5 replies to this topic

[mmiguel]

At a job interview, your potential employer presents you with the following test:

There are 2 buckets, and a bin with 50 green balls and 50 red balls.
He tells you he will leave the room, and that you must place the balls in the buckets.
When he comes back, he will randomly select a bucket (with equal probability), and randomly draw a ball from that bucket.
If he draws a green ball, you are hired.

Rules:
I. No bucket can be empty
II. Each of the 100 balls must be placed in one of the two buckets

What do you do?

Edited by mmiguel, 25 October 2012 - 05:52 AM.

[vistaptb]

Spoiler for Solution

One green ball in one bucket, the other 99 in the other bucket.

[bonanova]

Spoiler for Solution

One green ball in one bucket, the other 99 in the other bucket.

I would agree.

[hhh3]

i don't get it.. what if he goes for the one with 99 balls and gets a red ball?

[mmiguel]

I would agree.

So would I.

i don't get it.. what if he goes for the one with 99 balls and gets a red ball?

Spoiler for

Since all the red balls have to end up in some bucket, that the potential employer has the ability to choose, you cannot guarantee with probability 1 that he will draw a green ball.
The goal is to maximize that probability.
Using vistaptb's answer, you will get hired if the employer picks the bucket with the single green ball in it (he will pick that with probability 0.5). If he picks the other bucket, you will get hired with probability 49/99 = 0.49494949494949494949494949494949 ~ 0.5. He will pick this bucket with probability 0.5, so the net probability of you getting hired this way is 0.5*0.5 = 0.25.

Adding that to the 0.5 from the green ball, your chances of getting hired are approx 0.75.
When many people see this problem, the initial reaction is that it doesn't matter how you place them, it is always a 50% chance. This is not true, as shown above.

The intuition you should take from vistaptb's answer is that to minimize the effect of the red balls on probability, it is best to place as many green balls in the same bucket as the red balls, and that for other buckets, only a single green ball is needed to make the entire bucket a guaranteed win.

If we generalized the problem to n buckets, I haven't made a proof, but I think this strategy would also hold for any valid n.
The problem starts to seem different for n > 50 though, since at that point there would be some buckets with only red balls in them. Still given the situation, a strategy like the one above will still give the best of limited options I believe.

Edited by mmiguel, 25 October 2012 - 12:27 PM.

[hhh3]

hmmm.. cool.. thanks mmiguel..