Math exercise - What is my probability of winning?

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?  Show your work.  

(I will give the answer later this afternoon or early this evening.)

0

 

Showing my work......

 

Niners 30 Rams 13

 

 

1) if the Rams defeat (or tie) the 49ers tonight OR the combined total for both teams tonight is 46 points or more.

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

These two are unrelated.

Assuming chance of 2) going one way is 50%/50%.

It depends on what one assumes as a normal standard deviation from the line set, so what you provided lack information.

This is not my strong suit, but I'm assuming it's around 45-50%

I think FadeOnly has some good insight to the probability. I would guess he is close or it is a little more towards 40%?

I never hedge. 

Actually I am going to change my opinion, to lose BOTH of them is probably a fair amount under 50% so you probably are at least something over 55 to 60 to win 1 of them. 

Ed, great avatar, who is that girl?

Denise Milani.  Google her.  She's my favorite pinup gal.

Your chance of winning the pool is 33%.

You have 3 chances. 2 chances contain a goat and one chance contains a winner

You take a chance. Another chance is revealed to show a goat. You could stick with your first choice or Should you swap your pick? The answer is swap.

You have a one in 3 chance of winning or 33%. You have a 66% chance of picking a goat.

Once you take your chance and another is revealed to you. You now have a 50% chance of choosing the winner or the goat IN THEORY ONLY.



In reality you have a 66% chance of winning if you swapped . You already know where one goat is.

Its the Monty Hall problem

However, you are taking the goat so you have a 33% chance of hitting and a 66% chance of losing. Since you picked a goat, I suggest you swap picks and take SF.

I got about 41%

 

I took the odds of them winning a +165 ML , approx. 37.7%

then the odds of the game going OVER 46, approx. 45%

I then took the combination of both scenarios, basically 200% (added 37.7% and 45%) to make 82.7% (out of 200%) which gives me a little more than 41%

 

Did I do it right? Who knows lol

 

 

I totally fcked up. If theyre 45% to hit just the OVER how could they be only 41% to win the bet??

I'll say add the percentages

I'll say 82.7% lol

 

 

 

 

The lines offered by books do not reflect the probability of the event occur, they reflect the incoming money...

Ding, Ding Ding.  We have a winner.  That winner is... drum roll please... DWN (with his second post.)

The Rams are underdogs by 3.5 points.  This gives them an estimated probability of winning the game of .393.  (My source is https://wizardofodds.com/games/sports-betting/nfl/)

So I have a disadvantage here.

Since the Over/Under is 44, and yet I need 46 or more points, I also have a disadvantage here, albeit a slight one.  I have about a 44% chance to cover this.  (Source: https://wizardofodds.com/games/sports-betting/nfl/alternate-totals/)

Note that these two events are unrelated.

The only player still alive in the pool other than me needs BOTH events to win.  (As stated in the problem, I don't need both.  I need either/or.)  Thus, his overall chance is the PRODUCT of HIS two individual chances.  He has a .607 probability of San Fran winning, and a .56 probability of the game going under 46.

.607 x .56 = .3399.  Although his chances of EACH event are greater than 50-50, because he needs BOTH events to occur, his overall chance to win is about 34%.  

Thus, MY overall chances are about 66%.  (1 - .34)

(I decided ahead of time if you answered between 64.5% and 67.5% I would grade your answer as correct.)  

If you didn't know that a 3.5 underdog only wins approximately 39% of the time (or the percentage of the time ANY underdog wins, given the pointspread), then shame on you.  That information should be at the fingertips of everyone who wagers on NFL games - it's good information to know.

You also needed to know the chances of two independent events occurring is the PRODUCT of the chance of each individual event.  The chances of you flipping a coin twice and having it land tails both times (for example) is the product of each individual event.  1/2 x 1/2... or 1 out of 4.  The chances of you flipping a coin three times and having it land tails all three times is the product of each individual event.  1/2 x 1/2 x 1/2... or 1 out of 8.

FadeOnly's estimations of each individual event WERE good, but I don't need BOTH events to occur, I need EITHER event.  As shown, my overall chance is MUCH greater than 40 or 50%.

BlindBind said I did not provide enough information.  BUZZZZZ.  There was enough information there to solve the problem.

As I suspected, there will always be individuals like Train who have no interest in math or probability.  He didn't know the answer either, which is why he tried to get cute.

I'm well aware of the famous Monty Hall problem, and how and why you should always swap doors, but that problem has absolutely NOTHING to do with this exercise.  Absolutely nothing.  BUZZZZZ.  If anything, there are four "doors" in this problem and I win when three of them are opened:

Rams win a low scoring game (I win)
Rams win a high scoring game (I win)
49ers win a high very high scoring game (I win)
49ers win a low scoring game (I lose)

If the game were a pick'em and if my total was right on the Over/Under, you should be able to see with three out of four "doors" in my favor I would have a 75% chance of winning.  But since the 49ers are a slight favorite, and since I'm at a slight disadvantage with my 46 point total, my overall chances have to be less than 75%, and the above mentioned method is the only way to compute them.

Kudos to FadeOnly and Carpinteria who made a stab at solving the problem.

I find it interesting (but not surprising) that over 465 individuals read this thread, and only ONE of them provided the correct answer, which no one else took the time to confirm.

Thanks for the Denise Milani info - unreal and I think those boobs are real!! what a body and face!! And thanks also for the math lesson, your explanation was much more precise, mine was just a rough guess.  

I PREFER BOOBS OVER THE MATH

Ed-Collins this rule is accurate but keep in mind that this implies that the percentages are only suitable if and only if you bet 12 game considering there are almost 15 games a week that betting percentage you're given can be off a tad however this probability algorithm works well for getting for me!!!

Um... you lost me.

What rule are you referring to?  I mentioned no rule.
What does betting 12 games have to do with anything?

My post was nothing more than a math problem, with a football theme.

OR-probability is calculated as such: 

You lost me.  Are you disputing my answer?

According to your equation, my probability was zero:
Probability of A (.39) + probability of B (.45) - probability of A and B (.84) = probability of A or B

.39 + .45 - .84 = 0
That makes no sense.

probability of A and B means the probability of A and B occurring, not their sum. as stated above by you, it's simply their product.

this is more clear: P(A) + P(B) - P(A and B), where

He only needs one to win.

So he states it correctly when he says that his opponent has roughly 34% chance of winning because the independent events are .61 and .56.

This give him roughly 64% chance of winning.

Correct?