Math-heads: Probability problem

J4amieC

cool, thanks

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J4amieC

Thanks for the abridged explaination. It makes alot more sense now ive read the above link and thought about it a bit more logically. It seems you dont have to even stretch to subs. The actualy on pitch players (plus, say, the ref) make up the set needed.

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Roger Stoltz

Hmm, probability calculations are in fact my worst enemy. If you say it's a chance of 50% that two soccer players share the same birthday... But that's before they meet! After they've met and concluded that they share birthdays, the probability that they share birthdays increases to 100%... :confused::confused:

"It's supposed to be hard, otherwise anybody could do it!" - selfquote
"High speed never compensates for wrong direction!" - unknown

Don Miguel

Roger Stoltz wrote:

After they've met and concluded that they share birthdays, the probability that they share birthdays increases to 100%...

But if the conclusion is that tey didn't share birthdays, then the probability will be again 50%! :laugh::laugh::laugh:

Pete OHanlon

The probability is highest when they are twins.:laugh:

Deja View - the feeling that you've seen this post before.

Roger Stoltz

Don Miguel wrote:

But if the conclusion is that tey didn't share birthdays, then the probability will be again 50%!

Isn't the probability in that case 0%? :~ Or do you expect one of them to change his birthday? That brings another question on the board: what's the probability for anyone to change his birthday? :-D

"It's supposed to be hard, otherwise anybody could do it!" - selfquote
"High speed never compensates for wrong direction!" - unknown

Roger Stoltz

I would say that the probability for all 44 players to be twins is about 0%... ;P

"It's supposed to be hard, otherwise anybody could do it!" - selfquote
"High speed never compensates for wrong direction!" - unknown

Pete OHanlon

Except when it's the Intra Chapter TAMBA team. (Twins And Multiple Births Association).;P

Deja View - the feeling that you've seen this post before.

Don Miguel

Indeed, probabilities are complete insane.... ;P

darrellp

This is a standard question in probability, primarily because people find it counterintuitive. The first hint that it's a little odd is when you realize that if you have 366 people the odds are 100% that two of them have the same birthday. It's not quite so hard at that point to imagine the the 50% mark is around 30 or so. It's easier to ask what the chances of them all having different birthdays is and subtract that from one. So for two people, the chances that both birthdays are different is 364/365 (only a 1/365 chance that the second person's birthday matches whatever birthday the first has). For three people, the chances that all their birthdays are different is (364/365)*(363/365) - i.e. the second person's birthday must be different than the first's (364/365) and the third person's must be different than both the first two (363/365). Similarly, for four people, the chances are (364/365)*(363/365)*(362/365) and the pattern is obvious. I believe that this does, in fact, drop under 0.5 around 30 people or so and there you have the answer. Also, I believe that conventionally, oddes of 2:1 means the first alternative has a 2/3 chance and the second has a 1/3 chance. Hence, one to one odds means 50%. :)