|Posted by Paladin||04/23/09|
|Probabilities always make me stop and think. I haven't had any probability courses, but I can usually muddle through with the math I do know and get a decent shot at it. This was fun but very challenging. Well done!|
|Posted by Shadows||04/23/09|
|Well . . . I think I'll just look on the bright side. I got the first one! :lol:
I see you put quite some work into this. It's a nice teaser. When I saw it in proofreading, I rated it fun and hard.
|Posted by bradon182001||04/30/09|
|Waaaaay over my head.Excellent teaser. Thanks for posting. :o|
|Posted by gooberbaby1||06/11/10|
|I would have just put down an answer,NOT an answer & a HUGE explanation :oops: :oops:|
|Posted by racoonieboy||07/25/10|
|Actually, the explanation... well, explained a lot and I like it!|
|Posted by Zag24||08/04/10|
|I'm betting, goober, that you wouldn't have put the answer at all, since I bet you couldn't have come up with it.|
|Posted by ali_p||09/13/12|
|I don't think this is right...what accounts for the fact that you're more likely to match 2 games when you have 5 games per ticket instead of just having to match one game to another? I think you're using the wrong "P"...otherwise, what would account for it being less likely to match 3 out of 5 versus just 2 out of 5??|
|Posted by Zag24||02/15/13|
|all_p, sorry I didn't see your question until just now.
In the explanation, step 2, there are four fractions we are multiplying:
(5245785 * 5245784 * 5245783 * 5245782) / 5245786 ^ 4
That is, we are confirming that all 5 combos on the ticket are different. If we had only, say, three combos on one ticket, that step would only be
(5245785 * 5245784) / 5245786 ^ 2
and if there were only 2 combos on one ticket, it would be the same as the first step,
5245785 / 5245786
Remember that these are the chances that all the combos are different. The chances that there are two the same is 1 minus this.|
|Posted by masutra95||05/27/15|
|Problem reads like 6 numbers are drawn in order without replacement out of the 42 possible, but in the solution instead it seems to be assumed that 6 numbers are drawn with replacement and the order does not mater...|
|Posted by Zag24||06/13/15|
|masutra95: Sorry, but you're mistaken. The number of combinations if it were 6 numbers with replacement where order does matter, the number of combinations would be simply 42 to the 6th power.
If it were 6 choices without replacement, but order does matter, it would be
42 * 41 * 40 * 39 * 38 * 37 = 42! / (42-6)!
Think of drawing the balls in order. There are 42 from which to choose for the first one, then there are 41 from which to choose the second one, and so on down to 37 from which to choose the last one.
Now, if we want to make it such that order does not matter, we divide by number of ways you can arrange 6 different numbers. This is 6 * 5 * 4 * 3 * 2 or 6!
[42! / (42-6)! ] / 6! = 42! / 6! (42-6)! which is the formula I used above, also known as (42 choose 6).
This is, by the way, exactly how the (n choose m) formula was originally derived.|