# What are the odds?

1. Dec 4, 2007

### billiards

A friend recently observed that Manchester United have been drawn against Aston Villa in The 3rd round of the FA Cup 4 times in the past 7 draws. There are 64 teams in the 3rd round draw. What are the odds of this happening?

I believe I have the answer, but I need someone to verify my answer to ensure that my understanding is correct.

So, what are the odds?

2. Dec 5, 2007

### billiards

Okay, obviously this is too easy a question to bother most of you. My answer was 1/472,216. Is that right?

3. Dec 5, 2007

### Office_Shredder

Staff Emeritus
Actually, I think the problem is here

If you want to make sure your understanding is correct, the best way is surely to post what you think the correct method would be

4. Dec 5, 2007

### billiards

Okay.

Chance of Man U drawing Villa on one occasion A = 1/63
Chance of that not happening on same occasion B = 62/63

Chance of it happening 4 times in a row followed by 3 failures is AAAABBB=623/637.

This probability is equal to the probability of the initial condition being satisfied (that Manu draw Vill 4 times in 7), by each of the 7!/(4!*3!) = 35 different possible permeatations. Accounting for all possible solutions, the probability that the initial condition is satisfied, is 35 times 623/637 = 1/472,216.

5. Dec 5, 2007

### Office_Shredder

Staff Emeritus
Looks good to me

6. Dec 5, 2007

### CRGreathouse

I get 35*62^3/63^7 = 1191640/562711519881, but that's close enough to your answer.

Note that the chance of Man U drawing some team 4 times out of 7 (not just Villa in particular) is 1191640/8931928887, which at roughly 1/7495 is a fair bit more likely. The chance of it happening to at least one team (not just Man U) is about 1/242. The chance of at least one team pulling some other team at least 4 times is little changed at about 1/238.