A Puzzle
A speaker at the conference I was at last week in Vegas used the following puzzle to illustrate a point. I thought you might enjoy it. (Mostly, Keryn wanted to see if Samuel could come up with the answer.) As with all the blog puzzles, post a comment if you think you know the answer. In a few days, I will post a comment with the correct answer if it hasn't already been given.
In bucket A, you have 1000 red marbles. In bucket B, you have 1000 blue marbles. First, you randomly select 20 marbles from bucket A and put them in bucket B and then mix. Then, you randomly select 20 marbles from bucket B and put them in bucket A and mix. Repeat this three times. At the end of the experiment, which bucket will have more foreign colored marbles in it?
In bucket A, you have 1000 red marbles. In bucket B, you have 1000 blue marbles. First, you randomly select 20 marbles from bucket A and put them in bucket B and then mix. Then, you randomly select 20 marbles from bucket B and put them in bucket A and mix. Repeat this three times. At the end of the experiment, which bucket will have more foreign colored marbles in it?
posted by Bradley Ross @ 8:15 PM
8 Comments:
At 2/06/2007 9:10 PM,
Samuel said…
I admit I knew the answer even before I could explain the reason. Therefore, I didn't trust the answer that was in my head until I could fully convince myself that it is the correct answer. I'm not sure I want to explain my reasoning yet so as to allow others to figure it out without the easy way of looking at the comment.
I do want to post my reasoning for my answer, along with my answer, but I will wait until tomorrow (Wednesday) night. This should still give enough time for me to answer before Brad posts the 'correct' answer.
For now I will say this: there are only 2 buckets with only 2 total colors and each bucket must have 1000 marbles at the end of each of the 3 trading cycles. Until tomorrow when I will post my answer along with my reasoning.
At 2/07/2007 5:54 AM,
Jenafer said…
I admit, I'm not sure why, but instinctivly I say B. My thought process is: All of the marbles of the first twenty placed in the B bucket are the other color. After that, you would get a mix of both colors in the random picking, sometimes returning a color to it's original bucket. By virture of being the first to recieve a forgein color, bucket B would have the greatest percentage of the other color in it. I could be wrong, as I said, this is just instinct. (Please forgive my spelling!)
At 2/07/2007 5:10 PM,
Telima said…
I have to agree with Jenafer, for her exact same reasoning. That just makes sense to me. However, I look forward to reading Samuel's answer and explanation.
At 2/07/2007 8:43 PM,
Samuel said…
So here it is. The solution to the puzzle. The answer is that they will have the same number of foreign marbles in each bucket. Now for the explanation:
First, all the marbles are American and so they are all locals, not foreigners.
Actually, the best way to explain this is to do it in the way I see it. This is going to use algebra but I will explain it all to you each step of the way. At the end of the experiment, Bucket A will have:
Ar + Ab = 1000.
(Ar is the number of red marbles in Bucket A and Ab is the number of blue marbles in Bucket A.)
Similarly we have in Bucket B:
Br + Bb = 1000.
Notice that the second letter determines the color while the first letter tells the Bucket. In the end both Buckets will still have 1000 marbles.
Now let's adjust the second equation a little bit:
Br = 1000 - Bb.
This has to be true no matter what Bb and Br are.
Now look at the starting values. All the red marbles are in Bucket A. This means:
Ar + Br = 1000.
If we use the above form of Br and plug it into the above equation we get:
Ar + (1000 - Bb) = 1000.
Notice that I replaced Br with (1000 - Bb) as shown to be equal to each other above.
Get rid of the parenthesis and we see:
Ar - Bb + 1000 = 1000
Ar - Bb + 1000 - 1000 = 1000 - 1000
(Here I subtracted 1000 from both sides.
Ar - Bb = 0
Or
Ar = Bb!!!!!!!!!!!!!
ThThere must also follow that:
Ab = Br!!!!!!!!!!!!!
Thus we see that each bucket will have the same number of foreign marbles in it. -SKT
At 2/07/2007 8:56 PM,
Jenafer said…
HUH!!!!!!!!!!!!!!!!
At 2/07/2007 10:02 PM,
Telima said…
First of this is Tyler. Unlike Samuel I did not get it at first but did get it before his mathamatical solution. Simple put, with 1000 red marbles and 1000 blue marbles you could be given the chance to knowingly move any marble from one bucket to the other the trick is that both buckets have to end with 1000 marbles. Now think about it... any marble is either in one bucket or the other, therefore if I have any number of blue marbles in a bucket I must have 1000 minus that number in the other bucket and vice versa. Which adds up to opposites.
Insimpler terms: If I have 975 red marbles and 25 blue marbles in one bucket the other bucket must have the remaining 25 red marbles and 975 blue marbles. It is impossible to throw off the balance as long as you have only two buckets, two colors and both have to have the same number in the end.
At 2/08/2007 8:02 PM,
Jenafer said…
Tyler,
You explained it better then Samuel! I get it now
At 2/08/2007 8:06 PM,
Bradley Ross said…
Somehow I'm utterly unsurprised that Samuel was able to immediately determine the correct answer. When I was presented the puzzle, I made the some conclusions as Jenafer and Telima. But Tyler and Samuel's explanations should make the case that the number of foreign marbles will always be equal. The tricky math that jumbles up our brains is figuring out HOW MANY foreign marbles will be in one bucket--but that wasn't the question.
My first instinct (along with most others who've heard the puzzle) is to solve the wrong puzzle.
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