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    Odds & Probabilities




    My 8 year old grandson is fascinated by the NFL playoffs. We were looking at a website that showed the odds of winning the Super Bowl for each of 12 teams that made the playoffs. Seattle was 11:4, Denver 3:1, San Francisco 13:2, and so on down to San Diego at 75:1.

    He asked me what 11:4 meant. I started by explaining that Denver, at 3:1, had 1 chance in 3 of winning the Super Bowl. He asked if Seattle then had 4 chances in 11?

    That didnít seem right. Then I realized that 3:1 really means 1 chance in 4 (not 3). If you get 3:1 odds, you risk $1 to win $3. If you bet that way 4 times, you should break even, winning once (+$3) and losing 3 times (-$3). Is that right?

    I explained that to him and suggested that we normalize the odds so that they are all relative to 1. 11:4 becomes 2.75:1 and 13:2 becomes 6.5:1. So far so good?

    Then I suggested that we convert everything to percentages. Denverís 3:1 becomes 25%. Seattleís 11:4 becomes 26.67%. The formula is 1/(N+1) where N is the normalized odds.

    Hereís the whole table:

    Code: 
        Team        Conf  Odds    :1     %
      Seattle        NFC  11:4   2.75  26.67
      Denver         AFC   3:1   3.00  25.00
      San Francisco  NFC  13:2   6.50  13.33
      Carolina       NFC  10:1  10.00   9.09
      New England    AFC  10:1  10.00   9.09
      Cincinnati     AFC  16:1  16.00   5.88
      Philadelphia   NFC  18:1  18.00   5.26
      New Orleans    NFC  20:1  20.00   4.76
      Kansas City    AFC  20:1  20.00   4.76
      Green Bay      NFC  28:1  28.00   3.45
      Indianapolis   AFC  28:1  28.00   3.45
      San Diego      AFC  75:1  75.00   1.32
    Just for fun, I added up the percentages. The sum is 112.06%. I didnít really expect them to sum to 100%, but they were close.

    So, how did I do? Have I screwed this little guy up for life? He seems to have a good strong interest, so I want to make sure I give him good information.

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    Re: Odds & Probabilities

    Where did you get the "Odds" from?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Odds & Probabilities

    Gosh, I can't remember. I think I did a search on "Super Bowl Odds" or something like that. I just did another. Here's one:

    http://www.betvega.com/super-bowl-odds/

    I don't remember if it was the one I used.

    But my question is not about the accuracy or validity of the odds, but about the validity of my analysis.

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    Re: Odds & Probabilities

    Quote Originally Posted by Jennifer Murphy View Post
    Just for fun, I added up the percentages. The sum is 112.06%. I didnít really expect them to sum to 100%, but they were close.
    Yeah, they won't add up to 100%. So that you don't benefit from betting on each team. Also, the house needs to take a cut. This might not be a reason why the sum is >100%.

    Quote Originally Posted by Jennifer Murphy View Post
    So, how did I do? Have I screwed this little guy up for life?
    I think you did great. Football gambling today, stock investing tomorrow.
    All things are known because we want to believe in them.

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    Re: Odds & Probabilities

    Quote Originally Posted by Jennifer Murphy View Post
    Gosh, I can't remember. I think I did a search on "Super Bowl Odds" or something like that.
    I don't remember if it was the one I used.
    I think he's trying to entrap us. Just say we're looking up the odds for learning purposes.

    My above post may not be correct, because I am not a gambler. I've never even been to Los Vegas.
    All things are known because we want to believe in them.

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    Re: Odds & Probabilities

    In order to get the bookmaker's implicit estimated probabilities, you could proceed as follows:

    We start by denoting \text{Odds for team i} = Odds_i = O_i and then proceed by noting that

    \sum_{j=1}^N{\frac{1/(O_j+1)}{\sum_{i=1}^N{(1/O_i+1)}}}=1.

    If we make use of this methodology we can see that

    \hat{P}(\text{Team j wins}) = \frac{1/(O_j+1)}{\sum_{i=1}^N{1/(O_i+1)}}.

