Card Game Analogous to Monty Hall?

Date: 12/12/2000 at 08:10:55
From: Gabriel Lozano
Subject: New aspect of Monty Hall. 50% proof?

I hope you can help me on this new aspect of the good old Monty Hall 
problem. Here is the new problem: 

There are three cards in front of you, two that are kings, and one 
that's an ace. You are trying to correctly guess which one is the ace. 
Say you pick the far-left card, and next, the middle card is revealed 
to be a king. Before having the far-left card revealed, you are given 
the choice of switching to the far right card. Are you better off 
switching or staying, and how much better off are you?

I have come up with a way that makes it seem as if the answer is 1/2 
instead of 2/3. Please tell me what is wrong with the way I am 
thinking:

Assume that there are 2 people. At the beginning, you have 3 cards. 
You and a friend both pick a card - he picks the left card, you pick 
the right card. The middle card is shown to be a king. According to 
the classic answer that says that the chances improve 2/3, there is a 
2/3 chance that the other card I didn't pick is the ace, and a 1/3 
chance that the card I picked is the ace. To my friend, he has a 2/3 
chance of my card being the ace, and 1/3 chance of his own being the 
ace. Our chances of switching and getting the correct card are both 
equal, so shouldn't our chances be equal? Think of it this way:

To me, my card has a 1/3 chance of being correct. To my friend, it has 
a 2/3 chance of being correct. To me, my friend's card has a 2/3 
chance of being correct. To my friend, his card has a 1/3 chance of 
being correct. My chance of being correct is thus 1/3 + 2/3 = 3/3. His 
chance of being correct is 3/3. Thus, our chances are equal, at 50%.

ANOTHER CASE:

Someone offered the counter-argument that (here accepting the fact 
that the answer is 2/3), "If both people switch, 2/3 of the time, 
they're right." I say, however, that if one person switches 3/3 of the 
time, he is right only 2/3 of the time. Which means that if the second 
person switches whenever the other person does, he is only right 1/3 
of the time. Wouldn't this be a counter-example, that the odds dropped 
to 1/3 instead of rising to 2/3?

Thanks for your time.
Gabe

Date: 12/12/2000 at 11:14:04
From: Doctor TWE
Subject: Re: New aspect of Monty Hall. 50% proof?

Hi Gabriel - thanks for writing to Dr. Math.

You are correct in your assessment. In the situation you describe, 
each of you has a probability of 1/2 of having the ace. The difference 
between your scenario and the Monty Hall scenario is that Monty Hall 
knows beforehand where the prize (the ace) is, and deliberately 
reveals the remaining non-prize (a king). In your scenario, there was 
a 1/3 chance that you would turn over the middle card and reveal the 
ace - and neither of you wins. This won't happen in the Monty Hall 
problem because Monty NEVER reveals the prize. Here's how the two 
scenarios look on tree diagrams: (For simplicity, I'll assume that 
you'll pick the right door/card in both scenarios).

MONTY HALL SCENARIO:

Case I: You stick with your first choice
----------------------------------------

The prize is equally likely to be behind L, C, or R (you pick R):

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize (Ace):       L        C        R

If the prize is behind R, Monty is equally likely to reveal doors L or 
C, since they're both losers. (It really doesn't matter, since the 
outcome is the same in either case.) If the prize is behind L, Monty 
will always reveal C, since he knows that that's the other loser. 
Likewise, if the prize is behind C, Monty will reveal L. So our tree 
now looks like this:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize:             L        C        R
                      1 |      1 |   1/2 / \ 1/2
                        |        |      /   \
     Monty shows:       C        L     L     C

Since our strategy in this case is to stick with our original choice, 
the first two branches are losers (in which the prize was behind L or 
C), but the last two branches are winners (in which prize was behind 
R). Completing the tree:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize:             L        C        R
                      1 |      1 |   1/2 / \ 1/2
                        |        |      /   \
     Monty shows:       C        L     L     C
     Outcome:          (L)      (L)   (W)   (W)
     Probability:      1/3      1/3   1/6   1/6

So our probability of winning with this strategy is:

     P(W) = 1/6 + 1/6 = 1/3.


Case II: You switch
-------------------

Again, the prize is equally likely to be behind L, C, or R (you pick 
R). We'll make another tree diagram:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize (Ace):       L        C        R

If the prize is behind R, Monty is equally likely to reveal doors L or 
C, since they're both losers. If the prize is behind L, Monty will 
always reveal C, since he knows that that's the other loser. Likewise, 
if the prize is behind C, Monty will reveal L. So our tree now looks 
like this:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize:             L        C        R
                      1 |      1 |   1/2 / \ 1/2
                        |        |      /   \
     Monty shows:       C        L     L     C

This time, our strategy is to switch doors, so the first two branches 
are winners (in which we originally picked L or C, but switched to R), 
but the last two branches are losers (in which we originally picked R, 
but switched to either L or C.) Completing the tree:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize:             L        C        R
                      1 |      1 |   1/2 / \ 1/2
                        |        |      /   \
     Monty shows:       C        L     L     C
     Outcome:          (W)      (W)   (L)   (L)
     Probability:      1/3      1/3   1/6   1/6

So our probability of winning with this strategy is:

     P(W) = 1/3 + 1/3 = 2/3

twice as good as our chances in case I.


YOUR SCENARIO:
--------------

The ace is equally likely to be L, C, or R (you pick R, your friend 
picks L):

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize (Ace):       L        C        R

Now you reveal the card not chosen (C) and discover that it is not the 
ace. What you have done is to eliminate the middle branch from the 
sample space, so our tree now looks like this:

                           ------+------
                      1/3 /  1/3 |      \ 1/3
                         /       |       \
     Prize (Ace):       L        C        R
     Outcome:          (W)      (X)      (L)

Since the sample space is now 2/3 (because we've eliminated the middle 
branch), your probability of winning (without switching) is:

     P(W) = (1/3) / (2/3) = 1/2

This is a conditional probability. We can state it as P(W|C<>ace). 
This is read as "the probability of winning GIVEN THAT the center card 
(C) is not the ace."

In this scenario, you can easily see that your friend's probability of 
winning (without switching) is also (1/3)/(2/3) = 1/2, and indeed, 
switching gains you nothing.

I hope this helps. If you have any more questions, write back.


- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/