# Odds & Probabilities

#### Jennifer Murphy

##### New Member
[FONT=&quot]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.[/FONT]

[FONT=&quot]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?[/FONT]

[FONT=&quot]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?[/FONT]

[FONT=&quot]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?[/FONT]

[FONT=&quot]Then I suggested that we convert everything to percentages. Denver’s 3:1 becomes 25%. Seattle’s 11:4 becomes 26.67%. The formula is 1/(N+1) where N is the normalized odds.[/FONT]

[FONT=&quot]Here’s the whole table:[/FONT]

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
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
[FONT=&quot]Just for fun, I added up the percentages. The sum is 112.06%. I didn’t really expect them to sum to 100%, but they were close. [/FONT]

[FONT=&quot]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.
[/FONT]

#### Dason

Where did you get the "Odds" from?

#### Jennifer Murphy

##### New Member
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.

#### Mean Joe

##### TS Contributor
Just for fun, I added up the percentages. The sum is 112.06%. I didn’t 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%.

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.

#### Mean Joe

##### TS Contributor
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.

#### Englund

##### TS Contributor
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

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#### Englund

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

#### Jennifer Murphy

##### New Member
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.

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#### Jennifer Murphy

##### New Member
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.

Thanks

#### Englund

##### TS Contributor
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

#### StatsClue

##### Member
[FONT=&quot]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.[/FONT]

[FONT=&quot]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?[/FONT]

[FONT=&quot]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?[/FONT]
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)

#### Englund

##### TS Contributor
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.

#### Roy Seymour

##### New Member
I just love the prediction game before such huge matches. The emotions are high and people are excited. Its like a festival where people just enjoy every moment. I looked on various search engines and found that the odds are not updated. This time I am rooting for 49ers!

#### Nonlinear_Zero-Sum

##### Member
On the determination of probabilities from odds using a first-order approximation of each team's relative chances using only its own odds with the conventional linear-inverse model and then normalizing the agglomerate to 100%, all I can say is ... you're doing it wrong.

The results using the nonlinear-inverse odds-to-probability conversion model appear below (the odds listed are from the initial post about the 2014 NFL playoffs, eventually won by Seattle).

[Methodology has received a US patent.]

It should be noted that the distortion of probabilities using the linear odds-to-probs conversion, with the inevitable normalization, is quite small in events with multiple outcomes, compared to the oft-extreme distortion with events with only two outcomes (see link below).

Finally, on a related note ... Go, Patriots.

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#### Dason

Finally, on a related note ... Go, Patriots.
Desire to misuse mod powers and ban user intensifies

#### Nonlinear_Zero-Sum

##### Member
I'd be wary of that strange tingling you feel tugging at your judgment ... it could lead to emotions.

Additionally, I wouldn't be concerned with yet another boring Patriots run through postseason from a purely rational standpoint. Oddsmakers have them at only about a 5% chance of even winning their own conference (see below), let alone the darn Super Bowl again.

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#### Nonlinear_Zero-Sum

##### Member
In the interest of completeness and symmetry ... the implied probabilities, nonlinearly derived from their odds, follow for the 2020 NFC championship.

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#### Nonlinear_Zero-Sum

##### Member
I mean who had both extreme-longshots TEN and MIN to 'win' ..what the.. off the charts ... shows what those darn oddsmakers know ... wow, the excitement reaches 11.

Can't wait until next week. I bet that Tom feels that way a lot.

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