Warriors: Probability of the Warriors going scoreless for the last 4 minutes.

[da_suns_fan]

Who gives a ? All you need is an 3rd grade education (which it seems you don't have) to calculate probabilities. No one on planet earth would multiply occurrences to come up with a probability. Want to know what his re ed "formula" looks like when you apply it to dice roll odds?

.834 x .166 = 0.138444 or 13.8 percent.

Tell me with a straight face that the odds of rolling any one number with a die are 1/7?

His massively dumb way (which was why you're using it) essentially adds another face to a die

You really have no clue what youre talking about. He's talking about continuous probability and youre cherry picking events.

This one is in the ol deep freeze.

[midnightpulp]

You really have no clue what youre talking about. He's talking about continuous probability and youre cherry picking events.

This one is in the ol deep freeze.

And the way you calculate that is multiplying the number of favorable/total outcomes continuously. Rolling a 6 with a die is our favorable outcome, so we assign it a value of 1. The number of total outcomes is 6, ergo 1/6. If you want to know the continuous probability of rolling five 6s in a row, you do it: 1/6 x 1/6 x 1/6 x 1/6 x 1/6.

Let's see how that dumb 's formula calculates flipping heads twice in a row (which everyone on Earth knows is a .25 probability).

First we have to find the odds of flipping heads or tails, which this dumb calculates by multiplying the two outcomes against each other, so: .50 x .50=.25

Then we multiply .25 x .25=0.0625 or 6.25 percent.

His way produces a 1/16 probability of flipping heads twice in a row

His math is laughably wrong.

[da_suns_fan]

And the way you calculate that is multiplying the number of favorable/total outcomes continuously. Rolling a 6 with a die is our favorable outcome, so we assign it a value of 1. The number of total outcomes is 6, ergo 1/6. If you want to know the continuous probability of rolling five 6s in a row, you do it: 1/6 x 1/6 x 1/6 x 1/6 x 1/6.

Let's see how that dumb 's formula calculates flipping heads twice in a row (which everyone on Earth knows is a .25 probability).

First we have to find the odds of flipping heads or tails, which this dumb calculates by multiplying the two outcomes against each other, so: .50 x .50=.25

Then we multiply .25 x .25=0.0625 or 6.25 percent.

His way produces a 1/16 probability of flipping heads twice in a row

His math is laughably wrong.

You just dont get it.

Hes not multiplying the odds of a make and a miss against each other.

Lebron had both a make and a miss in a single possession. So he's saying the odds of Lebron missing a three (.66), heat getting an offensive rebound (.167) and Lebron making a three (.33) is .012 which is 1.2% (which sounds about right...in 1.2% of their possessions the Heat probably shot a three, missed, got the offensive rebound and hit a three).

Youre stupid "double headed die" analogy was both painful and hilarious. Its not that "he" is dumb. Its just that you didnt understand.

[midnightpulp]

You just dont get it.

Hes not multiplying the odds of a make and a miss against each other.

Lebron had both a make and a miss in a single possession. So he's saying the odds of Lebron missing a three (.66), heat getting an offensive rebound (.167) and Lebron making a three (.33) is .012 which is 1.2% (which sounds about right...in 1.2% of their possessions the Heat probably shot a three, missed, got the offensive rebound and hit a three).

Youre stupid "double headed die" analogy was both painful and hilarious. Its not that "he" is dumb. Its just that you didnt understand.

That's impossible.

There can't be two different offensive events in a single possession for individual players because a shot, a miss, or a turnover counts as a "possession used" in the NBA. You have to consider the events independently of each other, just like a coinflip. He's applying a different form a probability theory that is commonly used to analyze events with a load of more random variables, like trying to figure out the probability of how many years a steel beam will last under a load in a climate that is humid in the Summer. No one uses it for sports or gambling, since it's assumed the prior event doesn't influence a future event (in our steel beam example, the weather can influence the steel beam in some fashion).

See how SABR does it for baseball:

In this article, it will always be assumed that each at-bat, game, etc., is an independent chance event; for instance, that a batter's chance of getting a hit is the same in every at-bat.

As far as we are concerned, therefore, each time at-bat must be treated as a chance event, as far as our ability to predict or analyze the outcome is concerned. Even if it is conceded that the outcome is really determined in advance by complicated factors such as the precise configuration of batter's and pitcher's muscles, etc., it would not change things as long as the analysis of these factors remains impractical. The outcome of a roulette spin is also presumably determined by the laws of mechanics, but in a way too complicated to be useful. "Chance" is just a word we use for any combination of factors too complicated for us to control or analyze. In this sense, any particular time at bat, or any particular game, is a chance event whose outcome can only be predicted statistically.

