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You know the drill. You give us percentages on each game. For instance, you could say that FSU has an 80% chance of beating Maryland (0.80), or that they have a 50% chance to beat UF (0.50). We then add up those odds for you (you're welcome, humanities major). And you get a picture of a likely final schedule result.
If you think about it mathematically, you can assign a non-negative real number instead of a discrete 0 or 1. The result is that you can specify a level of confidence you have about a predicted win or loss. A win percentage more (less) than 0.50 means you're inclined to predict the game as a Nole win (loss) - just not every time. A win percentage of 1 (0) means FSU wins this game 100 (0) times out of 100.
This is your chance to show us how good or bad you think Florida State will be. It's simple. Ask yourself, "how likely is FSU to win in each game?"
Take the survey by clicking this sentence.
Please share your results in the comment section.
Here's a link to the spreadsheet if you want to look at yours' or others' numbers.
For those of you who are reading this, I'm already gone.
No, no, no. There's more quantitative fun. TN has its very own professor in statistics in csfuu, and he's done a great write-up on how to take your accumulated win percentages and turn them into probabilities on a final record. For instance, your win prediction poll says 9.5 wins. That's nice, but 9.5 is a non-integer so you're not going to finish any season with 9.5 wins. So what is the probability of a 9 win season given your WPP of 9.5? Or 10 wins? 11?
(For a complete read-through of good work, check out his whole article here. I'm just pilfering it and copying-and-pasting portions from it below).
I'm going to do the calculations using R, "a freely available language and environment for statistical computing and graphics," and I will provide an R function at the end of this post that you can use with your own probabilities. R can be downloaded from CRAN (for Windows you can just click here).
Here are arrdubs's win probabilities for the individual games, taken from his post. Note that a 1 is assigned to games we've already won and a 0 to games we've already lost. This is the format I used to enter them in R (it one should be just one line, but it wouldn't fit here that way and R won't care).
arrdub <- c(Sam=1,OU=0,BYU=1,WF=.9,UVa=.7,Miami=.4,BC=.8, NCSt=.65,UNC=.65,Clemson=.65,UMd=.85,UF=.4)
The labels for the games are not required and I don't use them for anything here, so I could have just entered:
arrdub <- c(1,0,1,.9,.7,.4,.8,.65,.65,.65,.85,.4)The expected number of wins is computed simply by summing the probabilities (this is a special case; calculation of expected values is generally much more complicated). For arrdub this is
sum(arrdub)which yields 8.
First cut and paste this function definition into the R window (you could also save it into a plain text file named wintotpr.R and load it into R through some menu or the other or with R's source command):
wintotpr <- function(p, as.pct=TRUE, cumulative = "none") {
n <- length(p)
outcomes <- matrix(rep(c(FALSE,TRUE), n), ncol = n)
outcomes <- t(expand.grid(data.frame(outcomes)))
row.names(outcomes) <- names(p)
probs <- rep(p, times=dim(outcomes)[2])
probs[!outcomes] <- 1 - probs[!outcomes]
dim(probs) <- dim(outcomes)
seasontot <- apply(outcomes, 2, sum)
seasonprob <- apply(probs, 2, prod)
res <- tapply(seasonprob, seasontot, sum)
res <- switch(cumulative,
left = cumsum(res),
right = rev(cumsum(rev(res))),
res)
if (as.pct) round(100*res,1) else res
}
Now running
wintotpr(arrdub)gives the following results (I've reformatted them and removed the probabilities for 0, 1, and 12, which of course are zero anyway)
3(0%), 4(0.4%), 5(2.7%), 6(9.6%), 7(21.4%), 8(29.4%), 9(24%), 10(10.5%), 11(1.9%)
So 8 wins is arrdub's most likely total, but he shouldn't be too surprised with any of 6, 7, 8, 9, or 10 wins.
Here is one way of plotting that distribution:
plot(0:12, wintotpr(arrdub,as.pct=FALSE), type = "h",
xlab="Number of Wins", ylab="Probability")
Share your graphic below!