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?

- Thread starter msmith
- Start date

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?

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.

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