His math is WRONG for this particular context.

[midnightpulp]

And here's a video from the Khan Academy, a great educational organization that has helped plenty of massively dumb people like yourself, calculating the odds of Lebron making 10 FTs in a row (guess what, they do it the exact same way as I and SABR do, considering each event an independent "chance" event):

[Stalin]

[midnightpulp]

You just dont get it.

Hes not multiplying the odds of a make and a miss against each other.

Lebron had both a make and a miss in a single possession. So he's saying the odds of Lebron missing a three (.66), heat getting an offensive rebound (.167) and Lebron making a three (.33) is .012 which is 1.2% (which sounds about right...in 1.2% of their possessions the Heat probably shot a three, missed, got the offensive rebound and hit a three).

Youre stupid "double headed die" analogy was both painful and hilarious. Its not that "he" is dumb. Its just that you didnt understand.

Furthermore, this should not be factored in. See the Khan Academy video. When figuring odds, you only count favorable outcomes (we're operating from the Heat's perspective, figuring out the odds they had of coming back). The odds of a Lebron make (.375) off a Heat o-board (.26) is about 10 percent, which is much more believable number than 1.2%. if the odds of scoring second chance points were that low, teams would only be scoring like 1 or 2 second chance points per game. The average amount of second chance points scored in the year's playoffs was about 12 per game.

Let's look at the math again, as well.

Lebron miss (.63), o-board (.26), Lebron made 3 (.37), Lebron miss (.63), o-board (.26), Ray Allen made 3 (.41)=0.00446704315 or 0.44%. Yeah, sounds about right. Despite another possession where the Heat missed a three, got the o-board, and then hit a 3, their percentage rate of that type of possession actually went down by more than 50%.

"I took discrete math in college."

[Kawhitstorm]

I failed Discrete mathematics in college.

FIFY

[midnightpulp]

FIFY

[midnightpulp]

And the Final nail:

But after a Spurs run, which culminated with a LeBron James turnover with 28 seconds remaining, the Heat trailed by five and had only a 1.5% chance of winning the game. In other words, with 28 seconds left, the Spurs had a 98.5% chance of winning the championship.

http://www.businessinsider.com/one-c...me-ever-2013-6

[140]

FIFY

[SquawkinHawkBigCock]

Is Da the new forum punching bag? Haven't seen Luva around in a while

[da_suns_fan]

That's impossible.

There can't be two different offensive events in a single possession for individual players .

Yes there can. Its called an offensive rebound. Were not determining individual usage percentages, were determining continuous probability.

Youre baseball analogy was just sad. A baseball player cant strike out and hit a home run on a single at bat, but he can strike out and then hit a home run on his next at bat. You can easily determine the probability that a given player will do these two things in that particular order which is the same as the source I used determining the probability that Lebron would miss a three and then make his next three.

Its clear to me that you realize you ed up. You actually claimed he was "multiplying the opposite probabilities against each other". He wasnt, you just didnt understand. You do this all the time.

[da_suns_fan]

And the Final nail:



http://www.businessinsider.com/one-c...me-ever-2013-6

Again, you dont understand.

But after a Spurs run, which culminated with a LeBron James turnover with 28 seconds remaining, the Heat trailed by five and had only a 1.5% chance of winning the game. In other words, with 28 seconds left, the Spurs had a 98.5% chance of winning the championship.

Thats the odds of Spurs winning using ANY sequence events. Youre confusing continuous probability with discrete probability.

You didnt determine the odds that Golden State would go four minutes without a point, you attempted to determine the odds of particular players missing particular shots and then giving up offensive rebounds in a given order.

All you do is expose your ignorance. Just remember I'll always be here to expose you when you "crunch some numbers".

[ambchang]

There are four distinct outcomes for the last sequence.

1) Lebron makes the three on the first try (37%)
2) Lebron misses the three on the first try, did not get offensive rebound (0.63 x 0.74 = 46.62%)
3) Lebron misses the three on the first try, got offensive rebound, Allen misses the next shot (0.63 x 0.26 x 0.59 = 9.7%)
4) Lebron misses the three on the first try, got offensive rebound, Allen makes the next shot (0.63 x 0.26 x 0.41 = 6.7%)

The total probabilities of these four events should add up to 1, and it does using MDP's method (see above in parantheses)

It doesn't using dsf's method.

In fact, if you use http://4.bp.blogspot.com/-xW5MJdDISp...ame+6+Math.jpg as the basis, just take the first scenario where Ginobili makes or misses the FT.

