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So I was watching the show Numb3rs, and the math genius was teaching, and something he did just stumped me.

He was asking his class (more specifically a student) on which of the three cards is the car. The other two cards have an animal on them. Now, the student picked the middle card to begin with. So the cards looks like this

+---+---+---+
| 1 | X | 3 |
+---+---+---+

The X Representing The Picked Card

Then he flipped over the third card, and it turned out to be an animal. All that is left now is one more animal, and a car. He asks the student if the chances are higher of getting a car if they switch cards. The student responds no (That's what I thought too).

The student was wrong. What the teacher said is "Switching cards actually doubles your chances of getting the car".

So my question is, why does switching selected cards double your chances of getting the car when 1 of the 3 cards are already revealed. I thought it would just be a simple 50/50 still, please explain why the chances double!

Marconius
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Michael Jones
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  • We have to know how the teacher picks the card he flips over. If he flips over one of the remaining cards at random, then you get no benefit from switching. However, in the usual Monty Hall problem, it is specified that the card flipped over/ door left open is chosen such that it is always an animal. This is a key assumption needed for switching to be better. – user264781 Sep 10 '15 at 10:44

3 Answers3

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This is the classic version of the Monty Hall problem.

Note that there is only one car, so the host is always able to reveal a goat behind one of the other two doors that were unselected. So the fact that he does this reveals no new information if the contestant does not switch.

So by not switching:

$$P_n = \frac{1}{3}$$

If on the other hand the contestant switches, he has inverted his probability of winning (wins when he would have lost if not switching, and vice-versa), so:

$$P_s = 1 - P_n = \frac{2}{3}$$

Marconius
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2

The situation your first intuition tells you happens is that if the host picks a card entirely at random then there's an equal chance of either other cards revealing a car.

However, the actual situation is that the host picks a card after you made the first selection.

So, you have either selected a car or not.   There's a $1/3$ chance you selected the car using unbiased random selection.

Now, if the host were to reveal a car it would be game over (you would be certainly wrong to either keep your choice or pick the remaining card.)   Still, we know this hasn't happened because the host did not reveal a car.   (Typically this never happens because the host is using foreknowledge to avoid this branch of the probability tree. However even if that's not the case, we still know a car was not revealed.)

  • If you selected a car, the host was free to flip either other card.   Then if you change your mind you are certainly wrong.

  • If you did not select a car, the card the host didn't choose is a car.   Then if you change your mind you are certainly right.

Thus there is a $2/3$ chance of you being right if you change your mind.

Graham Kemp
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1

The probability that you've picked the wrong card in the first place is 2/3. This happens to be the truth even when one of the cards is removed; it's just that 2/3 of the initial guesses are wrong.

So when you have the chance to change your mind, you do it, because in the "second round", 1/2 of the guesses are correct (assuming they are random).

Finally, chances when you keep the card are 2/3 times 1/2 that the guess is wrong. On the other hand, chances are 1/3 times 1/2 that the guess is wrong when you switch the cards. Comparing the terms gives the answer, I think.

David
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