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What are the odds (math nerds)

I Are Baboon

Run, Forrest, Run!
Aug 6, 2001
29,508
2,119
MTB New England
So you've got two grid sheets that contain 100 squares. Each row and column is randomly numbered 0-9. You pick one square on each sheet in different spots. Turns out you've got the same random column and row number on each sheet. What are the odds of that happening? Does my explanation make sense? Pay attention, god dammit.

A simpler explanation is this, for those of you familiar with Super Bowl grid pools: I've got the same numbers in two different pools. Colts 3, Saints 4. Not bad numbers to have. I want to know the odds of getting the same numbers.
 

HAB

Chelsea from Seattle
Apr 28, 2007
10,666
960
Seattle
So you mean if you pick one number on the first sheet, what are the odds of getting the same number in the same row and column on the second?


In that case, it's 1/1000. You've got a 1/100 chance of picking the same space, and there's a 1/10 chance that the number contained within that space is the same as the one you got the first time. 1/100*1/10=1/1000.
 

I Are Baboon

Run, Forrest, Run!
Aug 6, 2001
29,508
2,119
MTB New England
So you mean if you pick one number on the first sheet, what are the odds of getting the same number in the same row and column on the second?


In that case, it's 1/1000. You've got a 1/100 chance of picking the same space, and there's a 1/10 chance that the number contained within that space is the same as the one you got the first time. 1/100*1/10=1/1000.
My thought was 1 in 1,000, but it seems to me that the odds should be exponentially larger. I'm two beers in though and am not thinking clearly.
 

HAB

Chelsea from Seattle
Apr 28, 2007
10,666
960
Seattle
for any particular duplicate square, HAB's right.

for *that* particular duplicate square, you are right.
Right, I was taking the first one not to matter, all you have to do is match it on the second. If you have to hit a specific square on the first one, then match it and the number on the second, odds are 1/100,000. If you have to get a specific number on the first square too, 1/1,000,000.
 

I Are Baboon

Run, Forrest, Run!
Aug 6, 2001
29,508
2,119
MTB New England
Right, I was taking the first one not to matter, all you have to do is match it on the second. If you have to hit a specific square on the first one, then match it and the number on the second, odds are 1/100,000. If you have to get a specific number on the first square too, 1/1,000,000.
I am not trying to *match* anything. The squares were completely random. Imagine throwing a dart at two grids, and the numbers get filled in randomly after the fact.
 

RayB

Monkey
Jan 31, 2008
745
94
Seattle
Well...

For any of the 2 grids, there are 100 unique pairs. So you have a 1/100 chance of getting some random combination (in your case Colts 3-Saints 4). Layout is irrelevant.

Since the 2 boards are independent, you have an equal probability of that particular combination in both boards. Thus,

(1/100)^2 = 1/10,000 = .0001

is the probability of getting the exact same combination...twice.


http://en.wikipedia.org/wiki/Independence_(probability_theory)
 
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RayB

Monkey
Jan 31, 2008
745
94
Seattle
1/1000
HAB is not the same as BAH backwards


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[[[]][[]][

just sauing ...
Yeah, but how are you (or anybody) getting 1/1000?

This is basic probability......

Allow me to reiterate... there are 100 distinct ordered pairs for any board. (1,1),(1,2),(1,3),...,(1,n),...(m,n) for m, n in the set (0,1,...,9)

The chance that you get (3,4) is 1/100 which is the same probability for (4,3).

The two boards are independent. Whatever you got on the first board, doesn't affect the probability of you getting any one of the distinct 100 ordered pairs on the second.
 

vroterman

Chimp
Jan 31, 2010
24
0
So you've got two grid sheets that contain 100 squares. Each row and column is randomly numbered 0-9. You pick one square on each sheet in different spots. Turns out you've got the same random column and row number on each sheet. What are the odds of that happening? .
1/100 is correct:

the odds to hit the same numbet in row=1/10
the odds to hit the same number in column=1/10
and the odds to hit the same number in row AND column =
1/(10^2)...
 
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vroterman

Chimp
Jan 31, 2010
24
0
Well...

For any of the 2 grids, there are 100 unique pairs. So you have a 1/100 chance of getting some random combination (in your case Colts 3-Saints 4). Layout is irrelevant.

Since the 2 boards are independent, you have an equal probability of that particular combination in both boards. Thus,

(1/100)^2 = 1/10,000 = .0001

is the probability of getting the exact same combination...twice.


http://en.wikipedia.org/wiki/Independence_(probability_theory)
Seems u R mistaken,
1)you are correct : there are 100 unique pairs for each grid.
2)But!!! after you've chosen pair on 1st grid(the odds of choosing pair 1/1, equal 100%, it doesn't matter what pair you choosing, ) the chance that the pair on 2nd grade will match = 1/100, so the answer is not (1/100)*1(100), but 1*(1/100)
 

I Are Baboon

Run, Forrest, Run!
Aug 6, 2001
29,508
2,119
MTB New England
ok yo, so it turns out the WIFE also has Saints 4, Colts 3. Seriously. Soooo...what are the odds now of pulling the same numbers on THREE sheets. LOL...I'm not kidding.
 