The probability of him missing the FT, as noted in the graphic, was 0.73 x 0.27 = 19.77%
Then what is the probability of Ginobili making the FT? It would be 0.27 x 0.73, which is also 19.77%

What are the other probabilities?

The person who came up with that graphic is clearly ignorant of how statistics work. let's break down all the possible scenarios using his method:
1) Ginobili makes FTs = 19.77%
2) Ginobili misses FTs, Lebron misses 3 = 0.73 x 0.27 x 0.375 x 0.625 = 4.6195%
3) Ginobili misses FTs, Lebron makes 3, Kawhi makes FT = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 = 1.0732%
4) Ginobili misses FTs, Lebron makes 3, Kawhi misses FT, Allen misses 3 = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 x 0.41 x 0.59 = 0.2596%
5) Ginobili misses FTs, Lebron makes 3, Kawhi misses FT, Allen makes 3 = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 x 0.41 x 0.59 = 0.2596%

The total adds up to 25.9219%. What happens to the other 74% or so? What are the other possibilities?

[midnightpulp]

Yes there can. Its called an offensive rebound. Were not determining individual usage percentages, were determining continuous probability.

Youre baseball analogy was just sad. A baseball player cant strike out and hit a home run on a single at bat, but he can strike out and then hit a home run on his next at bat. You can easily determine the probability that a given player will do these two things in that particular order which is the same as the source I used determining the probability that Lebron would miss a three and then make his next three.

Its clear to me that you realize you ed up. You actually claimed he was "multiplying the opposite probabilities against each other". He wasnt, you just didnt understand. You do this all the time.

Jesus, you are just a stubborn individual who can never admit you are wrong.

My baseball analogy was apt, since events in this context are INDEPENDENT CHANCE EVENTS, meaning Manu's made freethrow has NO INFLUENCE on his next shot, so it should not be factored in. His make is also IRRELEVANT to the problem we're analyzing (Heat's odds of coming back).

Let's use in' dice as an example again. You're the Heat, I'm the Spurs. We're playing a dice game in which we each take turns rolling dice. Both of us get 5 rolls. We're using a special set of dice that has a "magic" side with a value of 20, but the odds of the die landing on this magic side are 1/100. All of my rolls are 6's, giving me total of 30. Your first 3 rolls give you a score of 3. So to beat me, you need to do the improbable and roll two consecutive 20's.

You do. I'm pissed and want to find out just how much a chance you had of doing that. I ONLY consider the odds of the magic side. Your rolls of 1 are irrelevant since it's a favorable outcome to me. I do this by obviously 1/100 x 1/100.

Now if I wanted to figure out the odds of your SPECIFIC 1, 1, 1, 20, 20 sequence, then I would factor in your previous rolls. But all I want to figure out is your comeback odds, so the 1,1,1 sequence is irrelevant. That's where you (the massive dumb ) and the dumb who made the graphic erred. Him factoring in Manu's make, Lebron's miss only gave the odds for those events happening in a specific sequence and not the odds of the what the Heat needed to go right in the final 28 seconds to tie the game.

I honestly don't know how I can explain it any more clearly. His math was right, but he was citing the probability of something completely different, which was the odds of the ALL the events happening in that specific sequence. Basically, any event that was in favor of the Spurs should not be factored in. We treat those events like the 1,1,1 dice rolls.

[midnightpulp]

Again, you dont understand.



Thats the odds of Spurs winning using ANY sequence events. Youre confusing continuous probability with discrete probability.

You didnt determine the odds that Golden State would go four minutes without a point, you attempted to determine the odds of particular players missing particular shots and then giving up offensive rebounds in a given order.

All you do is expose your ignorance. Just remember I'll always be here to expose you when you "crunch some numbers".

Discrete probability is what you use in this context.

Show me a white paper that examines continuous probability to calculate the odds of something like a player hitting 5 straight home runs or missing 8 straight jumpshots. So you're going to tell SABR they're wrong?

They treat an event in sports (homerun, error, missed jumper, offensive rebound, etc, etc) as an INDEPENDENT CHANCE EVENT, just like a roll of the dice, where the previous event has no in' influence over the next event. As I said, continuous probability is used in different applications.

Intuitively, a continuous random variable is the one which can take a continuous range of values—as opposed to a discrete distribution,

A jumpshot has a discrete, binomial (i.e. a yes/no event) distributions, just like a coinflip. There is only two outcomes (make/miss) and they can be statistically defined with percentages.

