This is an original problem which bugged me this morning. I suspect many of you will learn something by trying it.

Imagine that in a hypothetical season:

All Top 8 spots decided bar 8th, where A & B are the only possibilities.

Only 2 rounds left

A is 1 game ahead of B

A & B won't be playing each other.

Both have 50% chance of winning their remaining games. No draws.

If both end up equal 8th on wins then resolve with a coin toss.

What are B's chances of making the Top 8?

OK, here's what i've got so far:

Let X = win
Let O = loss

Possible outcomes for Team A:

1. X X (25%)
2. X O (25%)
3. O X (25%)
4. O O (25%)

Possible outcomes for team B to progress under scenario (all other outcomes result in failure):

1. 0%
2. X X (25%)
3. X X (25%)
4a. X O (25%)
4b. O X (25%)
4c. X X (25%)

However scenario's 2, 3, 4a & 4b result in a draw = coin toss

Possible outcomes for team B after coin flip (assuming they win the coin flip):

1. 0%
2. 0.25*0.5 = 12.5%
3. 0.25*0.5 = 12.5%
4a. 0.25*0.5 = 12.5%
4b. 0.25*0.5 = 12.5%
4c. 25%

Probability of qualification = 67.5%

__________________
I like Bing Lee.

Seems a bit high, so i'm thinking i've mde a mistake (any hints welcome).

__________________
I like Bing Lee.

18.75%

I think i have calculated the probability of team A qualifying. If that's the case then obviously team B would be 32.5%. Alternatively i'm just getting more and more confused.

__________________
I like Bing Lee.

Consider 1 of your positions

If A has an
X 0

And B has a
XX

Then it's up to a coin toss

So you have to multiply ALL 3 probabilities together: A, B, Toss

25% x 25% x 50%

So quite often your results are 4 times too much

Okey Dokey, i gotcha now.

XX = 0%
XO : XX = 0.25*0.25*0.5 = 3.125%
OX : XX = 0.25*0.25*0.5 = 3.125%
OO : XO = 0.25*0.25*0.5 = 3.125%
OO : OX = 0.25*0.25*0.5 = 3.125%
OO : XX = 25%

Probabilty of outcome = 37.5%

__________________
I like Bing Lee.

WRONG

Better complete this now seeing it's Census night.

I was hoping more would have tried this.

If Mad divides one more of his rows by 4 and sums, then he should get 18.75%. Which is what ATC and I get.

Anyone wanting to figure out how to try such problems should look up Probability Trees. Mad seemed to be using them without knowing about them, but he forgot about the initial branches.

But there seems to be a more powerful way of solving the problem (in one spreadsheet cell!) for any number of remaining Rounds and Handicaps!

If you look at how many more wins B gets than A then that seems to be a Binomial Distribution for 4 games.

n B-A
1 2
4 1
6 0
4 -1
1 -2

16 Total

Apply the 1 game handicap to B and

Probability = 1/16 + (4/2)/16 = 3/16 = 18.75%

Anyone prepared to try something akin to the current situation with Freo and Geelong?

4 Rounds
2 Handicap for Geelong


I think for now ATC should disqualify himself pending a testosterone test.

Nice to know i got an answer right JFC!
I dont think I would test positive to testosterone like Floyd but may register on the breathometer after a weekend on the bottle.
Gotta love your probability quizzes. I remember doing STATS 101 in my first year at Uni and coming across the Monty Python - 3 doors problem for the first time in the end of year exam. I ended up spending the whole 2 hours on the question and happened to be the only one that got it right! I think that day kick started my career as a statistician turned punter.

Cheers
ATC