    If we would use P(O_j) = 1/(O_j+1) we would get an estimated probability a little bit too high since, as already stated by Angry Joe, the bookmaker wants a positive turnover. They accomplish this by setting the odds a little bit lower than the estimated fair odds.

    Edit: You seem to be having a pretty cleaver grandson if he can follow your explanation
    Last edited by Englund; 01-07-2014 at 04:05 PM.

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    Re: Odds & Probabilities

    Note that the sum, \sum_{i=1}^N{1/(O_i+1)}=1.1206.

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    Re: Odds & Probabilities

    Quote Originally Posted by Mean Joe View Post
    I think you did great. Football gambling today, stock investing tomorrow.
    OK, thanks. Then I'll continue the discussion with him. We'll watch the odds change as the playoffs progress.

    I was not trying to get him into gambling, per se, but math. I got him a programmable calculator for Christmas. His dad says they have to take it away at night or he would never get any sleep.
    Last edited by Jennifer Murphy; 01-24-2014 at 09:54 AM. Reason: Typo

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    Re: Odds & Probabilities

    Quote Originally Posted by Englund View Post
    In order to get the bookmaker's implicit estimated probabilities, you could proceed as follows:

    We start by denoting \text{Odds for team i} = Odds_i = O_i and then proceed by noting that

    \sum_{j=1}^N{\frac{1/(O_j+1)}{\sum_{i=1}^N{(1/O_i+1)}}}=1.

    If we make use of this methodology we can see that

    \hat{P}(\text{Team j wins}) = \frac{1/(O_j+1)}{\sum_{i=1}^N{1/(O_i+1)}}.

    If we would use P(O_j) = 1/(O_j+1) we would get an estimated probability a little bit too high since, as already stated by Angry Joe, the bookmaker wants a positive turnover. They accomplish this by setting the odds a little bit lower than the estimated fair odds.

    Edit: You seem to be having a pretty cleaver grandson if he can follow your explanation
    I'm not sure he fully follows it, but he's definitely interested. It can't hurt to stretch him a little. He's in second grade, but they have him doing 4th and 5th grade math.

    Not if only I were clever enough to follow your explanation!

    Thanks

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    Re: Odds & Probabilities

    Quote Originally Posted by Jennifer Murphy View Post
    It can't hurt to stretch him a little. He's in second grade, but they have him doing 4th and 5th grade math.
    That's awesome, and it is true that encouraging him is also a good thing. As a hobby psychologist I know I've came across many sources that states that high expectations from family and friends is a good predictor for future success. As an anecdote I can tell you about when I was studying my second semester statistics (statistical theory course) I impressed my lecturer by solving a homework brilliantly and without asking for help (which everyone else did, at least that was what he said), and after that he seemed to have high expectations on me which made me study even harder and I ended up with 60 out of 60 points on that exam. Now I'm almost done with my Master's degree and I think that my lecturer's expectations on me have had a major influence in this

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    Re: Odds & Probabilities

    Quote Originally Posted by Jennifer Murphy View Post
    My 8 year old grandson is fascinated by the NFL playoffs. We were looking at a website that showed the odds of winning the Super Bowl for each of 12 teams that made the playoffs. Seattle was 11:4, Denver 3:1, San Francisco 13:2, and so on down to San Diego at 75:1.

    He asked me what 11:4 meant. I started by explaining that Denver, at 3:1, had 1 chance in 3 of winning the Super Bowl. He asked if Seattle then had 4 chances in 11?

    That didn’t seem right. Then I realized that 3:1 really means 1 chance in 4 (not 3). If you get 3:1 odds, you risk $1 to win $3. If you bet that way 4 times, you should break even, winning once (+$3) and losing 3 times (-$3). Is that right?
    If the odds of team A winning are 3:1, 3 victories to 1 loss are expected. The probability of winning is 3/4. Odds is not a probability (probability, or simply, chance, is a fraction). Odds is a ratio.

    Odds = probability of event divided by probability of non-event. (in your case: 3/4 divided by 1/4 = 3/1=3:1)

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    Re: Odds & Probabilities


    Quote Originally Posted by StatsClue View Post
    Odds = probability of event divided by probability of non-event. (in your case: 3/4 divided by 1/4 = 3/1=3:1)
    **** then my explanation above is slightly incorrect.

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