Probability Question For Our Mathematicians

[DragonFire]

For some reason probabilities have always tripped me up. I've been Googling for about 30 minutes but remain at low confidence that I can calculate this.

 

I am wanting to calculate the probabilities of any two teams out of five ending up on the same side of the bracket in a five team tournament random draw. So say you have teams A, B, C, D, E. A five team bracket is set up so that teams drawing position 4 & 5 play each other first, the winner advances to play team 1 in a semifinal, and the other semifinal matches teams 2 & 3. What is the probability that team A & B will be on the same side? They could meet directly by drawing positions 2 & 3, could meet directly drawing 4 & 5, or one could draw into position 1 and the other could draw into position 4 or 5.

 

I think I calculated it correctly by mapping out literally all the potential combinations (there are 48 times A & B are on same side out of 120 potential combinations if I did it correctly - so 40% chance), but I would like to know if I'm correct and also how to figure that out without having to jump through those crazy hoops.

 

Can anyone help me out?

[UKMustangFan]

Without calculating it completely out, I think you're correct.

 

There are 120 possible brackets.

[doomer]

Factorial. 5x4x3x2 = 120

2/5 x 120=48

48/120 = 40%

 

I’m pretty rusty, and I’ll think about it some more, but I get the same answer.

 

Or is it just as simple as 2/5 = 40%?

[UKMustangFan]

So hopefully this doesn't get too confusing, but I'm sure it will, as I have a hard time putting this kind of stuff into writing...

 

There is a 20% chance that A is in slot 1 (1 out of 5).

That leaves 4 remaining slots, so there is a 25% chance that B is in slot 2.

That means there is a 5% chance that B is in slot 2 if A is in slot 1, B will be in slot 2.

You can flip those and say there is also a 5% chance that A is in slot 2, if B is in slot 1.

That means a 10% chance both A & B are on the first side of the bracket.

 

On the other side of the bracket, the side with 3 slots, you have a 20% chance that A is in slot 3. Same probability applies to slots 4 & 5 as well.

If A is in one of those 3 slots, there is a 50% chance (2/4) that B is in one of the remaining 2 slots on that side of the bracket. So a 10% chance for each slot, thus 30% overall.

 

So 10% chance they're both on the first side of the bracket, and a 30% chance they're both on the second side of the bracket. So 40% in total.

[DragonFire]

I appreciate it you all. I consider myself pretty strong at math, but probability just always seemed to be something that gave me trouble.

 

Is the factorial the way you calculate combinations? So if it were 6 teams it would be 6 * 5 * 4 * 3 * 2 * 1 = 720?

 

And if I track the logic right, say it were six teams, with three teams on each side. There'd be a 1/6 chance for a slot on one side, then a 2/5 chance for them to be in one of the other two slots on that side, making it a 1/15 chance for them to be on the same side for a given slot. The odds would be the same for each of the six slots since there are three on each side, making it a 6/15 chance in total for two teams to draw on the same side of a six team bracket - still 40%. Right?

[doomer]

I am quickly thinking add a team and the random probiliry goes down to 2/6 or 33%.