I am an engineer by education and employment, not a statistician. My best credential for attempting this little exercise may be that I once read a book entitled "The Statistical Theory of Communication" for pleasure. One of the activities most common to my engineering days was to broad brush probabilities and examine a range of outcomes to establish worst and best case scenarios. These were rough guesses. So, without the use of any of the cool statistical programs that many TN resident experts use when crunching numbers here, I set out to satisfy my curiosity with Excel and see what I might calculate as FSU's chances to win it all this year. Again, I'm no statistician so please point out my errors in thinking. I used only simple Excel functions and I'm not totally confident my statistical reasoning is sound, but the results appealed to my intuition and passed a few simplistic reasonability tests, so here goes
Approach, assumptions, data and results to follow. Criticism welcome.
Step by step:
1.) I first assumed that all the games played were independent events.
2.) I claimed that the probability of FSU winning it all, P(winall), was equal to the probability of FSU getting to play in the game, P(play), times the probability of FSU winning the MNC game, P(winMNCgame).
3. I further broke down the first term:
P(play) = Probability of Getting to Game if Undefeated * P(undefeated) + P(playing with 1 loss) * P(1 loss) + P(2 or more losses) .
4, I assigned a 95% chance to play if we go undefeated, a 20% chance to play if we have one loss and no chance to play with 2 or more losses.
5. I assigned 50% to our odds of winning the NC if we got to play.
6. All that remained then was to calculate two probabilities, going undefeated this season and the chances of losing only one. I used the average value of the TN proportional win share survey for each game and arbitrarily gave us a 60% chance to win the ACC championship.
Here is the data where the first number is the TN average win share value and the last number is the calculated probability of losing only that game.
Team Win Loss Only Loss
ULM, 0.98, 0.02, 0.000415
CSU, 0.98, 0.02, 0.000415
OU, 0.44, 0.56, 0.025909
CLM, 0.62, 0.38, 0.012477
WFU, 0.88, 0.12, 0.002776
DUK, 0.93, 0.07, 0.001532
MRY, 0.81, 0.19, 0.004775
NCS, 0.76, 0.24, 0.006429
BC, 0.72, 0.28, 0.007917
UM, 0.70, 0.30, 0.008724
UVA, 0.85, 0.15, 0.003592
UF, 0.60, 0.40, 0.013571
ACC 0.60 0.40 0.013571
So, I think, P(winall) is equal to the product of the first column of 13 values = .020357. Assuming all outcomes are independent, we have about a 1 in 50 chance to go undefeated.
i calculated the odds of losing just one game as the sum of the odds (thirteen terms, last column) of losing each individual game.
P(L1) * P(w2) * P(w3).....* P(w13) +
P(w1) * P(L2) * P(w3)..... * P(w13) + etc.
The probability of exactly one loss based on the above came out to = .102104. (An aside, that implies that the probability of FSU losing 2 or more this year is nearly 90%. (.8775) ).
Here's what I found interesting, according to this approach, even with only a 20% chance of getting invited to the NC game with one loss, that route was about equally likely to produce a championship.
Winning it all by going undefeated = .020357 * .95 * .5 = .00967
Winning it all with one loss = .102104 * .20 *.50 = .01021
Odds of winning it all, total = .00967 + .01021 = .01988, which is not much different, about 1 in 50, than our chances of going undefeated in the 12 regualr games plus ACC championship?
Something else,looking at the odds if we had a 90% chance of winning each of the 13 games, the data says that...
1. We would only have a 25.4% chance of going undefeated.
2. We would have a 36.7% chance of losing exactly one game.
3, We would still have a 37.9% chance of losing two or more.
Math or reasoning errors?