## Probability/math question

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- brian
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### Probability/math question

I don't need this down to the exact number, but I am hoping someone can give me a pretty accurate estimate, but what would the approximate odds be of selecting the EXACT finisher correctly for each of the top ten positions in a NASCAR race? Would have 36 racers to pick from.

So the dumb odds (assuming all drivers were equal) would be 36*35*34*33*32*31*30*29*28*27, correct?

So the dumb odds (assuming all drivers were equal) would be 36*35*34*33*32*31*30*29*28*27, correct?

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- Walter Sobchak
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### Re: Probability/math question

Makes sense to me.brian wrote: ↑Mon Jun 24, 2019 4:00 pmI don't need this down to the exact number, but I am hoping someone can give me a pretty accurate estimate, but what would the approximate odds be of selecting the EXACT finisher correctly for each of the top ten positions in a NASCAR race? Would have 36 racers to pick from.

So the dumb odds (assuming all drivers were equal) would be 36*35*34*33*32*31*30*29*28*27, correct?

And his one problem is he didn’t go to Russia that night because he had extracurricular activities, and they froze to death.

### Re: Probability/math question

I think that sounds right, but can someone confirm why it wouldn't be 36 to the power of 10?

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### Re: Probability/math question

mister d wrote:Couldn't have pegged me better.

EnochRoot wrote:I mean, whatever. Johnnie's all hot cuz I ride him.

### Re: Probability/math question

Because the 2nd, 3rd, 4th...etc places don't have 36 cars to choose from. After the 1st it's 35 and descends by 1 each time.

mister d wrote:Couldn't have pegged me better.

EnochRoot wrote:I mean, whatever. Johnnie's all hot cuz I ride him.

### Re: Probability/math question

Ok. I was thinking that, I just needed to hear it to confirm my thoughts that Brian had it right.

My avatar corresponds on my place in the Swamp posting list with the all-time Home Run list. Number 45 is Andre "The Hawk" Dawson with 438.

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- Walter Sobchak
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### Re: Probability/math question

If you had to pick the winners of ten races in a row, and each race was random, then it would be 36^10.

And his one problem is he didn’t go to Russia that night because he had extracurricular activities, and they froze to death.

- brian
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### Re: Probability/math question

The reason I ask is because we have a partner who did a promotion for the Indy 500 where they were offering a $150K prize if a user picked all 33 spots of the race exactly right. I complained to my boss (who works with the corporate parent of that partner) that it was a stupid concept since the odds of picking all 33 right are so astronomical even to someone who doesn't comprehend basic math that not that many people are going to enter. (It turned out I was right).

So they're coming back for the Brickyard race in September and wanting different ideas so I suggest just make them pick the top 10 only but I lowballed in my head how hard even that is. (Again, the dumb odds here aren't really relevant since there's probably only about 15 or so drivers in a given week that have a chance to win, though of course the shitty teams can finish in the top ten, so if they run it by companies that insure these promotions I expect insuring only the top 10 is still going to be relatively cheap.)

That's a crazy industry -- the people who insure those billion dollar prizes for perfect NCAA brackets or making a half-court shot and stuff like that. I imagine the odds of someone actually getting even the top 10 drivers correct in exact order would mean the person guessing correctly is a time traveler.

So they're coming back for the Brickyard race in September and wanting different ideas so I suggest just make them pick the top 10 only but I lowballed in my head how hard even that is. (Again, the dumb odds here aren't really relevant since there's probably only about 15 or so drivers in a given week that have a chance to win, though of course the shitty teams can finish in the top ten, so if they run it by companies that insure these promotions I expect insuring only the top 10 is still going to be relatively cheap.)

That's a crazy industry -- the people who insure those billion dollar prizes for perfect NCAA brackets or making a half-court shot and stuff like that. I imagine the odds of someone actually getting even the top 10 drivers correct in exact order would mean the person guessing correctly is a time traveler.

### Re: Probability/math question

You're totally right with that. 1 through 10 would be a fair compromise. Hell, top 5 in this instance is 1 in 45 million, but that also assumes all things are equal. A person who is passively aware of top race car drivers could potentially nail the first five based on racing history and some simple research.

Look at horse racing's super-fecta. Even the most knowledgeable and savvy bettors don't get that and that's only 4 horses from a smaller field.

Look at horse racing's super-fecta. Even the most knowledgeable and savvy bettors don't get that and that's only 4 horses from a smaller field.

mister d wrote:Couldn't have pegged me better.

