for the cheaper prizes first saving the biggest until last?

#### hlsmith

##### Less is more. Stay pure. Stay poor.
This looks like a homework problem. What attempts have you made to solve it? What questions do you have?

#### msmith

##### New Member
No homework problem. It's been years (decades) since I've had homework. I belong to a club that does a yearly event with door prizes. We have always saved the biggest door prize for last, but this seems unfair.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
How are you selecting the tickets?

This isn't mathematical, but can you just do all of the little prizes, then put all of the pulled tickets back into the urn for the last grand prize selection?

#### msmith

##### New Member
Arg! I thought I had replied to this earlier, must have missed pressing a button.

Anyways... No, once your ticket is drawn you're done. I was just looking for a valid argument to show a difference between the probability of winning the first prize drawn for verses the last prize drawn for. I guess there is none. I'll quit thinking about this one.

Thanks for the replies.

#### hlsmith

##### Less is more. Stay pure. Stay poor.
How many people and how many prizes?

#### rogojel

##### TS Contributor
hi,
I think one could tackle this by calculating the expected gain at the last drawing: E=Sum(value_of_object_i*P(object i being the last)). If we assume N objects in the draw the probability of object i being the last must be 1/N. If the draw is run correctly there can be no preference for any of the objects. So the expected value for the last draw is Sum(value_i*1/N) which exactly equal to the expected value of a gain in the first draw.

So, the conclusion is that it does not matter in which order you draw, the procedure is always fair in the sense that the expected value of the gain is the same for each draw.

regards
rogojel