Money Talks Final, Fezzik vs. Blankenship, Matchbook should have the game up tomorrow

[QUOTE=MobileBandit;12358]Fez...What does the ^ mean in your equation? Thanks.[/QUOTE] That's an exponent. So 3 squared (nine) can be written as 3^2. 7 to the fifth power is 7^5.

That's What I Thought but [QUOTE=joelshitshow;12360]That's an exponent. So 3 squared (nine) can be written as 3^2. 7 to the fifth power is 7^5.[/QUOTE] the math doesn't work out to the answer Fez has of 52.25%... (.515 X .515 X .515)+(.515 X .515) X .485 X 3 = .585

i think he meant .485^3

Nah Fez had his math right, it might be a little tricky to understand. Basically you are adding up the probabilites that Fez wins 3-0 or 2-1 in the 3 trials. First, brush up on combinatorial math: [url]https://en.wikipedia.org/wiki/Combination[/url] so let's write n choose k as (n,k) and let's say x = .515 to make this easier Prob of 3-0 = (3,0) * x^3 Prob of 2-1 = (3,1) * x^2 * (1-x) Add those up and you get your 52.25 You can also calculate the probability of the other outcomes as follows: Prob of 1-2 = (3,2) * x * (1-x)^2 Prov of 0-3 = (3,3) * (1-x)^3

I think you're right but... [QUOTE=joelshitshow;12378]i think he meant .485^3[/QUOTE] He must have also meant "+" instead of "x" following the .515^2. But even that gives a different answer than 52.25%. Here's what I get... (.515 X .515 X .515)+(.515 X .515)+(.485 X .485 X .485)= 51.6% Thanks for your help on this. Just trying to understand...

spn, in the line giving the probabilities for winning 2 out of 3, you have to multiply it by 3, because there are 3 ways you can win exactly 2 games: A. winning B and C while losing A B. winning A and C while losing B C. winning A and B while losing C this will yield a 52.25% win rate.

Here's a good link to calculate this easily: [url]https://onlinestatbook.com/analysis_lab/binomial_dist.html[/url] N = number of trials (3) p = probability of succes (.515) and then click the greater than or equal to and enter 2 (since Fez would need to go at least 2-1) If you do it for 13 trials you have (in the same vein, to eliminate ties on 14 trials) Prob of a win: 54.39% No juice ML equiv: -119 Of course this is all based on Fez's estimate of 54% vs. 51%

Thanks Guys! [QUOTE=spn137;12388]Here's a good link to calculate this easily: [url]https://onlinestatbook.com/analysis_lab/binomial_dist.html[/url] N = number of trials (3) p = probability of succes (.515) and then click the greater than or equal to and enter 2 (since Fez would need to go at least 2-1) If you do it for 13 trials you have (in the same vein, to eliminate ties on 14 trials) Prob of a win: 54.39% No juice ML equiv: -119 Of course this is all based on Fez's estimate of 54% vs. 51%[/QUOTE] This helps me a lot Appreciate it!

[QUOTE=donniep;12387]spn, in the line giving the probabilities for winning 2 out of 3, you have to multiply it by 3, because there are 3 ways you can win exactly 2 games: A. winning B and C while losing A B. winning A and C while losing B C. winning A and B while losing C this will yield a 52.25% win rate.[/QUOTE] Yeah, maybe the "n choose k" function wasn't clear. (3,1) means "3 choose 1", which is calculated as: 3! / (1! * (3 - 1)!) -- where ! is a factorial Thus you have (3*2)/2 = 3 But yeah, what you said is correct as well and you can think of it that way for smaller numbers of trials. It's when you get to 13 trials that it can be confusing!

If you assume 55.5% instead of 54%, the number s get a lot higher... 54% was conservative......