(08-11-2015 09:28 AM)OWLmanz Wrote: WHAT?? I hear what you're saying, but I find it hard to believe.............. I'm probably too "simplisticly rational" I guess! 
Also, if we are truly favored in 9 games, why isn't that the highest probability result???????????? 
(08-11-2015 08:40 AM)baker-13 Wrote: (08-11-2015 07:46 AM)OWLmanz Wrote: 
OK I am "just" an "academ," but how can 9 and 10 wins be the most likely scenarios, since the data table appears to show 8 wins as the highest probability outcome, with 7 wins as the second highest probability????? 
Makes no sense.....
Of course, we learned even in Econ and Finance, that you can interpret statistics any way that you want...........
"Most likely scenario" = most specific exact combination of things. You get higher probabilities for 7 and 8 wins because there are more different combinations that lead to 7 or 8 wins than there are 9 or 10 wins, even if those individual combinations have a lower probability that the most likely 9- and 10-win combinations.
Even if we were favored in all 12 games, 12 wins wouldn't be the most likely result.
Let's call the games A, B, C, D, E, F, G, H, I, J, K, and L. We'll call the probability of winning each W(X) for game X. So, if A is against Wagner, then W(A) is close to 100%, so probably something like .99. For each game, the probability of losing, L(X), is 1 - W(X), so L(A) is about .01.
Since each game is DIFFERENT, then the probability of having a particular season is the product of the probabilities for each game. Kinda like the probability of flipping a coin heads and then tails is .5 * .5 = .25. Now, the probability of getting one heads and one tails is .5, since you get HT 25% of the time and TH 25% of the time.
So, the probability of winning all 12 games is:
W(A)*W(B)*W©*W(D)*W(E)*W(F)*W(G)*W(H)*W(I)*W(J)*W(K)*W(L)
The probability of getting 11 wins is not so easy to write, because there are 12 ways it could happen, with some much more likely than others (e.g., it's EXTREMELY unlikely we'll beat Texas and Baylor but lose to Wagner).
So, we'd take L(A)*W(B)*W©* ... * W(K)*W(L), the first way to win 11 games, then add
W(A)*L(B)*W©* .. *W(K)*W(L),
then add
W(A)*W(B)*L©*W(D)* ... *W(K)*W(L),
and so on. Add those all up to get the probability of winning 11 games.
For 10, 9, 8, and so on, there are a lot more terms that need to be added up.
In the end, games B, D, and H are the hardest games. W(H) is about .5, it sounds like. W(B) is about .20, and W(D) is something small. These are the ones where L(B) and L(D) are higher than W(B) and W(D), and W(H) is about the same as L(H).
Thus,
W(A)*L(D)*W©*L(D)*W(E)*W(F)*W(G)*L(H)*W(I)*W(J)*W(K)*W(L)
and
W(A)*L(D)*W©*L(D)*W(E)*W(F)*W(G)*W(H)*W(I)*W(J)*W(K)*W(L)
are the most likely seasons.
Now, looking at the other nine games, we're favored in all of them, BUT that doesn't mean it's likely we'll actually win all of them. It's like, roll a die nine times. Count how many times you roll higher than a 2. Each individual time you roll, you're probably getting higher than a 2, but the chances of ALL NINE rolls being higher than 2 are very small (about .00005).
So, the season
W(A)*L(D)*L©*L(D)*W(E)*W(F)*W(G)*L(H)*W(I)*W(J)*W(K)*W(L)
is less likely than the one where we just lose to Texas and Baylor, but if we take into account ALL of the other nine games (and LaTech is a coin flip), we're more likely to lose one or two of them than we are to win all of them.
