Probability puzzles can pose a challenge to even the most intelligent and educated in our society. Sometimes, intuition leads us to an incorrect answer; other times, overconfidence in our mastery of math can be our downfall. And on some occasions, we’re too stumped to decide which solution to go with, and why.

Let’s see if you can find the correct answer to these three brain teasers – the solutions will follow on the next page.

## Bertrand’s Box Paradox

I present you with three identical boxes. One box contains two gold coins; another contains two silver coins; another contains one gold and one silver coin. You have no way of knowing which box contains which coins.

Now, pick a box at random. Open the lid, slip your hand inside – don’t peak! – and withdraw one of the two coins at random. Look at the coin in your hand – it’s gold. What are the odds that the other coin in the box is also gold?

## The Three Prisoners Problem

Three prisoners have been sentenced to death: Alfred, Bart, and Charlie. The day before their execution, Alfred overhears the warden mention that one of the prisoners has been selected at random to receive a pardon. He asks who, but the warden says he is not allowed to say until execution day.

Alfred reasons with the warden – if he cannot say who *is* receiving the pardon, then he can at least reveal the identity of one of the other prisoners who will be executed. He says, “If Bart is to be pardoned, tell me Charlie will be executed. If Charlie is to be pardoned, tell me Bart is to be executed. And if I’m to be pardoned, then flip a coin to decide whether to tell me that either Bart or Charlie is being executed.”

The warden agrees to these terms and tells Alfred that Bart is to be executed. Alfred now believes his probability of being pardoned has risen from 1/3 to 1/2 – that it is now between him and Charlie. In his excitement, he tells Charlie, who becomes even happier, because Charlie believes that Alfred still only has a 1/3 chance of being pardoned, while his odds have risen to 2/3.

Is Alfred or Charlie correct? Are Alred’s odds of survival 1/3 or 1/2?

## The Monty Hall Problem

You’re on a game show. You must choose between three identical doors. Behind one door is a car; behind the other two are goats. If you pick the door with the car, you win the car.

At random, you make your selection. The game show host, who knows what is behind every door, opens one of the doors you didn’t select to reveal a goat. He then gives you the choice to switch your selection to the other unopened door, or keep your initial choice. Should you switch?

*Have you solved each of these brain teasers? Click through to the next page to find out if you’re correct!*

## Bertrand’s Box Paradox

What are the odds that the other coin in the box is also gold?

You may have guessed 1/2, since there are only two boxes that contain gold coins, and you had equal odds at selecting either. If you picked the double gold box, then you’ll find another gold coin; if you picked the silver and gold box, you won’t find a gold coin. 50 percent chance, right?

Not quite. The trick here is that, while the initial odds of selecting any of the boxes were equal, the fact that we know we are holding a gold coin changes everything. The question isn’t, “What are the odds of randomly selecting a given box?” Instead, it’s, “What are the odds that this gold coin came from each of these boxes?”

There are six coins in total. Each are equally likely to be selected (1/6 odds). The gold coin you selected cannot be from the double silver box, and it can only be one of the two coins from the gold and silver box. This eliminates three possibilities, leaving us with three: it is either the gold coin from the gold and silver box, or one of the two coins from the gold and gold box. Since each of these outcomes is equally possible, in 2 out of 3 cases, the coin will come from the gold box, meaning in 2 out of 3 cases, the other coil will also be gold.

## The Three Prisoners Problem

Is Alfred or Charlie correct? Are Alfred’s odds of survival 1/3 or 1/2?

Since the Warden said Bart is to be executed, then there are only two prisoners remaining who can receive the pardon, meaning Alfred’s odds of survival are 1/2, right?

Wrong. Let’s examine the possibilities. If the Warden says that Bart is to be executed, then it is either because Charlie is to be pardoned, or because Alfred is to be pardoned **and the Warden’s coin flip made him pick Bart.** That last point is key. The different outcomes can be represented in the following table:

If Charlie is to be pardoned (1/3 odds), then the warden can say nothing but “Bart will be executed” (1/1 odds). However, if Alfred is to be pardoned (1/3 odds), then there’s a 50 percent chance that the warden will say “Bart will be executed” (1/2 odds). This means that when the warden says that Bart is to be executed, Charlie’s odds are twice as good as Alfred’s, making his odds 2 out of 3 to Alfred’s 1 out of 3.

## The Monty Hall Problem

Your instinctual response to this question may be to haughtily declare that it makes no difference – that switching doors would not be beneficial, because each door has a 50 percent chance of being the correct one. I know that was my first response. Now, let me explain why so many people get this one wrong.

When you initially select a door, the odds that you selected the correct door are 1 in 3. That means that the collective odds that one of the other two doors is the correct door are 2 in 3. When the host opens one of those two doors, the collective odds remain the same – 2 in 3 – but we now know with certainty that the other door is not a winner. Therefore, the odds that the remaining door is the winner are 2 in 3, making it always beneficial to switch when the host gives you the option.

Don’t buy it? Then check out this graphic with the various possibilities. The graphic assumes that you selected Door 1 (the choice is irrelevant).

As you can see, factoring in all possibilities, changing your initial selection will see you win in 2 out of 3 cases.

If you got this one wrong, or are still trying to wrap your head around it, don’t worry – this problem has stumped tens of thousands of people, including thousands with PhDs. Bertrand’s Box and the Three Prisoners problem are closely related and considered precursors to the Monty Hall problem, and hopefully by presenting them to you in this order, you were able to follow the increasing scale of complexity while getting a better understanding of probability theory. There are multiple solutions to each, some more intuitive, others more mathematical, so don’t worry if you arrived at the correct answer through a different means.

Now go win a game show.

*How many of these did you get correct? Did any stump you? Sound off in the comments!*