EnochRoot wrote:I mean, whatever. Johnnie's all hot cuz I ride him.

- Rams Fanny
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### Re: Probability/math question

We can confirm Brian's answer with gambling:

If there is a six horse race and you trifecta boxed all horses your $1 cost would be $120 as there are 120 possible combinations (6x5x4) for top three spots. And my Dad said the only thing I would learn from gambling was how to lose money!

If there is a six horse race and you trifecta boxed all horses your $1 cost would be $120 as there are 120 possible combinations (6x5x4) for top three spots. And my Dad said the only thing I would learn from gambling was how to lose money!

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- A_B
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### Re: Probability/math question

And also why trifecta are often not available on six horse races!Rams Fanny wrote: ↑Mon Jun 24, 2019 5:09 pmWe can confirm Brian's answer with gambling:

If there is a six horse race and you trifecta boxed all horses your $1 cost would be $120 as there are 120 possible combinations (6x5x4) for top three spots. And my Dad said the only thing I would learn from gambling was how to lose money!

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- degenerasian
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### Re: Probability/math question

or hockey.Rams Fanny wrote: ↑Mon Jun 24, 2019 5:09 pmWe can confirm Brian's answer with gambling:

If there is a six horse race and you trifecta boxed all horses your $1 cost would be $120 as there are 120 possible combinations (6x5x4) for top three spots. And my Dad said the only thing I would learn from gambling was how to lose money!

I think it would be the same odds as a team up 3-0 giving up 4 power play goals on the same 5 minute major.

(sorry..)

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- brian
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### Re: Probability/math question

Learned about Benford's Law yesterday and that is some crazy, crazy, crazy, crazy shit.

Not sure if it makes me maybe actually believe in the idea of intelligent design or if it refutes it somehow.

Not sure if it makes me maybe actually believe in the idea of intelligent design or if it refutes it somehow.

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### Re: Probability/math question

Yeah, I saw an episode ofbrian wrote: ↑Wed Sep 16, 2020 2:09 pmLearned about Benford's Law yesterday and that is some crazy, crazy, crazy, crazy shit.

Not sure if it makes me maybe actually believe in the idea of intelligent design or if it refutes it somehow.

*Connected*about that and my mind was blown.

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### Re: Probability/math question

I don't get this ... wouldn't it be completely logical on property addresses because the 100s sequentially come way before the 900s and thus occur more often?

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### Re: Probability/math question

That makes sense but the stuff that’s crazy is when the distribution is the same for the distance from the Earth to all known galaxies regardless of the measurement (light years, miles, kilometers, yards, etc and stuff like that.

### Re: Probability/math question

In my mind that still makes sense because distance counts up so if you take distance from earth, you'd cycle past 100 miles before 900 then right up to 1,000 and etc, etc where 1 is always preceding 9. I know I'm missing something because I don't solve laws on the fly, but I can't wrap my head around why this isn't the expectation.

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### Re: Probability/math question

It might make sense for it not to be a true random distribution (i.e. 11.1 percent each), but the disparity in distribution is interesting. And even if it's explainable the fact that it can be used to identify truly random data sets and fraudulently random data sets is also interesting.mister d wrote: ↑Wed Sep 16, 2020 2:46 pmIn my mind that still makes sense because distance counts up so if you take distance from earth, you'd cycle past 100 miles before 900 then right up to 1,000 and etc, etc where 1 is always preceding 9. I know I'm missing something because I don't solve laws on the fly, but I can't wrap my head around why this isn't the expectation.

### Re: Probability/math question

Yeah, I still don't follow even there. If you listed out every transaction over the course of a month, wouldn't you still expect higher frequency on 1 because you spend $100 and $1,000 more than $900 and $9,000? Transactions would naturally skew to the lower end of the spectrum once you control for outliers like recurring payments or salary distribution.

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### Re: Probability/math question

I think the law remains true even if you cap the maximum value in the data set though. From what I read if you took the population of every city in the world between 10,000 and under 9,999,999, it still follows Benford's law.mister d wrote: ↑Wed Sep 16, 2020 3:03 pmYeah, I still don't follow even there. If you listed out every transaction over the course of a month, wouldn't you still expect higher frequency on 1 because you spend $100 and $1,000 more than $900 and $9,000? Transactions would naturally skew to the lower end of the spectrum once you control for outliers like recurring payments or salary distribution.

### Re: Probability/math question

I still don't see why that wouldn't be expected. Even with the cap, wouldn't one expect more cities of 1MM than 9MM?

