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# A question of odds Answered

Hello math gurus, here's one for ya. I'm one of the 50 finalists for the Green Science Instructable competition ( yay me - a pleasant surprise to get this far! ) , there are a total of 18 prizes, what are the odds of me winning a prize? If there was just one prize even I could work out that I'd have a one in 50 chance, but I don't know how to figure the complexity of multiple prizes. I'd appreciate an answer, please show your workings :-) Cheers, Bosherston

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## 18 Replies

westfw (author)2008-08-16

Um, not to be pedantic or anything, but since selection of finalists for prizes is NOT a "random chance" occurrence, "odds" doesn't really apply...

bosherston (author)2008-08-29

I beg to differ. Only in the the abstract can you get true, I stress TRUE "random chance" occurrence. Odds can be applied to all else surely? If we approach each model with the aim of setting the odds and appreciate all known data , can we not then start to get a handle on the likely outcomes of an event? Ok it's a given that the foibles of a judging team make it an extremely complex model....

westfw (author)2008-08-29

Fine. But in that case, only in the abstract can you use the simplified mathematics people have been talking about to calculate the odds, and you become a bookmaker instead of a mathematician. Mathematical statistics is based on the abstract... (which ought to make you really suspicious of most statistics, of course!)

whatsisface (author)2008-06-16

I think it's as simple as 18 in 50, which reduces to 9 in 25 or 36%

Patrik (author)2008-06-16

Actually, there are 28 prizes, including 10 DISCOVER magazine T-shirts and 10 books.

But yes, if each contestant can only win one prize at a time, 28 out of 50 contestants will win a prize, so your chance of winning something would be 28/50 = 56%

Patrik (author)2008-06-16

Actually, I count 51 finalists, so 28/51 = 55%.

Flumpkins (author)2008-08-15

Wow.

stinkymum (author)2008-06-18

I counted only 49 finalists today! So now the odds (whatever they are) must be greater?

bosherston (author)2008-06-16

Your right, I could have sworn that wasn't there the other day, good luck to them. I was too late for the early entry prizes so didn't count them.

Patrik (author)2008-06-16

Hm... and some finalist instructables were submitted by the same person. So depending on whether it's "one price per instructable", or "one price per instructabler", the odds could be slightly different.

Flumpkins (author)2008-08-15

NERD!!!!!!!!!

Flumpkins (author)2008-08-15

jk :D

bosherston (author)2008-06-16

I think I get it, was initially thinking along the lines of 1 in 50/18 and then my brain dribbled out of my ears.

PKM (author)2008-06-16

1 in 50/18 is the same as 18/50. It's hard to express in text, because it needs stacked fractions, but (1 divided by (50 divided by 18)) is the same as 18 divided by 50.

Easy :

1) - you have more chance to win a price than someone who does not have entered the contest.

2) - you have more chance to win one of the 18 prizes than you have chance to win at your national lottery.

3) - you have more chance to win the first price than the French soccer team has to go to the Euro2008 final.

In mathematical language :
2spis-1sin(pi×s×0.5)×a_greek_letter(1-s)×an_other_weird_greek_letter(1-s) = again_this_wierd_greek_letter(s)

And no, I don't know what it means ... just consider that your chances are greater than 0. That's simpler.

bosherston (author)2008-06-16

Euro 2008?, at least France qualified.

Kiteman (author)2008-06-16

England didn't, but when they were selling the TV rights, apparently somebody thought we would - it's on both main national channels, and a game that I find boring anyway is now rendered doubly so because it can't even appeal to the Jingoist in me.