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

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North Carolina 65%
@ Vanderbilt 90%
Miss. State 90%
West Virginia 65%
Tennessee 55%
@ Florida 30%
McNeese St. 99%
@ Auburn 45%
Alabama 45%
UL-Monroe 99%
Ole Miss 60%
@ Arkansas 60%

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Tennessee has no QB and no offensive line. No Eric Berry. It better be 100 % LSU wins this game.

I said I was giving up, but I can't help myself.

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To arrive at 51%, 25%, 99%, etc "probabilities", statistical references, interconf records, common matchups and other measurements would be used to speculate an outcome. This would have nothing to do with "chance".

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This is not the argument. One can use any number of methods to arrive at a probability of winning a game. There is no way to know who is right or wrong, because as repeated ad nauseam the games will only be played once. All we can do is make educated guesses and see what happens.

But it doesn't matter how accurate one's predictions are to convert individual game win likelihoods into a predicted final win-loss record because the two are one in the same. If I say LSU will finish 9-3, I could just as well say they have a 75% chance of winning each individual game. This is exactly the same as a basketball player who is a 75% free throw shooter. Out of every 12 free throws, on average he makes 9 of them. In other words, typically he scores 0.75 points for each free throw attempted. If the guy just got divorced or something and I think he will be off his game, I could either say I think his FT% will drop to 50% this game, or I could say if he takes 12 FT's, he will only make 6 of them. Either way, it's the same thing. For every shot he takes, I will give him half a point. I may be right, I may be wrong... it's all speculation. All I'm trying to say is you can convert a series of win probabilities into a probable win-loss record (or made-missed free throws, or whatever).

Now let's try the blackjack example. You deal yourself a couple of 10s. That leaves 50 cards, with only the 4 aces giving you 21. In this scenario standing at 20 isn't an option because it's akin to forfeiting a football game. You have to take the hit, resulting in either a win or a loss. Well, 4/50 gives you an 8% (ie. .08 probability) chance of winning. Now give yourself a hit 12 times (not adding them all together, just take each individual hit and see if it's an ace), reshuffling in between. Add together .08 + .08 + etc etc 12 times (or, you know, just multiply) gives you .96, just less than 1. So of those 12 hits (ie. football games) you will probably win once. Of course, you may not win any, or you may win all of them. I'm sticking with my guess of 1 though. Now, if I wanted to predict that you would win 6 out of those 12 hits, I would have to load the deck with aces. In fact, of the 50 remaining cards I would have to stick 25 aces in there to make my prediction probable. As you can see, this changes your probability of a win to 25/50 = .50 = 50%. Putting any other number of aces in the deck just wouldn't make sense. So here comes what I think is your leap of faith: if I load 2 decks, 1 with 30 aces (out of 50 total cards) and one with 20 aces (out of 50 total cards), which corresponds to 60% and 40% wins for you, and deal you 6 hits from each deck, I would still expect you to win 6 of the 12 hits. .60 + .60 + .60 + .60 + .60 + .60 + .40 + .40 + .40 + .40 + .40 + .40 = 6. Think about it... I might as well just shuffle the 2 decks together, which would put 50 aces in a 100 card deck, giving you a .5 probability of winning each hit. The predicted win-loss record is directly related to the probability of winning each individual game. Saying anything else just doesn't make sense.

Now, I don't know if it's the fact that you can convert between those 2 different ways of saying the same thing that has you confused, or if it's the math when varying probabilities are involved. Please tell us exactly what it is about the whole thing that you don't believe or comprehend.

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Flawed. Each of those is a chance. 4/50 for an ace and 1-4/50 for bust, 1/1 holding pat, dealing only yourself, with a fresh deck of course.
3 seperate chances, you decide which to take.

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No, holding pat was not one of the options my question examined. I was asking about the predicted outcomes of hitting, and there are two: an ace or a bust. According to your earlier formula, that means there is a 50% chance of either occurring.

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Thankyou! Refer back to definition of chance above.
To arrive at 51%, 25%, 99%, etc "probabilities", statistical references, interconf records, common matchups and other measurements would be used to speculate an outcome. This would have nothing to do with "chance".

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So you are saying that when there are only two outcomes, there cannot be a 60% "chance" of one of them occurring because anything other than 50/50 would be "probability" and not "chance"?

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No, holding pat was not one of the options my question examined. I was asking about the predicted outcomes of hitting, and there are two: an ace or a bust. According to your earlier formula, that means there is a 50% chance of either occurring.

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Sorry, I missed the no pat part. But in short, no. That would involve two inequal probabilities that would disqualify the term chance, by classical definition.

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So you are saying that when there are only two outcomes, there cannot be a 60% "chance" of one of them occurring because anything other than 50/50 would be "probability" and not "chance"?

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Correct ,60%= 3/5, 6/10 etc. Check back 10 or 15 posts and it's quoted verbatim. I only repeated it.
I still say as in previous post, it's ok to say I think, have a hunch, gotta good feeling on a game, but there's very vague, unexact measures to try and put percentage points on.
Just where did these numbers come from? To actually numerically quantify each advantage teams have over each other is exhaustive and still very short of exact.

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Check back 10 or 15 posts and it's quoted verbatim. I only repeated it.

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I saw that. But what you quoted was not a definition of the word "chance", but rather a use of the word in the phrase "theory of chance", a term of art used in statistics and some probability study. The essay from which you quoted was expounding on a specific application of LaPlace's mechanics of infinitesimal calculus. That is a methodology for reasoning the probability of past unwitnessed events having actually occurred based upon witnessed events. While that usage of the word "chance" is meaningful in that area of probability studies, it is by no means the only meaning of the word and is not close to the usage employed in this conversation. For more relevant meanings of the word "chance," you can see here.

If Merriam Webster is too plebeian for you, you can try this:

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Probability is a number representing the likelihood that a particular event will occur by chance.

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Three definitions of Probability

* A priori: rational analysis as in the examples.
* Relative frequency: if there are many experiments of the same kind, the probability of a favorable event would be
. . .
*Intuitive Probability : An evaluation of the probability based on informal criteria.

In statistics all three definitions are used.

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where "chance" is used to describe a determining factor of probability, which includes intuitive probability where informal criteria (as opposed to true randomness) are applied.

The meaning of "chance" is this discussion is more particularly that expressed in definition 4(a) of the Merriam Webster:

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the possibility of a particular outcome in an uncertain situation; also : the degree of likelihood of such an outcome <a small chance of success>

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The fact that the science of statistics often manages that notion with the term "probabilities" instead of "chance" does not change the fact that the term "chance" does indeed fit.

However, on a more academic note, I think your initial analysis might still be flawed, because in fact the sample space of a football game is not two outcomes. Even though Texas beating Akron by 40 and Texas beating Akron by 4 are both Texas wins, they are in fact two different outcomes that are simply grouped into the same event (Texas wins). One could suggest that there are infinity outcomes in each event (with the two teams theoretically capable of scoring any two unequal integers), making each as likely as the other to occur randomly. But in fact there are not. There are physical limitations on the amount of scoring that can even theoretically occur (it will always take some time for points to be scored, and the clock will eventually run out). As such, there is a finite number of possible outcomes. And since those physical limitations are inherently both varied and non-random, the outcomes cannot be assumed to be equally distributed between the two events.

Again, that is all irrelevant to the discussion because you used the wrong meaning of the word chance in the first place.