### Re: Probability/math question

Examining a list of the heights of the 60 tallest structures in the world by category shows that 1 is by far the most common leading digit, irrespective of the unit of measurement (cf. "scale invariance", below):

**Example**Examining a list of the heights of the 60 tallest structures in the world by category shows that 1 is by far the most common leading digit, irrespective of the unit of measurement (cf. "scale invariance", below):

Here's a good one. The link shows 58 structures, from 249 feet to 2,722 feet. Logically, the higher you go, the more rare it would be. So, rounded, your lead numbers are:

**2**00s

**3**00s

**4**00s

**5**00s

**6**00s

**7**00s

**8**00s

**9**00s

**1**,000s

**1**,100s

**1**,200s

**1**,300s

**1**,400s

**1**,500s

**1**,600s

**1**,700s

**1**,800s

**1**,900s

**2**,000s

**2**,100s

**2**,200s

**2**,300s

**2**,400s

**2**,500s

**2**,600s

**2**,700s

So why would anyone be surprised 1 is 1st and 2 is 2nd?

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### Re: Probability/math question

I'm with Mr. D here. Makes total sense to me logically.

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### Re: Probability/math question

*Benford's law tends to apply most accurately to data that span several orders of magnitude. As a rule of thumb, the more orders of magnitude that the data evenly covers, the more accurately Benford's law applies. For instance, one can expect that Benford's law would apply to a list of numbers representing the populations of UK settlements. But if a "settlement" is defined as a village with population between 300 and 999, then Benford's law will not apply.*

I just read this and now I'm confused at how this is even a law.

### Re: Probability/math question

I’m pretty much with you, but if you expect rarity as you go up, could someone that has no clue about the real-world distribution think that the 300s far outnumber most of the 1000s combined?

he’s a fixbking cyborg or some shit. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

### Re: Probability/math question

Also, I would expect the number of torque wrenches to be greater than the number of tea towels in any real man’s house. Oh wait, that’s Binford’s Law.

he’s a fixbking cyborg or some shit. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

### Re: Probability/math question

Yes, but in that instance wouldn't you expect the 100s to far outnumber the 300s?

(Also I assume there's a reason they picked something weird like structures versus something weird but not fitting like "most FT attempts in an NBA season". This all seems very stupid.)

Last edited by mister d on Wed Sep 16, 2020 3:42 pm, edited 1 time in total.

### Re: Probability/math question

Your list started in the 200s but yeah

he’s a fixbking cyborg or some shit. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

### Re: Probability/math question

As expected

holy fuckbAllZ, what a ducking nightmare. Holy shot. Just, fuck. The

### Re: Probability/math question

Distribution of the first digit of the 60 tallest buildings in NYC (in feet).

1: 23.3%

2: 0.0%

3: 0.0%

4: 0.0%

5: 0.0%

6: 0.0%

7: 36.7%

8: 25.0%

9: 15.0%

Fuck you, Benford.

1: 23.3%

2: 0.0%

3: 0.0%

4: 0.0%

5: 0.0%

6: 0.0%

7: 36.7%

8: 25.0%

9: 15.0%

Fuck you, Benford.

### Re: Probability/math question

I enjoy the occasional reminders that I'm one of the dumbest people on this site.

Muh.

### Re: Probability/math question

Yeah this seems logical to me as well. Every time you go up one base ten unit (ie 10s to 100s), you're going to have a whole set of ones and eventually as climbing this list (likely) gets progressively harder, the 1s would be expected to dominate and 9s would be expected to be crowded out.

"Enjoyed this one," Gunpowder told the Burlington Observer. "But it's back to work tomorrow."

### Re: Probability/math question

This also ignores that this using a base 10 system is likely to produce such an effect as we model the very numbers used to measure natural and manmade things with a thing starting with 1. It might be different if you used, like, octal numbers or something weird.

"Enjoyed this one," Gunpowder told the Burlington Observer. "But it's back to work tomorrow."

### Re: Probability/math question

The wikipedia page covers that. It apparently happens in other bases as well, for the same reasons. I think it has more to do with the way we create numbers. If we wrote or arranged them differently (not sure what that would be), it might not happen.Gunpowder wrote: ↑Thu Sep 17, 2020 1:00 pmThis also ignores that this using a base 10 system is likely to produce such an effect as we model the very numbers used to measure natural and manmade things with a thing starting with 1. It might be different if you used, like, octal numbers or something weird.

Totally Kafkaesque