Help Settle and Argument re: Chance of Failure in Fantasy Football

#1
Hi, all!

I'm new, so please accept my apologies in advance for... whatever.

So here's the question. In a fantasy football league, there are two divisions, each with 6 players.

In the entire league, there are 3 brothers (each other's brothers!). 2 of the brothers are in division A. The third brother is in division B.

What is the probability that any of the brothers comes in last in his division?

My friends and I have worked this out in several different ways, but we are unable to come to an agreement. Please help!
 
#2
FWIW this is what I think...

P(A) is the probability that 1 of the 2 bros in Division A will come in last. So P(A) = 2/6
P(B) is the probability that the brother in Division B will come in last. So P(A) = 1/6


Since P(A) and P(B) are mutually non-exclusive (non-mutually exclusive? all inclusive? what's the right term here?)...

P(A or B) = P(A) + P(B) - P(A and B) = 2/6 + 1/6 - (2/6)(1/6) = 44.4_%


Is that accurate? Did I blow it? It's been a looooooong time since undergrad.
 
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Dason

Ambassador to the humans
#5
I get the same answer but went about it slightly differently. First I considered the question "What is the probability that none of the brothers comes in last" which is (4/6)*(5/6) and then said well the probability at at least one comes in last is the complement of the event that none of the brothers comes in last so the probability is 1 - (4/6)*(5/6) = 4/9
 

Dason

Ambassador to the humans
#9
Seems like more work Dason, having to include the "1". Though, thanks for sharing!
It also used the numbers 4 and 5 which are clearly larger than 1 and 2. This approach may cause your computer to overheat from working with numbers so close to infinity.
 
#10
Who was right or more right - you or your friend?
At first I was right, then I changed my mind, and then I went back to my original answer (which apparently is right).

So the real question is, "Who was right - me or myself?" God bless Us, Every One!
 
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