## Thursday, November 12, 2009

### What Kind of Hat Do You Look Best In?

You are presented you with 3 identically looking hats. In the 1st hat there are 2 white balls, in the 2nd hat there is 1 white ball and 1 black ball, and in the 3rd hat there is 2 black balls. You are blindfolded and (without peeking) you reach into one of the hats and pull out a ball.

If the ball is black, what are the chances of you reaching back into the same hat and finding that the remaining ball in the hat is black.

1. I'm so bad at these kinds of questions, but I THINK it would be a 50% chance. If the first ball is definitely black, then the next one to come out will either be white or black, because we've eliminated the hat with two white balls...

Right?

2. But we haven't eliminated one option, we've eliminated three...

If you think about it in terms of balls, not hats, there are now three equally likely situations that we are in:
1) we've pulled out the only black ball in the black/white hat
2) we've pulled out the first black ball in the black/black hat
3) we've pulled out the second black ball in the black/black hat

Situation 1) means that the next ball out *has* to be white.
Situation 2) means that the next ball out *has* to be black.
Situation 3) means that the next ball out *has* to be black.

So, do you still think that the probability of the next ball being black is 50%? (Yes, I'm being teacher-y :) ).

3. I think the answer is 33%.

Reason being, after drawing the first black ball, I still have three hats in front of me and I can draw from any of the hats.

So I have to reach the same hat and draw the second black ball.

So probablity of reaching one hat from three available is 33%.

www.guessthelogo.blogspot.com

4. Jon,

I'm not the best with stats, but as soon as we've pulled one of the black balls out of the hat, we've selected our situation.

Situation 1 could exist, after we pull out a black ball, we could pull a white ball out next.

But after we pull a black ball out, can both situations 2 + 3 exist? Haven't we selected our situation (2 or 3, but not both) to compete with situation 1?

So I'm thinking we pull out a black ball and of course the other ball is white or black. But the other ball can't be white or black or black can it?

I think the odds would be like you said before picking out the ball, but not after. But I could be on crack, too.

Fly Safe

5. I agree with Tabitha and the pilot, 50%.

6. The answer is 50%.

The other balls floating around have no impact at all on the hat selection.

Consider the analogous situation: We have a white hat with a white ball in it, a white hat with a black ball in it and a black hat with a black ball in it and the ball we have chosen is black... What is the probability that the hat we chose from is black?

7. Pilot (& commenter immediately before Pilot):

As with many of these probability problems, it depends on exactly how you parse the problem situation.

I was reading it as:
You pick a hat, and remove a ball from it.
That ball is black.
What is the probability that the remaining ball in that hat is also black?

This probability is not 33% or 50%, but 2/3 (which is approximately 66%):

P(2nd ball is black given 1st ball is black) = P(2nd AND 1st balls are black) / P(1st ball is black) = (1/3) / (1/2) = (2/6) / (3/6) = 2/3.

(this is using the formula for calculating conditional probability: P(A|B)=P(A&B)/P(B) )

To argue about the individual probabilities in this:
1) the probability that both balls are black is 1/3, as there is 1 out of 3 equally likely hats with two black balls.
2) the probability that the first ball is black is 1/2, as there are six balls that we could pick out, each equally like, and exactly half of them are black.

8. Jon,

It does not at all depend 'on exactly how you parse the problem situation'. The probability would only be 2/3 if after choosing a black ball first, one combined the remaining ball in the hat you picked from with the other hat that had black balls in to start with. If the Pilot's post didn't help, read anonymous' post above, the situations are analogous.

The answer is 1/2.

9. This reminds me of the Monty Hall Problem. Do you know this one, Mike? Probably the most counter-intuitive statistical problem ever conceived, in terms of fooling the most number of people.

10. Actually, after actually reading the problem and some of the knee-jerk reactions (particularly Anonymous there, tsk tsk...careful that you know what you're talking about before you go calling out other readers), this problem is isomorphic to the Monty Hall problem. The probability is 2/3. This is a classic appeal to people's innate sense of equal distribution of probability (they assume that probability is equally spread out among all remaining unknowns, even when one unknown is more likely than some others). Here, there's a 2/3 chance that the black ball you drew actually came out of the hat with two black balls. People have a really hard time believing this (as evidenced by the comments). Good problem :)

11. This is a version (as Abe points out) of the Monty Hall Problem. And for those of you who want to know, I started this blog by posting just that problem (http://statspuzzles.blogspot.com/2005/08/game-show.html)

The answer is 2/3 and it's almost impossible to convince anyone that it's true if they don't believe it. The only way I've found is for someone to run the simulation. There are simulations out there on the web, so go take a look. Become a believer! ;-)

12. After further review...

I do think I'm onboard with 2/3. But I'm still scratching my head.

Jon's math is right. P(B|A) = P(A & B)/P(A). Which means the probability of event B happening after event A has already happened is defined as the probability of events A and B divided by the probability of event A.

Probability (A) = drawing black first = 3/6
Probability (A and B) = drawing black then black = 2/6

Hence, 2/6 divided 3/6 = 2/3!

But what is screwy is my other twisted way of looking at it. If you number the balls, 1+2 white, 3 black, 4 white, 5+6 black and then draw a black ball.

You could have only drawn 3 or 5 or 6. In order to draw two black balls, 3 won't work, only 5 or 6 would. (That's where the 2/3 comes from, because 5 or 6 will work). BUT if you have drawn a black ball from one of the hats, you either have 3 or (5 or 6). If you have 3, no luck. If you have 5 you're a winner, and since you have already drawn 5, you can't have drawn 6. The other black ball can't be 5, because you're holding 5. Dang numbers.

13. Well, yeah, of course the answer is 2/3.

Couldn't this whole thing be more easily explained by pointing out that: If you've picked that initial black ball, you're twice as likely to have picked it from the hat with two black balls?

14. Apologies Jon (and others for that matter), it turns out that it was a bad day that was made worse by me being rude and wrong, but mostly worse solely because I was rude.

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