Three Doors with one prize

I don't think this is off topic at all.

I think the last radio caller has the best chance. This is based on the probability of getting the prize is the probability of picking the correct remaining number multiplied by the probability that the prize has not yet been won (i.e. 1 - the probability of success of the previous caller). I think the probablility of each caller selecting the right number increases more rapidly (1 in 10, 1 in 9, 1 in 8 etc ) than the probability of the prize still remaining decreases.

Just to illustrate, the probability of the first caller winning the prize is the probability that the prize has yet to be won 1 by the 1 in 10 probability that s/he picks the correct number. For the 10th caller the probability of winning the prize is the probability that the prize has yet to be won .63 by the probability that they pick the right number 1 (there's only one number left).

Hi,

Very few of you got it right. Best marks to gus who not only got it right but provided the clearest explanation.

If don't think statistics of small numbers are relevant to astronomy, you might consider phenomena where you get a handful of counts, e.g. neutrinos and very high energy cosmic rays.


If you doubt the result, I suggest you do the experiment with paper cups.

John

Gus --

Well done! That's the first analysis of Monty's Paradox that didn't make my brain hurt. (And it even arrived at the right answer.)

-- Jeff.

You're welcome

I think the last radio caller has the best chance.

If you work out the odds for each caller to win, at the start of the competition you may be surprised, if my calculations are right.

Glaven!

I ... er ... forgot to carry the 1 ... mmm ... hay.

If you work out the odds for each caller to win, at the start of the competition you may be surprised, if my calculations are right.

I got .1 for each of the first two callers Gus, 1 in 10 for the first and 9/10 by 1/9 for the second. The tenth comes in at a healthy .63 probability.