Somebody call the cops!

Permalink Reply by Karrie (The Original Bookworm) on September 9, 2012 at 8:26pm

Wow. This is official.

And thanks Jason, coming through in the clutch, as always. (Somebody should give that man an award!)

It's most likely because most things are attached to some sort of metal, which may be weighted to make it lean in a certain direction, sort of like how some chair's have one leg a little to short, which makes it lean one way.

But in a circular manner. :)

Permalink Reply by Phil (Duckbilled Baronyx) on September 10, 2012 at 11:08am

Looks good after the first spin of the modified wheel. I must disagree with Rhett that the secret spots would have been hit by now, because a 1/8 chance doesn't mean all eight options will be hit in eight spins. Just like a die, you can roll a three three times in a row and still have the same 1/6 chance of rolling any number.

K-ModPermalink Reply by Gumbo123 on September 10, 2012 at 11:35am

There are 24 spaces and there were 26 spins before today - - so, in a perfect world, every space should have landed once by now. 6 out of 26 on the "sweet spot" goes way beyond the odds of random selection.

Permalink Reply by Phil (Duckbilled Baronyx) on September 10, 2012 at 12:13pm

Yeah, but there is no perfect world, so all of the space would not have been spun.

Permalink Reply by Lilian Brandon (Tachyonic Raven) on September 10, 2012 at 1:45pm

The key is the sequence of events - the individual odds are multiplied in that case. The odds of not landing on a special spot on any given day is 7/8th (87.5%). But the odds of not landing on a special spot two days in a row is (7/8)*(7/8)=(~77%). The odds of not landing on a special spot 26 times in a row is (7/8)^26 = ~3% -- which is pretty small. Small enough that I think Rhett is justified in saying that "we should have hit one by now". The odds of hitting one tomorrow are still 1/8, but the odds of now going 27 days without hitting one is small. With each passing day it becomes more likely that we will hit one than it is that we'll have gone the entire time without hitting one.

Following similar logic, the odds of any particular space not having gotten hit at all by this point (say, the spot to the left of the red space) are (23/24)^26=33%, so your comment about that is fair. Once they get to 50 episodes, there's still an 11% chance that that space won't have been hit yet.

And what about the odds of hitting that sweet spot 6 times out of the 26 spins? The (unordered) probability of that happening is (26!/(6!*20!)*((1/24)^6)*((23/24)^20) = 0.05%. As Gumbo said, beyond the odds of random selection. (If I wanted to, I could look up the exact points in the sequence that the sweet spot was hit to get more accurate odds of it happening, but the point is sufficient).

Permalink Reply by Donna (peachgeek) on September 10, 2012 at 1:53pm

well done -- thanks

Permalink Reply by Lilian Brandon (Tachyonic Raven) on September 10, 2012 at 2:17pm

I should clarify that it's a binomial distribution, so unless you're looking at all failures or all successes (as in the first 2 paragraphs), you need to include a coefficient to account for the various ways of placing the successes within the total number or events (as in the last paragraph). Consult good ol' Viki for details: Binomial Distribution

Permalink Reply by Phil (Duckbilled Baronyx) on September 10, 2012 at 3:22pm

Thanks, Raven. My other answer probably just displays my ignorance.

K-ModPermalink Reply by Gumbo123 on September 10, 2012 at 2:04pm

yeah, like I said > > > 0.0000002% < HIGHLY UNLIKELY

Thanks for the math refresher course. I tried my "luck" on a virtual spinner and even then with 26 exactly even spins (power / speed / drag/ etc.) you don't hit every spot on the wheel - - but still no space was hit more than twice. http://www.mathsisfun.com/data/spinner.php

NOTE: the results shown don't add up to the 26 spins taken because 24 spaces on the chart won't all fit the screen, so results for spaces A,B&C are not shown.

Then, just for grins & giggles, I reset the spin counter to 200 and let it run its course:

Permalink Reply by Lilian Brandon (Tachyonic Raven) on September 10, 2012 at 2:14pm

Well, not 0.0000002% -- it's a binomial distribution, so unless you're looking at all successes or all failures, you need a coefficient in front of the multiplied probabilities to account for the different ways of placing the X number of successes. Which is how I got 0.05%, which is still pretty dang small.

K-ModPermalink Reply by Gumbo123 on September 10, 2012 at 2:21pm

AHA! you edited and changed your initial number posting (which was 2 / 10,000,000 ) while I was spinning the wheel and creating images. I just assumed you were right the first time. So much for accuracy, huh?

Permalink Reply by Lilian Brandon (Tachyonic Raven) on September 10, 2012 at 2:41pm

Ah, yeah, I did -- I realized I'd forgotten the coefficient and deleted my original response while I figured out the right percentage.

Somebody call the cops!

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