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        riddles >> easy >> Pick 2 cards...
        
(Message started by: Noke Lieu on Aug 29^th, 2012, 12:56am)

Title: Pick 2 cards...
Post by Noke Lieu on Aug 29^th, 2012, 12:56am

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?

Title: Re: Pick 2 cards...
Post by rmsgrey on Aug 29^th, 2012, 4:14am

It depends how we choose. If I'm allowed to pick freely, I can force 0, 0 and 0...

Title: Re: Pick 2 cards...
Post by Noke Lieu on Aug 29^th, 2012, 6:34am

They're face down and well shuffled... ;)

Title: Re: Pick 2 cards...
Post by towr on Aug 29^th, 2012, 10:31am

Assuming random picks without replacement:

[hide]
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
[/hide]

Title: Re: Pick 2 cards...
Post by Noke Lieu on Aug 29^th, 2012, 4:58pm

I'm finding it hard to disagree. But here goes. ::)

There are 15 different combinations that player one can choose.
Of those 15, [hide]
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) =[/hide] 8/90

That leaves [hide]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.[/hide]

Like I say, hard to disagree. ;)