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I have a potential prop bet in the making where my opponent has to pick 7 NFL teams not to make the playoffs for this upcoming season. If all 7 of their teams miss the playoffs, they win, if at least 1 of their teams make the playoffs, I win.

I've already gone ahead and calculated the vig free odds for the worst 7 teams to make the playoffs based on the odds off of one of the top online sports books:

Colts: 9.06% Browns: 9.50% Vikings: 11.37% Jaguars: 11.69% Rams: 13.22% Buccaneers: 18.01% Redskins: 18.01%

Now my question is how do I figure out what the odds are of at least 1 of those above teams making the playoffs in order for me to win my bet? Thanks.

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The opposite of at least one in is none in. – Frenzy Li Aug 14 '12 at 5:14
Easy if we assume independence, but it is hard to know whether independence holds even approximately. – André Nicolas Aug 14 '12 at 5:18
@AndréNicolas Yep, I've provided an answer based on independence which is not often met in races. – Frenzy Li Aug 14 '12 at 5:21
I see. If that's all the data you have, then the problem may not be solvable better than assuming they are independent. I should also warn you that a defensible strategy, according to a football fan I know, would be to try to pick the worst team in each division, ignoring what they see as the most competitive division. The Colts and Jaguars are in the same division, so the probability that both miss may be lower than what you'd expect from independence. Of course on paper assuming independence that would be better for you, but in practice I don't know. – Logan Maingi Aug 14 '12 at 6:51
Of course independence is not going to hold. It is likely some of these teams will play each other. The winner will then have an increased probability of making the playoffs while the losers chances will decrease. – Michael Chernick Aug 14 '12 at 11:03

If we view whether one team will or won't make the playoffs is independent from the performance of other teams, then the chances that all of them not making the playoff is

$$\operatorname{Pr}(\text{None Making}) = \operatorname{Pr}(\text{Team 1 Not}) \operatorname{Pr}(\text{Team 2 Not}) \cdots \operatorname{Pr}(\text{Team n Not})$$

So $\operatorname{Pr}(\text{None Making}) = (1-9.06\%)(1-9.50\%)(1-11.37\%)(1-11.69\%)(1-13.22\%)(1-18.01\%)(1-\mathbf{18.01\%}) \approx \mathbf{37.58\%}$

So $\operatorname{Pr}(\text{At Least One Made}) = 1 - \operatorname{Pr}(\text{None Making}) \approx \mathbf{62.42\%}$.

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How to get that annoying percent sign show up in $ $? Thanks. – Frenzy Li Aug 14 '12 at 5:22
so in other words, if I am offering them even odds on the bet and assuming those odds are correct and they take the teams with the 7 worst odds which might not be the case, I am +EV? – Harley Aug 14 '12 at 5:26
@Harley Problem is that their performance will usually not be independent, and this depends on the specific system of winner-choosing of Football. If two teams played against each other then there performance will be dependent. I don't know the impact because I'm not familiar with football. So I cannot say for sure, which means that a more accurate math model will give a different probability. I guess it's near? – Frenzy Li Aug 14 '12 at 5:30
for the last part of your equation, you did 1-8.01% instead of 18.01% but ya thanks I got the jist – Harley Aug 14 '12 at 17:29

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