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December 5th, 2018, 08:25 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 639 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, 01:13 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,759 Thanks: 696 
$\frac{3\times 2\times 1}{7\times 5\times 3}=\frac{2}{35}$. I'll leave it to you to figure out how.

December 5th, 2018, 10:30 PM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315  Quote:
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 Last edited by romsek; December 5th, 2018 at 10:33 PM.  
December 6th, 2018, 06:51 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
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, 01:48 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,759 Thanks: 696 
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, 06:44 PM  #6 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 639 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, 08:33 PM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315  Quote:
if you have $N$ pairs, i.e. $2N$ numbers it will be $p = \dfrac{(N)(N1)\dots (Nk) \dots 2}{(2N1)(2N3) \dots (2N2k+1)\dots 3} = \\ \prod \limits_{k=1}^{N1}\dfrac{Nk+1}{2N2k+1}$  

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making, numbers, pairs, probability 
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