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EdCollins
EdCollins
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October 13th, 2014 at 3:25:10 PM permalink
Here's a math exercise for those of you who enjoy this type of thing.

I will win our office football pool this week if the Rams defeat (or tie) the 49ers tonight OR the combined total for both teams tonight is 46 points or more.

Assume the current line is SF -3.5 with an over/under of 44.

Given the above information, what is my estimated probability of winning this week's pool?

(I will give the answer later this afternoon or early this evening, although I'm pretty sure with this crowd, I don't need to. I belong to a couple of other forums and most of those guys are clueless to this sort of thing.)
Ayecarumba
Ayecarumba
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October 13th, 2014 at 3:48:41 PM permalink
Here is the Wizard's formula for converting spreads to probabilities:

Quote: Wizard


My general formula for the probability of an x-point underdog winning is e-0.13176*x/(1+e-0.13176*x).

Simplicity is the ultimate sophistication - Leonardo da Vinci
EdCollins
EdCollins
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October 13th, 2014 at 3:53:04 PM permalink
I just use his "Probability of Winning in the NFL by Point Spread" chart.

http://wizardofodds.com/games/sports-betting/nfl/
sodawater
sodawater
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October 13th, 2014 at 4:01:54 PM permalink
Editing my post because I didn't notice your needed total was different from the posted total.


Depending on how you calculate the chances of winning over 46 when the posted total is 44, your overall chances of winning are around 65.5%
Ayecarumba
Ayecarumba
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October 13th, 2014 at 4:20:19 PM permalink
I get 41.2% on the back of my napkin. Probability of winning the straight bet ~38.6%, Probability of winning the over ~43.8%. Add them up and divide by two = 41.2%

I think the over is a much better bet than the Rams covering. I think the total will go over 50.
Simplicity is the ultimate sophistication - Leonardo da Vinci
sodawater
sodawater
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October 13th, 2014 at 4:38:12 PM permalink
Quote: Ayecarumba

I get 41.2% on the back of my napkin. Probability of winning the straight bet ~38.6%, Probability of winning the over ~43.8%. Add them up and divide by two = 41.2%

I think the over is a much better bet than the Rams covering. I think the total will go over 50.



This is doing it the wrong way.

You need to take the chances of winning the straight bet = 38.6%. Chance of not winning: 100 - 38.6 = 61.4%

Chances of winning the over = 43.8% Chance of not winning: 100-43.8% = 56.2%

Chances of not winning either are .562 * .614 = 0.345068

1 - 0.345068 = 0.654932

OVERALL CHANCES OF WINNING OFFICE POOL = 65.4932%
EdCollins
EdCollins
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October 13th, 2014 at 4:42:10 PM permalink
Quote: sodawater

This is doing it the wrong way.

You need to take the chances of winning the straight bet = 38.6%. Chance of not winning: 100 - 38.6 = 61.4%

Chances of winning the over = 43.8% Chance of not winning: 100-43.8% = 56.2%

Chances of not winning either are .562 * .614 = 0.345068

1 - 0.345068 = 0.654932

OVERALL CHANCES OF WINNING OFFICE POOL = 65.4932%



Ding Ding Ding. We have a winner. Good work. (I calculated them to be almost 66%.)

So far, over 418 members have read my post in that other forum, and they still haven't gotten it right!
sodawater
sodawater
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October 13th, 2014 at 5:02:22 PM permalink
Quote: EdCollins

Ding Ding Ding. We have a winner. Good work. (I calculated them to be almost 66%.)

So far, over 418 members have read my post in that other forum, and they still haven't gotten it right!



Yeah, assuming no effect of correlation. I can't decide which way the correlation goes in my head, so just say it's 0
mustangsally
mustangsally
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October 13th, 2014 at 5:39:13 PM permalink
Quote: sodawater

You need to take the chances of winning the straight bet = 38.6%.

Chances of winning the over = 43.8%

I agree

I will do this another way from Counting 101 - Inclusion/Exclusion

we can first add the probabilities but do not divide by 2 as was done earlier

= 38.6% + 43.8% - (38.6% * 43.8%)
try it (I know you will)
math is fun!

same concept as rolling at least one 6
with 2 6-sided dice

1/6 + 1/6 - (1/6 * 1/6) because we can get both 6,6 and we already counted them once
(so we subtract their union)
2/6 - 1/36 = 12/36 - 1/36 = 11/36
with probabilities we can only add them if they are mutually exclusive, meaning that they both can not happen at the same time
If they can happen at the same time (the OP example of he wins the game and wins the over)
we have to subtract out that probability of both happening

of course 1-(5/6)^2

this Inclusion/Exclusion method works well for more complex problems too

Sally
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Ayecarumba
Ayecarumba
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October 13th, 2014 at 10:35:03 PM permalink
Thanks folks. I am always eager to learn (or relearn) things. I don't think I'll ever integrate the multiplication of probabilities over their sum. For some reason, it doesn't hit me as "logical", especially when the problem is presented as an, "or", (A win in Event X or Event Y results in a win) instead of a "both" (You must win Event A and Event B).

Did anyone confirm that the probabilities I used were appropriate?
Simplicity is the ultimate sophistication - Leonardo da Vinci

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