Math exercise regarding tonight's NFL game. 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?

(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.)

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).

I just use his "Probability of Winning in the NFL by Point Spread" chart.

https://wizardofodds.com/games/sports-betting/nfl/

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%

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.

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%

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!

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

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

of course 1-(5/6)^2

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

Sally

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?

Quote: Ayecarumba

Did anyone confirm that the probabilities I used were appropriate?

Yes, the were fine.

You said the probability of winning the straight bet was approximately 38.6% and the probability of winning the over was approximately 43.8%. Those numbers were fine for my estimations.

Quote: EdCollins

Yes, the were fine.

You said the probability of winning the straight bet was approximately 38.6% and the probability of winning the over was approximately 43.8%. Those numbers were fine for my estimations.

Thanks. At least I got that part right.

Edit: And congrats on your win Ed!

Quote: Ayecarumba

Thanks. At least I got that part right.

Edit: And congrats on your win Ed!

Thanks. I got lucky! As it turned out I needed that pick six!

Initially when the Rams went up 14 to zip in the first quarter, I was looking real good. The Rams are leading (the team I picked) and we're on pace for a 56 point game. Can't ask for more than that.

I knew once the Rams scored 23 points or more, I couldn't lose. (If San Fran scores enough to overtake this total, they win the game but the game would of course go over 46, so I win on the tie-break.)

But at some point somewhat late in the 3rd quarter it was only 17-14 San Fran, and I wasn't looking so good. I still needed two touchdowns and a field goal. I was fortunate enough to get it. I felt bad for anyone who took the Under.