Applications of continuous probability:

The concept of the probability distribution and the random variables which they describe underlies the mathematical discipline of probability theory, and the science of statistics. There is spread or variability in almost any value that can be measured in a population (e.g. height of people, durability of a metal, sales growth, traffic flow, etc.); almost all measurements are made with some intrinsic error; in physics many processes are described probabilistically, from the kinetic properties of gases to the quantum mechanical description of fundamental particles. For these and many other reasons, simple numbers are often inadequate for describing a quan y, while probability distributions are often more appropriate.

Simple numbers aren't inadequate in our case. Now if you wanted to find the probability of Lebron's shooting percentage after he ate pizza on rainy day, then you probably use continuous probability.

Just stop digging.

[midnightpulp]

There are four distinct outcomes for the last sequence.

1) Lebron makes the three on the first try (37%)
2) Lebron misses the three on the first try, did not get offensive rebound (0.63 x 0.74 = 46.62%)
3) Lebron misses the three on the first try, got offensive rebound, Allen misses the next shot (0.63 x 0.26 x 0.59 = 9.7%)
4) Lebron misses the three on the first try, got offensive rebound, Allen makes the next shot (0.63 x 0.26 x 0.41 = 6.7%)

The total probabilities of these four events should add up to 1, and it does using MDP's method (see above in parantheses)

It doesn't using dsf's method.

In fact, if you use http://4.bp.blogspot.com/-xW5MJdDISp...ame+6+Math.jpg as the basis, just take the first scenario where Ginobili makes or misses the FT.

The probability of him missing the FT, as noted in the graphic, was 0.73 x 0.27 = 19.77%
Then what is the probability of Ginobili making the FT? It would be 0.27 x 0.73, which is also 19.77%

What are the other probabilities?

The person who came up with that graphic is clearly ignorant of how statistics work. let's break down all the possible scenarios using his method:
1) Ginobili makes FTs = 19.77%
2) Ginobili misses FTs, Lebron misses 3 = 0.73 x 0.27 x 0.375 x 0.625 = 4.6195%
3) Ginobili misses FTs, Lebron makes 3, Kawhi makes FT = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 = 1.0732%
4) Ginobili misses FTs, Lebron makes 3, Kawhi misses FT, Allen misses 3 = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 x 0.41 x 0.59 = 0.2596%
5) Ginobili misses FTs, Lebron makes 3, Kawhi misses FT, Allen makes 3 = 0.73 x 0.27 x 0.375 x 0.625 x 0.633 x 0.367 x 0.41 x 0.59 = 0.2596%

The total adds up to 25.9219%. What happens to the other 74% or so? What are the other possibilities?

Thank you. I'm think I'm finally done here.

[da_suns_fan]

There are four distinct outcomes for the last sequence.

1) Lebron makes the three on the first try (37%)
2) Lebron misses the three on the first try, did not get offensive rebound (0.63 x 0.74 = 46.62%)
3) Lebron misses the three on the first try, got offensive rebound, Allen misses the next shot (0.63 x 0.26 x 0.59 = 9.7%)
4) Lebron misses the three on the first try, got offensive rebound, Allen makes the next shot (0.63 x 0.26 x 0.41 = 6.7%)

No, there is an infinite number of outcomes.

Texas education system.

[midnightpulp]

A continuous variable is a variable whose value is obtained by measuring. A random variable is a variable whose value is a numerical outcome of a random phenomenon. A discrete random variable X has a countable number of possible values. Example: Let X represent the sum of two dice.

Another Khan Academy video to help our math illiterate friend here:



Continuous variable: "Something that can take on a whole set of values, from 0-infinity."

How many values does a jumper have? 2. A make or a miss.

[da_suns_fan]

Discrete probability is what you use in this context.

No its not. Im glad youre starting to TRY to learn though.

Never take Monospulp to Vegas, thats for sure.

[da_suns_fan]

Another Khan Academy video to help our math illiterate friend here:



Continuous variable: "Something that can take on a whole set of values, from 0-infinity."

How many values does a jumper have? 2. A make or a miss.

Now youre panicking huh?

Glad I could educate you.

[midnightpulp]

Now youre panicking huh?

No. Trying to educate you in the hope you'll stop being a Massive Dumb .

[ambchang]

No, there is an infinite number of outcomes.

Texas education system.

Yes there are, but the example you used is taking the assumption that each outcome is an either or, and I am taking that exact approach.

If you are using the infinite outcome approach, the onus is on you to come up with a way to calculate it rather than rely on a graphic that most definitely based on a FINITE set of outcomes.

[da_suns_fan]

No. Trying to educate you in the hope you'll stop being a Massive Dumb .

It burns, doesnt it?

You should be thanking me. Ive introduced you to some stuff, havent I?

You should go back to school if youre really interested.