- View SourceHi Veeresh,

This looks like a the Monty Hall problem - however one critical item is

missed - when you say C is opened - was it chosen (out of B and C)

randomly? Or did someone who knew it was empty opened it? The situations

differ - the latter is the Monty Hall problem.

Suppose C is chosen randomly

----------------------------

Then, effectively we have a situation where out of three boxes, one is

randomly eliminated. The probability of coin being in either A or B is the

same - viz 1/2, so no point switching.

Suppose C is chosen the Monty Hall way

--------------------------------------

Now, C was chosen out of (B and C) by someone who knew that C is empty. As

someone who knew the coin is made this choice, the odds are better than even

that B has the coin. In fact, in this case probability works out to what

Reuben suggested - switching wins with probability 1/3.

Both the above situations can also be rigorously demonstrated with

simulation or proven with conditional probability arguments.

regards,

Rino.

-----Original Message-----

From: veereshad@...

To: probability@yahoogroups.com

Sent: 5/31/01 7:02 PM

Subject: [probability] 3 boxes with a coin in one!

Hello,

I have an example-riddle about probability. Please help me to find

answer for the riddle.

It goes:

There are 3 closed-similar-boxes, say A, B, C, one and only one of

them has a coin in it. Now I am asked to point the one having the

coin. Randomly I opt for one, say A. Then, when one of the other two

boxes is opened, say C, found to be not having the coin. Now I am

again have the chance of changing my selection of box.

Now do I stick to the old choice A or switch to B? Which one gives me

more chance of winning?

Any body, please answer me, which one? And Why? What's the

probability of winning? Not 50 %?

Regards,

Veeresh - View SourceThis is is the situation I described in the random experiment as "My box is

not chosen"You have make very clear that the probability here is 1/2

Of course information given by random esperiment may change our initial

assignments of probability. That is the whole point of conditional probability

and expectation, aside from Bayesian decision.

Please, take a look at the problems I sent yesterday. I really need them

solved.

Pablo

"Rino Raj(Taco)" wrote:

>

> Well, looks like the case where an empty box is knowingly selected seems to

> meet agreement. In the other case, perhaps I can use a formal argument like

> Pablo...

>

> One of boxes A, B, C contain a coin. A is chosen by player. Let 1 denote

> coin and 0 empty state.

>

> Now, for X = A, B, C

> p(X=1) = 1/3 and p(X=0) = 2/3

>

> alternately, each of the following cases have equal probability

> A | B C

> ========

> 1: 1 | 0 0

> 2: 0 | 1 0

> 3: 0 | 0 1

>

> Random situation

> ----------------

> C is randomly opened and found to be empty.

> Now p(A=1 | C is randomly opened and found empty)

>

> = p(A = 1 and C is randomly opened and found empty)/ p(C is randomly opened

> and found empty)

>

> p(C is randomly opened and found empty) = (1/2)*(2/3) = 1/3

> p(A = 1 and C is randomly opened and found empty) = (1/3)*(1/2)*(1) = 1/6

>

> Thus p(A=1 | C is randomly opened and found empty) = (1/6)/(1/3) = 1/2

>

> Hope Reuben agrees that additional information may often influence/ improve

> the chances of locating something. There is nothing sacrosanct about the

> first box chosen and its probability of 1/3. You open, say a box B, and see

> the coin, that probability definitely changes. Now if you don't see a coin

> also, the probability may change, as shown above.

>

>

>

>

>

> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/