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Duplicate Lottery Picks

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#44355
Fun:** (1.91)
Difficulty:*** (2.96)
Submitted By:Zag24Aus****!!
Corrected By:Zag24




In the Massachusetts Megabucks lottery, six different numbers from 1 to 42 (inclusive) are selected. When you buy a ticket, you can ask for a "quick pick" in which the computer chooses the numbers for you, and you can purchase up to five games on a single ticket. We'll assume that the computer's random number generator is fair, giving each possible combination an equal probability of being chosen.

1. If I "quick pick" for two games, what are the chances that the two games have the same combination of numbers?

2. If I "quick pick" for five games (one five-game ticket), what are the chances that there are two games on that ticket with the same combination?

3 (The toughie). How many five-game quick-pick tickets would I have to buy in order to have a greater than 50% chance of having at least one ticket with two games on it that match exactly?

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Apr 23, 2009

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!
Apr 23, 2009

Well . . . I think I'll just look on the bright side. I got the first one!

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.

Good job!
Apr 30, 2009

Waaaaay over my head.Excellent teaser. Thanks for posting.
Jun 11, 2010

I would have just put down an answer,NOT an answer & a HUGE explanation
Jul 25, 2010

Actually, the explanation... well, explained a lot and I like it!
Aug 04, 2010

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.
Jul 21, 2011

THe answer was well explained. Good job!
Sep 13, 2012

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??
Feb 15, 2013

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.
May 27, 2015

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...
Jun 13, 2015

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.

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