Top 8 Probability Quiz
 

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%

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

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.

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%

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

ATC,

While both of us seem to find this stuff not only fascinating but also wealth-optimising, apparently we are in the vast minority.

The problems are relevant, not all that hard, yet reasonably challenging. And whether you get them right or wrong, you'll probably end up learning something useful.

But since there appear to be no other takers, you may undisqualify yourself.

Once these are resolved, I'll pose a follow up that comes even closer to real AFL life.

Restating Problem 2:

Under same terms as before,

4 Rounds left

B is 2 games behind A

What is the probability of B overtaking A?


Problem 3:

The general solution.

# Rounds Remaining = Column A
# Games B is behind A = Column B

In Column C give the formula for B overtaking A.

Problem 2:

What is the probability of B overtaking A?

8.9844%

Will leave the general solution for someone else.

8.9844% = Agreed

Problem 3 can keep for now.

Round to 9% as we move to Problem 4:

This should be quicker and easier than the others, so therefore there should be an avalanche of responses.

And the answer seems counter-intuitive to me.

Problem 4:

4 Rounds remaining.

Team G is 2 games out of the 8.

It is directly below 5 Teams all equal on wins. A, B, C, D, & E.

G has an equal 9% chance of finishing ahead of any one of the other 5.

The only 2 possibilities are:

- G fails to make the 8
- G makes the 8, kicking out 1 of A - E

So what is the probability of G making the 8?

In case I'm being too subtle, this hypothetical problem is moving quite close to the current real AFL one.

No sign of the avalanche so i will fire in my answer.

If G has a 9% chance of finishing ahead of each of A,B,C,D & E and assuming independance then the probability of G NOT making the 8 is 0.91^5= 0.624 or 62.4%. Therefore the probability of G finishing ahead of at least one of A,B,C,D & E and hence making the top 8 will be 1-0.624 = 0.376 = 37.6%.

Agree on 37.6% = $2.66 Implied Odds

So for A, B, C, D, E

Prob - Odds
92.5% $1.08

Before I tried this hypothetical exercise I'd thought Geelong's chances were as hopeless as Anthony Mundine's of getting elected to parliament, but fortunately this has now stopped me making really dumb bets on the 8.

----


Problem 3:

The general solution.

# Rounds Remaining = Column A
# Games B is behind A = Column B

In Column C give the formula for B overtaking A.

My Answer

=BINOMDIST(A1-1-B1,2*A1,0.5,TRUE) + BINOMDIST(A1-B1,2*A1,0.5,FALSE)/2

I wonder whether anyone will trouble to paste that formula in and try it.

This (or adaptations) should have a number of applications, presumably particularly for in-play Golf.

But judging from the lack of interest about such a powerful yet simple technique,"Most Punters Must Enjoy Losing Money".