Probability/math question

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brian
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Probability/math question

Post by brian » Mon Jun 24, 2019 4:00 pm

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?

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Re: Probability/math question

Post by Steve of phpBB » Mon Jun 24, 2019 4:10 pm

brian wrote:
Mon Jun 24, 2019 4:00 pm
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?
Makes sense to me.
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Re: Probability/math question

Post by L-Jam3 » Mon Jun 24, 2019 4:13 pm

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

Post by Johnnie » Mon Jun 24, 2019 4:18 pm

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

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Re: Probability/math question

Post by Johnnie » Mon Jun 24, 2019 4:21 pm

L-Jam3 wrote:
Mon Jun 24, 2019 4:13 pm
I think that sounds right, but can someone confirm why it wouldn't be 36 to the power of 10?
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.
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Re: Probability/math question

Post by L-Jam3 » Mon Jun 24, 2019 4:23 pm

Ok. I was thinking that, I just needed to hear it to confirm my thoughts that Brian had it right.
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Re: Probability/math question

Post by Steve of phpBB » Mon Jun 24, 2019 4:31 pm

Johnnie wrote:
Mon Jun 24, 2019 4:21 pm
L-Jam3 wrote:
Mon Jun 24, 2019 4:13 pm
I think that sounds right, but can someone confirm why it wouldn't be 36 to the power of 10?
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.
If you had to pick the winners of ten races in a row, and each race was random, then it would be 36^10.
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Re: Probability/math question

Post by brian » Mon Jun 24, 2019 5:00 pm

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.

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Re: Probability/math question

Post by Johnnie » Mon Jun 24, 2019 5:07 pm

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.
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Re: Probability/math question

Post by Rams Fanny » Mon Jun 24, 2019 5:09 pm

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!
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Re: Probability/math question

Post by A_B » Mon Jun 24, 2019 5:20 pm

Rams Fanny wrote:
Mon Jun 24, 2019 5:09 pm
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!
And also why trifecta are often not available on six horse races!
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Re: Probability/math question

Post by degenerasian » Mon Jun 24, 2019 5:22 pm

Rams Fanny wrote:
Mon Jun 24, 2019 5:09 pm
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!
or hockey.

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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