Can You Solve The Latest Math Puzzle Circling The Internet?

We all love a mathematical brainteaser here at IFLS. Monty Hall? Done it mate. Two circles shoved into a quadrilateral of unknown size? Been there, solved that. Some Singaporean kids’ seemingly impossible homework? Debunked it, found the real answer. It’s kind of our thing.

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So, when we saw a new probability puzzle making the rounds online this week, we couldn’t help ourselves.

“There are three boxes,” begins the puzzle, posted on Reddit this past weekend and no doubt everywhere else at some point too. “Each box contains two balls. One box contains two gold balls, another box contains two silver balls, and the final box contains one gold ball and one silver ball.” 

So, that’s the setup. Now for the challenge: “You pick a box at random,” the post says. “You put your hand in and take a ball […] it’s a gold ball. What is the probability that the next ball you take from the same box will be silver? Note: you can’t see into any of the boxes.”

Any guesses?

If you’re like the Redditor who posted the puzzle, then you might have said 50 percent. After all, the question says you took a gold ball first – meaning you’re dealing with only box A or box B. With one gold ball gone, that leaves either another gold ball (if you chose box A) or a silver ball (if you chose box B) – that’s a one in two chance of getting what you want. Hence: 50 percent.

It’s a pretty smart answer, but it’s not quite right. Why? It all comes down to our old friend conditional probability – the very same concept that makes the Monty Hall problem so tricksy.

First, some good news: your instinct to ditch box C immediately is correct. You chose a gold ball first, so fact is, you’re not dealing with that box. So, what next?

Now, here’s where the previous answer attempt went awry. Instead of thinking about what’s left inside each box, you need to consider the three potential sequences of events – and yes, we mean three.

Here’s the thing: you don’t know what box you’re dealing with. We labeled ours for ease of explanation, but you pick it at random in the question – the contents are a mystery to you. That means that, if you draw out a gold ball, there are actually three potential balls left over: a silver ball in box B, a gold ball in box A, or the other gold ball in box A.

That means that the chance of picking a silver ball is now one out of three – just 33 percent. Neat, huh?

It’s a very similar question to the so-called “Boy-Girl Paradox”, or “Two Child Problem”: “Mr Smith has two children,” it runs. “One of them is a boy. What is the probability that both are boys?”

Again, the temptation is to say 50 percent – but the answer is in fact one in three, or 33 percent. 

If you’re still scratching your head over these puzzles, though, don’t worry – even the greatest academic legends have run afoul of conditional probability.

“I told the [Monty Hall] problem to the late Paul Erdős, one of the most famous mathematicians of the century, when he visited my home in 1995,” recalled the mathematician Andrew Vazsonyi back in 1999. “In a conversation about the use of probability theory in decision making, I mentioned the goats and Cadillac problem and the answer to Erdős, full[y] expecting us to move onto the next subject.”

“But, to my surprise, Erdős said, ‘No, that is impossible’,” he explained. “‘It should make no difference’.”

“I mentioned Bayes, and showed Erdős the decision tree solution I used in my undergraduate course. I reminded him that probability is not a fixed, static thing; it changes as time goes by,” Vazsonyi wrote. “To my amazement this didn’t convince him. He wanted a straightforward explanation with no decision trees.”

“I gave up at this point,” he concluded, “because I have no commonsense explanation.”