a weekly roundup of cartoons, links, and nefarious probability brainteasers
Like the rest of the math internet, I recently fell in love with this rascal of a problem:
- I roll a die until I get a 6. What is the expected number of throws?
- When I tried this, I happened to roll only even numbers prior to getting the 6. Knowing this, what is the expected number of throws now?
I’ll let you think. See the end of the post for a solution.
By the way, I’m back in Heidelberg this week, working on the official blog of the illustrious Heidelberg Laureate Forum, which brings together Nobel-level scholars in math and computer science with young researchers just starting this career.
It’s the coolest thing I do all year. Let me know if you’ve got questions I should ask the honored laureates and/or German locals.
Martin Hellman – a pioneer of public key encryption, and one of the laureates here in Heidelberg – poses a fun problem about n-dimensional spheres.
Spoiler: n-dimensional objects are super counter-intuitive.
Speaking of linguistic precision, Rogers George spotted an interesting linguistic error in a recent roundup post: I referred to a “four-millennia-old Babylonian tablet,” when I should have called it a “four-millennium-old Babylonian tablet.”
Put them side by side, and my ear prefers the corrected version, even if I can’t explain why or elucidate the underlying rule. I find that kind of instinct fascinating. I suspect it’s more common in one’s native language than in any other domain of thought (since language is a complex rule-governed area in which we all have copious daily experience from a young age) but I feel like I have similar experiences in math from time to time. Sometimes a student only needs their eye drawn to an error to recognize it as wrong, even without knowing why.
Okay, solution to the dice problem:
I roll a die until I get a 6. What is the expected number of throws? This is a pretty standard question about the “geometric distribution.” When you’re waiting for an event with probability p, it takes 1/p attempts on average. So the answer is 6.
When I tried this, I happened to roll only even numbers prior to getting the 6. Knowing this, what is the expected number of throws now? A tempting approach: assume that this is like rolling a 3-sided die, and so the answer is 3.
Tempting, but quite wrong.
Here’s an intuitive illustration. Imagine that the three odd sides of the die have been labeled “RESET.” You count the rolls it takes until you get a 6, but landing on “RESET” sends you back to 0. This is equivalent to the original problem, and clearly not equivalent to a 3-sided die, because your count keeps resetting. You expect a lower value than 3.
But what value? Well, a truly elegant solution comes via the original post (well worth checking out):
Rephrase the problem as “Roll a die until the first time you get a 1, 3, 5, or 6.” You’re waiting for an event with probability 2/3, so on average it will take you 3/2 attempts. And the expected number of rolls remains the same no matter which terminating value (1, 3, 5, or 6) happened to come up. Knowing that the terminating value was a 6, your answer is still 3/2, or 1.5 rolls.
I find probability a fertile source of intuition-busting puzzles like this. (See also: Monty Hall, the two daughters problem, and others). It’s a circus trick that no other branch of math can quite match: elementary, easy-to-state problems; elegant, accessible solutions; and still, the ability to stump a mathematically sophisticated thinker.