re: Percentage chance that you give LSU to win in the Regular Season?

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This is obviously not the case, as it's pretty apparent that you would be predicting a 6-6 record.

No way. By using methods other than chance, I predict 10-2.

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I said you would be predicting a 6-6 record if you gave LSU a 51% chance of winning each individual game. By predicting 10-2, you are giving LSU an 83% chance of winning each game on average. It's as simple as that.

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There is no flaw in the way Nuts4LSU did the math.

I believe Nuts even predicted a possible 4 win season in one of these threadss. He keeps on the sunny side.

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He didn't predict that, he said that's a possibility, probably based on somebody else's win probabilities. It's not about having a "sunny" attitude, it's about doing basic math to convert win probabilities of individual games into a predicted record, and vice versa.

I really don't see what's so hard about this.

UNC 80%
Vandy 95%
Miss State 90%
West Virginia 70%
Tennessee 65%
Florida 55%
McNeese 99.99%
Auburn 70%
Alabama 60%
ULM 99.9%
Ole Miss 75%
Arkansas 75%

Wild guesses recorded for posterity!

Btw, my actual season prediction (which is not based on probabilities) is 11-1.

OK, here is my last attempt, and if this doesn't convince chilge and mayhawman, then there is no hope.

Let's do this with a 4 game season for simplicity.

Assume these win probabilities:

Game 1: 100%
Game 2: 100%
Game 3: 100%
Game 4: 100%

Obviously this will result in a 4-0 record.

Now change 2 games:

Game 1: 100%
Game 2: 100%
Game 3: 0%
Game 4: 0%

2-2, right?

Whoa now it gets tricky:

Game 1: 100%
Game 2: 100%
Game 3: 50%
Game 4: 50%

Games 1 and 2 are definitely wins, but what about 3 and 4? Well since they are both coin flips, odds are you will win 1 of them, right? RIGHT? So this will probably result in a 3-1 record, but it could also be 2-2, or 4-0. The math to determine the most likely record is simple: 1.00 + 1.00 + 0.50 + 0.50 = 3.00. This is the most probable number of wins...nothing more, nothing less.

A little trickier:

Game 1: 100%
Game 2: 100%
Game 3: 25%
Game 4: 25%

Applying the same math as before, 1.00 + 1.00 + 0.25 + 0.25 = 2.5. WHOA NOW HOW CAN YOU WIN 2 AND A HALF GAMES? Obviously, you can't, but 2.5 wins and 1.5 losses is still the best prediction assuming those win probabilities are accurate.

This is not to say that going 2-2 and 3-1 are equally likely, or that the odds of winning 3 or 4 games is just as likely as winning only 2 games. In fact, you are 56.25% likely to win just 2 games, losing both games 3 and 4 (calculated by multiplying 0.75 X 0.75, the probability of NOT winning each of those games). On the other end, your odds of winning all 4 games are only 6.25% (calculated by multiplying 0.25 X 0.25), leaving only 37.5% chance of winning exactly 3 games.

Now multiply .5625 X 2 wins = 1.125. And .375 X 3 wins = 1.125. And .0625 X 4 wins = 0.25. Adding up 1.125 + 1.125 + 0.25 = 2.5! This is not a coincidence. This is the average number of wins in that series of 4 games, if they could be played many times. It doesn't matter that they will only be played once each.

Why are you guys harping on the fact that the games will only be played once, anyway? The fact that you're assigning percentages to each game implies that you believe LSU would win X out of 100 games, if they could play 100 times.

North Carolina 50%
@ Vanderbilt 75%
Miss. State 85%
West Virginia 80%
Tennessee 60%
@ Florida 30%
McNeese St. 99%
@ Auburn 45%
Alabama 45%
UL-Monroe 99%
Ole Miss 80%
@ Arkansas 50%

North Carolina 60%
@ Vanderbilt 85%
Miss. State 85%
West Virginia 80%
Tennessee 65%
@ Florida 35%
McNeese St. 99%
@ Auburn 50%
Alabama 50%
UL-Monroe 99%
Ole Miss 80%
@ Arkansas 50%

[quote]Game 1: 100%
Game 2: 100%
Game 3: 25%
Game 4: 25%
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What you have there are 2 forfeits or disqualification by the LSU opponents, and LSU playing 2 games that are not cfb since there are 4 possble results to the contest.
Pr is defined as a # between 0 and 1, with 0 impossible and 1 certain.
I don't make the rules, I was just stating what I remember from school about "chance".

From dusty stat book:
The theory of chance consists in reducing all the events of the same kind to a certain number of cases equally possible, that is to say, to such as we may be equally undecided about in regard to their existence, and in determining the number of cases favorable to the event whose probability is sought. The ratio of this number to that of all the cases possible is the measure of this probability, which is thus simply a fraction whose numerator is the number of favorable cases and whose denominator is the number of all the cases possible.

– Pierre-Simon Laplace, A Philosophical Essay on Probabilities

(no message)

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No, playing 2 games with a 25% chance of winning in each game gives you a 43.75% chance of winning at least one of them, with 6.25% of that being the chance of winning both, so there's a 37.5% chance of winning exactly one.

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yes...I didn't quite word my point properly, which was to get the third win you're going to have to win a game that you have a 25% chance of winning. If you lose the first one you don't have a 37.5% chance of winning the second.

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WTF are you talking about?

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If by quoting your dusty stat book you are trying to imply that the probability of a team winning is always .5 since there are only 2 possible outcomes, you are way off target.

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Yeah, he's on some weird track like that. Apparently, he believes that any event that can be reduced to two possible outcomes must automatically have exactly the same chance of either outcome occurring. Though he hasn't responded to my question about hitting in blackjack with 20 (which has two possible outcomes: Ace or bust). I don't know if he's drunk or just trying to make some bizarre point about theoretical uncertainty, but so far it is complete nonsense (and really bad probability theory).

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When you play a football game there are 2 possible outcomes (let's keep it simple) but the outcomes are not equally likely.

Jesus Christ.

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Amen.