# Not able to understand the solution of this puzzle

1. May 23, 2012

### musicgold

Hi,

I am not able to comprehend the solution of a problem from a puzzle book.

Your friend shows you an urn and tells you that there is one marble in the urn. The marble is either black or white. (Assume that the marbles are identical in shape and any two marbles with the same color are indistinguishable.)

She then drops one black marble in the urn. Then sticks her hand in the urn and takes out one marble randomly, which turns out to be black. Now she asks you what are the chances of drawing a black marble from the urn.

According to the book the probability of drawing a black marble is 66.6%.

My answer was 50%. Here is my logic.
1. Assume that initially there was a black marble (b1) in the urn. So when the friend drops another black marble (b2), the urn has b1 + b2. Now when she takes out one black marble , the probability of drawing another black marble is 100%.

2. Now assume that initially there was a white marble (w1) in the urn. So when the friend drops another black marble (b2), the urn has w1 + b2. As she takes out a black marble, it has to be b2. So the probability of drawing a black marble is 0%.

Therefore, my answer is the average of the 100% and 0% chances.

2. May 23, 2012

### tamtam402

nevermind :D

3. May 23, 2012

### Whovian

I think the problem is you're assuming there's a 50% chance of having a black marble and a 50% chance of having a white marble at the start of the experiment. You're not given any probabilities, you're just given that your friend draws a black marble.

This is related to the concept of in Deal or no Deal (at least, I think that was the game show), it's a better idea to pick the other door.

4. May 23, 2012

### DonAntonio

Well, in fact the solution is 0.75 under the assumption that the probability of the first marble in the urn being black is 0.5...!

This problem's solved pretty easily with a probability tree: first there's a prob. of 0.5 to have a black marble in the urn and,

of course a prob. of 0.5 the marble is white. After that, we ll know the second marble added to the urn is black, so now

we have the branches:

B --> N , with probability $\,\displaystyle{\frac{1}{2}\frac{1}{2}=\frac{1}{4}}\,$ , and the only other relevant branch

which is $\,N --> N\,$ , with probability $\,\displaystyle{\frac{1}{2}\cdot 1=\frac{1}{2}}$ , so the final prob. to draw a black marble is $\,\displaystyle{\frac{1}{4}+\frac{1}{2}=\frac{3}{4}}$ , as

the book correctly states.

DonAntonio

5. May 23, 2012

### Whovian

Actually, the book states it's 2/3, not 3/4. Maybe the book's rubbish?

6. May 23, 2012

### DonAntonio

Perhaps, or perhaps it contains an error, or perhaps I made a mistake. It'd be interesting to know what book is that...?

DonAntonio

7. May 23, 2012

### DonAntonio

Rats! Re-reading the OP I realize now what part I misunderstood: the question is "what is the probability that the second marble in

the urn (the one that's left after the first one is drawn) is back"..! They don't mean the first marble already drawn!

Well, then: we havet thus 3 possible cases:

1) The urn contained a white marble, a black one wass added which was then first drawn: probability to draw again a black marble: 0

2) The urn contained a black marble (say, N1) and another black one (say, N2) was added. Probability to draw again a black marble after N1 was first drawn: 1

3) As in (2), but now we want the prob to draw a black marble after N2 (care here!) was first dranw: 1

As we can see, in two out of the three cases above we'll have a black marble left in the urn after the first one was drawn and, thus

the prob. is 2/3 = 0.6666, as the book says.

And btw, this indeed reminds the game Whovian mentions, which I recall as being Monty Python's contest game (one can even google this).

DonAntonio

8. May 23, 2012

### musicgold

This is not clear to me.

9. May 23, 2012

### viraltux

The book is right,

One way to see this is as follow, imaging she does not draw any marble and ask you about the chances to draw a black one, the you could say you have two scenarios;

1 - W B
2 - B B

Then you would say 3 B out of 4 possibilities, so the chances to get a black marble will be 3/4.

Now, you friend takes one B away and you have:

1 - W
2 - B B

or

1 - W B
2 - B

So your chances to get now a black marble are 2/3. If you don't see it this way just repeat the experiment over and over, then you have

she takes B1 from B1W
she takes B1 from B1B2
she takes B2 from B1B2

So in the long run you have W,B2,B1 so, 2B and one W, and therefore the probability is 2/3.

10. May 23, 2012

### musicgold

I think my question is not clear.

Initially the urn has only one marble. There is a 50% chance that it is black and a 50% chance that it is white. The friend adds one black marble to the urn.
Then she randomly draws one marble for the urn, which happens to be a black marble. Now she asks : what is the probability of drawing a black marble from the urn now?

11. May 23, 2012

### viraltux

2/3

12. May 23, 2012

### Staff: Mentor

there is a related problem of a princess facing marriage to a king that she hates. The king suggests a game to make it fair. If she can select the white stone from the pot then she can go free but if she selects the black stone then she must remain as his queen.

She notices that the king has cheated and placed two black stones in the pot. What should she do?

13. May 23, 2012

### viraltux

Kick his nuts and run.

14. May 23, 2012

### viraltux

All right musicgold,

In fact I find more interesting why people don't see the solution that the problem itself.

In this case your mind is going around and around the fact that there is only one marble left, that marble is either white or black, and therefore the probability for it to be black is 1/2, right? Well... wrong!

Think about this experiment, imagine a room full of people in which 2/3 are women and 1/3 are men. Then I pick one randomly and I ask you about the probability of being female.

Would you say, "well, that person is either male or female, therefore 1/2"? No, you wouldn't because you can see clearly here that the experiment favors females with 2/3. Well, so does the girl in your problem, she favors black marbles when she places one in the urn, then the probability to draw a black marble is 3/4, and when she draws one black marble she still favors black but not so much, this time 2/3.

You need to think how the experiment favors one case or the other, not just paying attention at the final layout; just the same way you would not say "well that random person must be male or female therefore 1/2" you cannot say "well that random marble must be black or white therefore 1/2"

Clear now?

Last edited: May 23, 2012
15. May 23, 2012

### Whovian

Take a black stone, and eat it (or put it out of view some other way.) Suggest to the audience that by looking at the remaining stone, they can see what colour the original one was. To note that the other one was also black is to admit cheating.

But this is getting off topic.

16. May 23, 2012

### cosmik debris

I think you mean Monty Hall, although Monty Python's would be a lot funnier.

17. May 23, 2012

### DonAntonio

Of course...lol. So I recalled not accurately, as well.

DonAntonio

18. May 23, 2012

### DonAntonio

The 3rd possibility is: we have two black marbles in the urn but this time the first one drawn wasn't N1 but N2...these are DIFFERENT, identifiable

black marbles for this problem's sake.

DonAntonio

19. May 23, 2012

### Staff: Mentor

Good answer except she gets brought up on assault charges and beheaded.

The best solution was to pick a stone and drop it quickly then say well you can what stone I picked by whats left and of course the king wont say he cheated cause it isn't kingly and so she gets to leave peacefully.

20. May 24, 2012

### moonman239

This.

As I recall, there was a story about a king who had taken some people captive. One of their wise men faced the king, who gave him challenges and told him he and his people could go if he completed the challenges. In the last game, the king said he would write or have written "Stay" on one piece of paper and "Go" on the other. If the guy picked the one that said "Go," he and his people could leave.

The wise man, however, was not fooled. He knew there'd be no paper that said "Go." Both of them would say "Stay." So he took a piece of paper and ate it. He turned over the remaining one and said, "This one says 'Stay,' so the one I ate must have said 'Go.'"

Anyways, OP, maybe running a Monte Carlo simulation would help.