For those of you that are good at statistics - If a football team has a .7 chance of winning each game this season: What is the odds they go undefeated over a 16 game season? Go 15-1? Win exactly 8 games? Please include your work because I have no idea how to do it…

a) 0.7^16 b) 0.7^15 x .3 c) 0.7^8 x 0.3^8

Can’t agree with your b or c. will elaborate after i get the numbers.

Excel gives me these answers, from building formulas - 0-4 games = 0% 5 games = .115% 6 games = .516% 7 games = 1.819% 8 games = 5.1% 9 games = 10.888% 10 games = 16.791% 11 games = 21.834% 12 games = 19.828% 13 games = 13.123% 14 games = 6.877% 15 games = 2.464% 16 games = .573% Thats with 2000 trials. How can I do this with a stat formula?

You said: a) 0.7^16 b) 0.7^15 x .3 c) 0.7^8 x 0.3^8 Thats the math that came to me first but the answers didnt make sense. B cant be right. It would make 15 games less common than 16. C cant be right, it would make winning 8 games extremely uncommon.

THinnk you may need to use the permuattion formulas of the form n factorial divided by (n-p)factorial. So 16!/8! for c? and 15!/14! for b? or smething? maybe you need to use the combo formulas instead. i will have to think about it.

W L % Chance 16 0 0.33% 15 1 2.28% 14 2 7.32% 13 3 14.65% 12 4 20.40% 11 5 20.99% 10 6 16.49% 9 7 10.10% 8 8 4.87% 7 9 1.85% 6 10 0.56% 5 11 0.13% 4 12 0.02% 3 13 0.00% 2 14 0.00% 1 15 0.00% 0 16 0.00%

Formula to use is: n!/(k!*(n-k)! * [chance to win]^(n-k) * [chance to lose]^k Where n is the number of games and k is the number of times you lose. Edit: You can visualize this as a huge binomial tree with each route as a unique way to reach an end-of-season record. needhelp was right with his first calculation of 0.7^16 because in this particular question, there is only one way to get to a 16-0 streak, which is is win 16 times in a row. However, for a 15-1 record, there are multiple ways to reach that result. For instance, you can win 15 games and then lose 1 game OR you could lose the first game and win the next 15. We need to account for these alternative routes, which is what the above formula does, while still accounting for the probabilities. Believe it for not, there are 2^16 ways to get to the end-game in this problem.

This is Binomial Distribution. n!/((k!(n-k)!)*p^k*(1-p)^(n-k) let n = 16 k = 15 p =.7 16!/15! * 0.7^15 * (1-.7)^(1) = 16*.7^15 * 0.3 You win 15 games and not win one game. ie 0.7^15 * 0.3^1 but this is for one outcome (ie lose the 1st game & win the rest, OR lose the 2nd game and win the rest…OR etc etc etc). There is 16 different ways to win 15 games and lose one. thus you multiple by 16. (15 choose 16)

I’m pretty sure this was on level 1. But i think you should use the binominal probability formula… http://www.mathwords.com/b/bernoulli_trials.htm (p^k)(q^(n-k)) p = probability of success q = probability of failure or (1-p) k = number of success n = number of trails

ymmt beat me to it.

ConvertArb Wrote: ------------------------------------------------------- > ymmt beat me to it. Heh, but my formula was incorrect. It’s really been too long since LI…

also this is not statistics… it’s probability.

Thanks ymmt. I’m impressed. I need to brush up on the stats!

This was on L1? Jeez its all been forgetten.