Probability of a Game Going Over a Certain Prop #

[Grizzly Joe]

I'd appreciate any help on this...

There are two teams, each with a calculated 60% chance of going OVER on a certain prop bet (based on previous game performances, and whether they're playing on grass/turf etc).

The prop is Longest Field Goal Made By Either Team 44.5 yds.

How do you find the COMBINED probability of one or both of the teams going OVER the prop? What's the formula?

Is the answer a COMBINED 60% chance of going OVER? 64%? 84%?

What is it, and how do you find the answer for other team percentages, such as 60/70, or 80/80? What's the formula?

[crabman]

to much activity by my brain early in the morning to comprehend what u r asking

[Grizzly Joe]

Lol, okay 15000...

Two teams are figured to each have a 60% probability of going OVER a 44.5 yds Longest Field Goal proposition bet in a game.

What's the combined probability that at least one of them will kick a field goal OVER 44.5 yds?

[Bisket]

wouldn't it be 60%?

if 2 guys in basketball both shot 60% from the free throw line, the probability of either one making a free throw is still 60%, no matter who goes to the line.

[Grizzly Joe]

Originally Posted by Bisket23

wouldn't it be 60%?

if 2 guys in basketball both shot 60% from the free throw line, the probability of either one making a free throw is still 60%, no matter who goes to the line.

It's hard to argue with that logic Bisket. Thanks.

[Kramer]

Say you take a coin, which has two possibilities: heads or tails.
For whatever reason, you want to get a tails. (This is your "out".)
When you flip the coin, you have a 50% chance of getting tails.
So let's say you're going to flip it TWICE, and you want to know your odds of getting tails at least once. What's the answer? Well, you have a 50% chance of getting tails the first time, and a 50% chance of getting tails the second time.

If you add those up, that's 100%. But of course, you know it CAN'T be 100%, because there's a chance you flip heads twice in a row. So what's the deal? This is EXACTLY like poker.

The turn card is the first "coin flip" and the river is the second. But you can't just add your chances of making on the turn and then the river... for the same reason you can't add 50% to 50%. Now... the ANSWER to our little puzzle is that your chances of getting tails at least once is 75%.

You can break it down like this into four possible outcomes:

Flip 1: Heads -> Flip 2: TailsFlip 1: Tails -> Flip 2: TailsFlip 1: Tails -> Flip 2: HeadsFlip 1: Heads -> Flip 2: Heads That's all your possibilities. The first three all include a tails, but the fourth does not. So three out of four times you'll get tails: 75%. How do you do this the "mathematical" way? It's actually easy.

All you have to do is multiply the chances AGAINST you the first time by the chances AGAINST you the second time. Then subtract that number from one.

So... your odds of heads the first time equals 1/2 and your odds the second time equals 1/2. 1/2 x 1/2 = 1/4 Then subtract it from one: 1 - 1/4 = 3/4 And that's it! Don't you just feel SMART right now?

[Cubbiefan16]

The probability is greater than 60% that one will do it. But with sports it's not really worth it to figure those types of percentages because there's so many different factors involved (who they're playing, wind, injuries, etc). There isn't a fixed probability of an outcome; it's not like rolling a dice or flipping a coin.

[Bobtheicon]

First off - I'm pretty f*cking drunk right now (yes, I know its Sunday, I'm on vacation).

First you have to realize that in order to achieve the goal in this example, a team would have to have possession of the ball. Now, figure that a rough estimate would be that a single team would have possession of the ball for 50% of the game (of course that is not always the case, but before the game starts that would be a fair guesstimate).

So, if a team has a 60% chance of hitting the target for possessing the ball for 50% of the game, then the odds would logically double were they to possess the ball 100% of the time (which is really what we are talking about now since one of the two teams is always going to be in possession of the ball).

I don't know the formula, but it seems to me in my drunken state that it would be an 80% chance.

[Bobtheicon]

Okay, I just read Kramer's post. If my logic holds and I understand what Kramer said, then:

4/10 x 4/10 = 16/100

1 - 16/100 => 100/100 - 16/100 = 84/100

84%

Hell, I was close.