Just add up the numbers
60 percent equals .6
50 percent equals .5
1.1 wins
You're a math major right? Help a kid out if you can.
What I did: go through the schedule of a given team (we'll go with the Vols), and assign to each game a probability of winning.
What I want to do with that: using the numbers I just came up with, figure out the chances that the team ends up winning [X] number of games. If we're talking the Vols, I might be looking to see what their chances of winning 8 games are, and if they're more likely (again, given my numbers) to win 6 games or 7 games.
That's it. I feel like this shouldn't be terribly hard, but I have absolutely no idea how to do it.
Balls deeep...
Just add up the numbers
60 percent equals .6
50 percent equals .5
1.1 wins
Not a math guy, but I think you multiply them together. The problem is you will get tiny, tiny numbers as you are not just calculating the number of wins, but rather the chances of winning all of the games. Which isn't what you are asking so this isn't really much help.
What may be more help is to look into something called Bayesian probability. That is going to get you where you want to go, but the journey will be boring and the results, while what you want, will likely not have the rigor or value you would expect (I distrust many applications of Bayesian logic and probability).
If you multiply them together you get the likelihood of every event taking place. I don't want that. I know the odds of UT going 12-0 are roughly the same as the odds of me having unprotected sex with Emma Watson live on national television.
I want to find out the chances of 6 wins, 7 wins, 8 wins, etc.
I have looked into Bayesian Probability, and a few other concepts I don't quite understand, and, well, I don't quite understand them. The closest I've come is Binomial distribution, but that requires that each event have the same probabilit, and so it does me no good.
Balls deeep...
The simplest way is the sum of the probabilities divided by the number of events.
If you want to do something more, do a Z score and compare to the z value for a 95 or 99 Percent distribution.
But you'll need to define a mean number of wins.
Can you elaborate a little bit more.
Simple example with 5 games.
Game 1: .40
Game 2: .15
Game 3: .95
Game 4: .35
Game 5: .50
Sum: 2.35
So when you say divide by number of events, you can't possibly be referring to the number of events for which i'm trying to get an accompanying percentage, because that wouldn't work here.
For example: 2.35/3 = .78. And clearly, given the listed percentages, there isn't a 78% chance of winning three games.
So obviously you are saying divide by the total number of games. Here: 2.35/5 = .47 So from there can I go back and just assign a .47 to each game and then use the binomial theorem? Because that actually doesn't sound like a terrible idea.
Balls deeep...
Can't he just run the standard deviation formula on it?
If I then use the binomial theorem to compute what I need, are my numbers going to be correct? In other words, for multiple independent events with differeing probabilities for each event, it's okay to just find the mean probability, assign it as the probability for each event, and proceed accordingly?
That almost seems too easy.
Balls deeep...
No, I would, if you had an average, use a z score.
They are used frequently in writing a warranty, which is kinda similar to what you are looking for.
For example, if a hair dryer has a mean life of 4 years, and a standard deviation of a year, a z score can be calculated to determine the probability of a dryer failing at 5 years. Or 6 years. And then based on that, the warranty period can be determined. If you find that 85% of your dryers fail by 5 years, and it would be detrimental to replace that many dryers, you set the warranty at 4 years. Or 3. Whatever you feel best, and whatever the market can bear.
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