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Topic: Pick 2 cards... (Read 2417 times) |
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Noke Lieu
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pen... paper... let's go! (and bit of plastic)
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Pick 2 cards...
« on: Aug 29th, 2012, 12:56am » |
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I have 6 cards labelled 1-6.
You choose 2, then I choose 2.
What's the probability that our cards have the same total?
What if you choose 2, and then I choose 3, and vice versa?
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a shade of wit and the art of farce.
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rmsgrey
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Re: Pick 2 cards...
« Reply #1 on: Aug 29th, 2012, 4:14am » |
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It depends how we choose. If I'm allowed to pick freely, I can force 0, 0 and 0...
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Noke Lieu
Uberpuzzler
pen... paper... let's go! (and bit of plastic)
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Re: Pick 2 cards...
« Reply #2 on: Aug 29th, 2012, 6:34am » |
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They're face down and well shuffled... 
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a shade of wit and the art of farce.
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: Pick 2 cards...
« Reply #3 on: Aug 29th, 2012, 10:31am » |
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Assuming random picks without replacement:
14,23
15,24
16,25 16,34 25,34
26,35
36,45
= 7 working combinations
x2 because first and second player are interchangable
= 14
6!/2!^3
= 90 possible combination in total
probability first two and next two cards have same total = 7/45
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Wikipedia, Google, Mathworld, Integer sequence DB
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Noke Lieu
Uberpuzzler
pen... paper... let's go! (and bit of plastic)
Gender:
Posts: 1884
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Re: Pick 2 cards...
« Reply #4 on: Aug 29th, 2012, 4:58pm » |
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I'm finding it hard to disagree. But here goes. 
There are 15 different combinations that player one can choose.
Of those 15,
4 yield a dead game {(1,2);(1,3);(4,6);(5,6)}
Of those 15, 8 combinations have a 1/6 probability of player 2 mathcing the totals.
So (8/15) x (1/6) = 8/90
That leaves 3 combinations that have a 1/3 of player 2 mathcing them.
So (3/15) x (1/3) = 3/45
8/90 + 3/45 (or 6/90) = 14/90 or 7/45.
Like I say, hard to disagree.
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