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FAI Cup problem

Shedite27

So the last 3 finals have been Cork City V Dundalk, looking likely to be the final again this year. Wondering if anyone could help me figure out the chances that they could both have been in that many draws each year and avoid each other.

So it's an open draw.
First Round - 32 teams
Second Round - 16 teams
Quarterfinal - 8 teams
Semifinal - 4 teams

So firstly, what is the odds that City/Dundalk avoided each other in a given year.

And then, the odds of that happening for 4 years straight.

Thanks.

FYI, I was trying to do it..
First round - 1/31 chance of getting anyone apart from self so is 3%
Second Round - 1/15 so 6%
QF - 1/7 so 14%
SF - 1/3 so 33%
Am I right to then multiple all those giving me 0.01% of avoiding a given team?
And multiply that by itself 4 times to get 4 years?

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Ciaran

Assuming they win all their games against any other team, the chance of them avoiding each other until the final is 16/31 = 51.6% in any year. The percentages you're multiplying above are the chances that Cork play Dundalk in each round. If you multiply 96.7% by 93.3% by 85.7% by 66.7% you get the percentage above.

I like to think of it by picturing the whole draw done out in a bracket. If two teams are in the opposite halves of the draw, they can't meet until the final. If you put one team in the top half, there are 16 spots in the bottom half out of 31 total for the other team to avoid them until the final.

OldMrBrennan83

Ciaran wrote: »

Assuming they win all their games against any other team, the chance of them avoiding each other until the final is 16/31 = 51.6% in any year. The percentages you're multiplying above are the chances that Cork play Dundalk in each round. If you multiply 96.7% by 93.3% by 85.7% by 66.7% you get the percentage above.

I like to think of it by picturing the whole draw done out in a bracket. If two teams are in the opposite halves of the draw, they can't meet until the final. If you put one team in the top half, there are 16 spots in the bottom half out of 31 total for the other team to avoid them until the final.

I don't think the FAI cup is predetermined like that. It's an open draw after each round.

justshane

Patww79 wrote: »

I don't think the FAI cup is predetermined like that. It's an open draw after each round.

Yeah it's an open draw after every round.

Ciaran

Patww79 wrote: »

I don't think the FAI cup is predetermined like that. It's an open draw after each round.

I'm aware of that but the effect is the same.

BOHtox

So Dundalk and cork have avoided each other for the fourth year in a row until the final.

Can anyone smarter than me do the percentage likelihood of that happening? 1%?

Edit: 32 teams in each cup

Pherekydes

Remember when calculating odds like these that not only do they have to avoid each other in each draw but they also have to win each tie, a not insignificant layer of probability to be added to the overall.

BOHtox

Ciaran wrote: »

Assuming they win all their games against any other team, the chance of them avoiding each other until the final is 16/31 = 51.6% in any year. The percentages you're multiplying above are the chances that Cork play Dundalk in each round. If you multiply 96.7% by 93.3% by 85.7% by 66.7% you get the percentage above.

I like to think of it by picturing the whole draw done out in a bracket. If two teams are in the opposite halves of the draw, they can't meet until the final. If you put one team in the top half, there are 16 spots in the bottom half out of 31 total for the other team to avoid them until the final.

51.6% for one year. Do you divide by 4 to get the probability of that happening for 4 years?

Yakuza

If an event has a 51.6% chance of happening, then for it to occur 4 times (assuming events are independent and the same probability applies) you (as the OP correctly states) multiply it by itself 4 times.

What I mean by independent is that one event has no bearing on the next, for example the odds of getting 4 heads in a row from an unbiased coin flip are .5 * .5 * .5 * .5 as one throw doesn't affect the next.

However, the odds of dealing 4 hearts in a row from a standard deck are (13/52) * (12/51) * (11/50) * (10/49) as getting a heart on the first go will affect the number of hearts left in the deck (as well as the number of cards to choose by) - these events aren't independent.

In the context of this question, it means that one year's league has no bearing on the next. This might be relevant if there are promotions / relegations to consider (which might mean a the chances of two teams meeting would be zero if they're in different leagues!)

Ciaran

BOHtox wrote: »

51.6% for one year. Do you divide by 4 to get the probability of that happening for 4 years?

Raise it to the power of 4. ~7% chance

BOHtox

Cheers lads

Cape Clear

No need to look at this from the point of view of probability. Return on investment is where its at. In order for the FAI to get the maximum return from the competition they have kept the 2 teams apart until the final. Simple formula MROI= CC+D≥C
being
MROI Maximum return on investment
CC Cork City
D Dundalk
C Corruption