RayB

Monkey
Jan 31, 2008
745
94
Seattle
Seems u R mistaken,
1)you are correct : there are 100 unique pairs for each grid.
2)But!!! after you've chosen pair on 1st grid(the odds of choosing pair 1/1, equal 100%, it doesn't matter what pair you choosing, ) the chance that the pair on 2nd grade will match = 1/100, so the answer is not (1/100)*1(100), but 1*(1/100)
Dude... WHAT?!

It's funny that you guys are trying to tell me I'm wrong......

The answer is [again]: (1/100)*(1/100) = .0001


THE SHEETS ARE INDEPENDENT EVENTS.
(If you don't understand what this means, read the Wiki I linked above)

This is a concept taught at the intro-level of probability theory. I would know...

;)

ok yo, so it turns out the WIFE also has Saints 4, Colts 3. Seriously. Soooo...what are the odds now of pulling the same numbers on THREE sheets. LOL...I'm not kidding.
The answer is: (1/100)*(1/100)*(1/100) = .000001

(Very, very improbable.)
 

limitedslip

Monkey
Jul 11, 2007
173
1
Ray,
You'd be right if he was asking the chance of getting colts 3, saints 4 on two of two sheets. He's just asking about the chance of two matching sheets (.01).

Calpoly Statistics > Berkeley Statistics
 

RayB

Monkey
Jan 31, 2008
745
94
Seattle
Ray,
You'd be right if he was asking the chance of getting colts 3, saints 4 on two of two sheets. He's just asking about the chance of two matching sheets (.01).
So you've got two grid sheets that contain 100 squares. Each row and column is randomly numbered 0-9. You pick one square on each sheet in different spots. Turns out you've got the same random column and row number on each sheet. What are the odds of that happening? Does my explanation make sense? Pay attention, god dammit.

A simpler explanation is this, for those of you familiar with Super Bowl grid pools: I've got the same numbers in two different pools. Colts 3, Saints 4. Not bad numbers to have. I want to know the odds of getting the same numbers.
How am I mis-interpreting this?

"Two matching sheets?" I don't see any mention of this in the OP's post. So as far as I'm concerned, I'm right, and just about everybody else has been throwing out garbage excuses for "math".

I interpreted the above in bold as: "What is the probability of picking the pair (3,4) in two independent draws?"

Calpoly Statistics > Berkeley Statistics
This, http://grad-schools.usnews.rankingsandreviews.com/best-graduate-schools/top-mathematics-programs/statistics

and my current salary beg to differ. ;)


But hey, you guys definitely have better mountain biking--that's FOR SURE.



Done with this thread. If ya'll don't believe me, that's your beef. But I've provided the correct solution(s). Cheers, and enjoy the 'Bowl!! :)
 
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limitedslip

Monkey
Jul 11, 2007
173
1
So you've got two grid sheets that contain 100 squares. Each row and column is randomly numbered 0-9. You pick one square on each sheet in different spots. Turns out you've got the same random column and row number on each sheet. What are the odds of that happening? Does my explanation make sense? Pay attention, god dammit.

A simpler explanation is this, for those of you familiar with Super Bowl grid pools: I've got the same numbers in two different pools. Colts 3, Saints 4. Not bad numbers to have. I want to know the odds of getting the same numbers.
Ray,

The answer to the bolded question is .01. There isn't any reason to think IRB meant (3,4) on two sheets, except for the passing mention of those two numbers in the example (which obviously confused you).

Calpoly doesn't have a grad program EDIT: and you didn't go to grad school anyway.

Congrats on the salary.
 
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vroterman

Chimp
Jan 31, 2010
24
0
Dude... WHAT?!

It's funny that you guys are trying to tell me I'm wrong......

The answer is [again]: (1/100)*(1/100) = .0001


THE SHEETS ARE INDEPENDENT EVENTS.
(If you don't understand what this means, read the Wiki I linked above)

This is a concept taught at the intro-level of probability theory. I would know...

;)



The answer is: (1/100)*(1/100)*(1/100) = .000001

(Very, very improbable.)
Man if u have degree in statistics you still should try to think...
To simplify - the question is"what is the chance that pair in 1st table will be the same as pair in 2nd" - you have one event!!!, not two >>>>what is 2nd event man?? where do u see it? or it is in wiki?)))