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 December 5th, 2018, 09:25 AM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 635 Thanks: 85 Probability When Making Numbers Into Pairs If the numbers 1 through 8 are made into four pairs, what's the probability that each pair will have one even number and one odd number?
 December 5th, 2018, 02:13 PM #2 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 $\frac{3\times 2\times 1}{7\times 5\times 3}=\frac{2}{35}$. I'll leave it to you to figure out how.
December 5th, 2018, 11:30 PM   #3
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Quote:
 Originally Posted by mathman $\frac{3\times 2\times 1}{7\times 5\times 3}=\frac{2}{35}$. I'll leave it to you to figure out how.

Pick a number w/prob 1.
Pick it's pair member. This pair will have different parity w/prob 4/7

Pick a number for pair 2 w/prob 1
It's pair member will have different parity with prob 3/5

Pick for pair 3 w/prob 1
Different parity w/prob 2/3

Pair 4 automatically has different parity elements if the other 3 pairs do.

P[all pairs have different parity] = 4/7 3/5 2/3 = 24/105

I realize that probably looks like garble but it's creating the unordered pairs of unordered numbers and selecting those in which all pairs have different parity elements.

There are 105 total groups of 4 pairs and 24 were selected giving

p = 24/105 as expected from above.

Can you double check your formula. Could be I'm wrong and I'd like to know why if so. Thanks
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Last edited by romsek; December 5th, 2018 at 11:33 PM.

December 6th, 2018, 07:51 AM   #4
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Quote:
 Originally Posted by romsek Are you sure about this? Pick a number w/prob 1. Pick it's pair member. This pair will have different parity w/prob 4/7 Pick a number for pair 2 w/prob 1 It's pair member will have different parity with prob 3/5 Pick for pair 3 w/prob 1 Different parity w/prob 2/3 Pair 4 automatically has different parity elements if the other 3 pairs do. P[all pairs have different parity] = 4/7 3/5 2/3 = 24/105 I realize that probably looks like garble but it's creating the unordered pairs of unordered numbers and selecting those in which all pairs have different parity elements. There are 105 total groups of 4 pairs and 24 were selected giving p = 24/105 as expected from above. Can you double check your formula. Could be I'm wrong and I'd like to know why if so. Thanks
I too came up with this result.

The number of possible pairs with unlike parity to be chosen first is (4 * 4). The number possible on the second pick is (3 * 3), etc., for a total of (4!)^2.

But we don't care about the order of picking. So

(4!)^2 / 4! = 4! = 24.

The number of ways to pick the first pair is 8 * 7. But we don't care about order so it is 8 * 7 / 2. The number of ways to pick the second pair is 6 * 5 / 2, etc. So we get a total of 8! / 2^4. But we don't care about the order in which the pairs are picked. So

8! / (2^4 * 4!) = (8 * 7 * 6 * 5) / (2^4) = (16 * 7 * 5 * 3) / 16 = 105.

24 / 105 = 8 / 35.

I went nuts last night trying to figure out mathman's answer. I decided that I had drunk too many martinis. Thanks romsek.

 December 6th, 2018, 02:48 PM #5 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 Sorry I made a mistake, it should have been $\frac{4\times 3\times 2}{7\times 5\times 3}=\frac{8}{35}$.
 December 11th, 2018, 07:44 PM #6 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 635 Thanks: 85 Can the formula be generalized to: x!/[(2x - 1)(2x - 3) and decreasing by 2 at a time until you reach 3] For 12 items in 6 pairs with one even and one odd in each pair, would it be 6!/[11x9x7x5x3] = 16/231?
December 11th, 2018, 09:33 PM   #7
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Quote:
 Originally Posted by EvanJ Can the formula be generalized to: x!/[(2x - 1)(2x - 3) and decreasing by 2 at a time until you reach 3] For 12 items in 6 pairs with one even and one odd in each pair, would it be 6!/[11x9x7x5x3] = 16/231?
don't use $x$ for indices

if you have $N$ pairs, i.e. $2N$ numbers it will be

$p = \dfrac{(N)(N-1)\dots (N-k) \dots 2}{(2N-1)(2N-3) \dots (2N-2k+1)\dots 3} = \\ \prod \limits_{k=1}^{N-1}\dfrac{N-k+1}{2N-2k+1}$

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