My Math Forum Probability of picking a card that is between two other cards.

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 May 22nd, 2015, 04:53 AM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra More or less, yes. Select the three cards first and worry later about what order you take them in. So the bottom card can be picked in 10 x 4 ways, the top card in 4 ways and the middle card in 8 ways. Then you consider how many of the six orders of selection meet your criteria. Doing it this way, you don't even have to worry about double-counting because the fixed gap between top and bottom ensures that each set of selections is distinct. Thanks from p4r4gon Last edited by v8archie; May 22nd, 2015 at 05:24 AM.
 May 22nd, 2015, 04:54 AM #12 Newbie   Joined: May 2015 From: England Posts: 7 Thanks: 0 And again sorry if I'm being dim, but how does 1408/132600 reduce to 176/39075 They give different results first one 94.17:1 and second 222.01:1 Bit confused at moment, thought I was getting to grips with it.
 May 22nd, 2015, 05:14 AM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra Oh, well that's an error then. Blame the internet fractions calculator I was using. My phone says that $${1408 \over 132600} = {176 \over 16575} = 0.0106 \;\text{(to 3 s.f.)}$$ I have no reason to think that the result you'd get from the other approaches is flawed. It may well yield the correct answer. There is usually more than one way to think about a combinatorial problem. I just think that in this case mine is rather more straightforward. Edit: oops! I just noticed a rather significant transposition of words ("top" and "middle") in my previous post - now corrected. Thanks from p4r4gon Last edited by v8archie; May 22nd, 2015 at 05:26 AM.
 May 22nd, 2015, 05:20 AM #14 Newbie   Joined: May 2015 From: England Posts: 7 Thanks: 0 Ok that's ok, thought I might have been missing something your way. Would/should both methods/ways result in the same probability/odds?
 May 22nd, 2015, 05:27 AM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra Yes. If they don't then one of them is wrong! Thanks from p4r4gon
May 22nd, 2015, 06:27 AM   #16
Math Team

Joined: Oct 2011

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Quote:
 Originally Posted by p4r4gon Denis; If the first card has already been choosen and is a 7 would it not then be 8/51 for the second card to provide excatly 1 gap, as could have 4 5's as well as 4 9's
Yes, of course...whassa matta me

9/13 of cases: 8/51 * 4/50 : 9/13 * 8/51 * 4/50 = 48/5525

4/13 of cases: 4/51 * 4/50 : 4/13 * 4/51 * 4/50 = 32/16575

48/5525 + 32/16575 = 176/16575 = .0106184...

If wrong, I don't have an excuse that rivals Archie's

 May 25th, 2015, 04:37 AM #17 Newbie   Joined: May 2015 From: England Posts: 7 Thanks: 0 Sorry been at work over busy weekend. Thanks everyone for all your help. Well Denis your answer is the same as Archie's so all good there. Last thing can you help me understand the adding of probabilities together (prob the easiest bit, but looking it up I don't think I quite get it. As I really cant see how 48/5525 + 32/16575 = 176/16575 Can you treat me like the idiot I'm being and give me it in nice simple terms thanks again everyone.
 May 25th, 2015, 06:04 AM #18 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038

 Tags card, cards, picking, probability

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### probability of getting excatly 7 tasks out of 10 when we are picking 12 tasks out of 27

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