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.
I was going to say the probability is exactly the same in both problems, on the principle that it’s probably not a magical evens-only die. Just a regular die that happened to land on a bunch of evens before coming up as six. I guess I was also assuming “knowing this” meant knowing after, not knowing before.
That is, I wasn’t assuming that pre-roll person got information that the die would only roll evens before rolling a six, and adjusted their odds accordingly. I was assuming that a person who had just rolled the die went, “huh, I only got evens before six came up. I guess the probability for this particular die was different all along!”
Is that also a valid answer? And if not, what am I missing?
It’s a good question! I think you’ve got the right framing of the problem; the question is whether “I only got evens before six came up” can reasonably be followed by the conclusion “I guess the probability for this particular die was different all along.”
My feeling is no! For example, if I get 6 on the first roll, I wouldn’t assume that I have a weird die, just that I happened to roll a 6.
It’s important to remember also that while bear markets are bad for some, they are good for others: that’s what “market” means. Historically, a bear-skin jobber was someone who was said to sell the bear’s skin before the bear was even caught, what we now call selling short: you sell a stock you don’t own on the assumption that when you need to actually make the settlement (about five days later), the price will have dropped and you can buy it quick. This is actually so successful that selling short into a falling market is banned on stock exchanges, because it can push the price of particular stocks into a tailspin: everyone assumes that the heavy selling means the stock is going to go down, and so it does. As Keynes said, playing the stock market isn’t about buying the best stocks or even the most popular stocks: it’s about buying what other people think are the most popular stocks, or even what other people think that everyone else thinks are the most popular stocks, and so on.
We can best see the special use of singular nouns in English compounds by looking at noun-noun compounds like rat-eater. A rat-eater is something that eats rats, not just one specific rat, and yet we use the singular. This is equally true with explicit numbers: a tablet that is four millennia old is a four-millennium-old tablet. I can call myself either six foot one (the traditional form) or six feet one, but I am definitely not six-footed. Exceptions arise when the noun is irregular (mice-eater is much better than *rats-eater, though mouse-eater is better yet) or when the emphasis is on the individuals in the group (an enemy list would be a list belonging to the enemy most probably, whereas an enemies list is a list of specific enemies). In the related languages, units of measurement are always accompanied by the singular: in German you speak of ein Fuss (a foot, the kind attached to a leg) or zwei Fuesse (two feet), but its length is vierundzwanzig Zentimeter (24 centimeter, not centimeters).
I love that etymology for bear market!
I was chatting with a Wall Street pal recently and I remembered to ask him my nagging question, which is “Now that most stocks don’t pay dividends, doesn’t that sort of un-tether the asset’s value from the company’s performance, and turn it into a guessing game of which direction you think the stock price will go?” I think his answer can be boiled down to, “That’s the *least* of the ways that finance is a guessing game of anticipating others’ behavior,” which is a pretty good match for your Keynes quote.
On the millennia/millennium example: try using more familiar units of time like month or year: “This is my four-month-old son” v. “This is my four-months-old son.” Your ear won’t have to struggle with that one. But it takes some time (for me, at least), to “hear” “four-millennium-old tablet” as correct v. “four-millenia-old tablet.” Happily, Grammerly knows that the latter is wrong.
And so does “Grammarly.” ;^)
Yeah, the more standard plurals that take an -s make a good step-ladder for my ear. At this point “four-millennia-old” sounds quite wrong, but obviously it slipped through my filter in the moment!
And I assume Grammerly is Kelsey Grammer’s personal site, which would certainly have accurate grammatical information. 😉
I was fooled on you Expected value problem until I worked through the math.
E[# rolls until 6] = 1/6 Σ n(5/6)^n = 6
E[# rolls until 6|all rolls are even] = 1/6 Σ n(2/3)^n = 3/2
Yeah, it’s a good one! I totally fell for the “it’s equivalent to a three-sided die” heuristic right away, as did a probability-loving friend I showed it to.
Interestingly, I showed it to two friends who are quantitative professionals (a mathematician and a stock analyst) but who don’t spend a lot of time thinking about probability puzzles. Both made more halting beginnings (e.g., wondering if the expected value might be > 6) but both avoided the pitfall I fell into and eventually hit on the right insight.
So I wonder if this problem is especially dangerous to people with experience solving probability puzzles; we’re especially liable to leap to the wrong idea.
Doug M,
Sorry, but the formulas in your post are incorrect. First of all:
Assuming the summation always goes from n = 0 to n = infinity,
1/6 Σ n(5/6)^n = 5 (not 6)
1/6 Σ n(2/3)^n = 1 (not 3/2)
So now maybe you’d want to improve the formulas to
E[# rolls until 6] = 1/6 Σ (n+1)(5/6)^n = 6
E[# rolls until 6|all rolls are even] = 1/6 Σ (n+1)(2/3)^n = 3/2
The first calculation is now well-defined and correct.
However, there is still a problem with the definition of the second calculation. Although it happens to lead to the correct outcome of 3/2 (in this particular case), you can’t actually justify how the formula is assembled — where did the factor of 2/3 in the formula come from? Well, my guess is that you still replaced the 6-sided die by a 3-sided die, which is not the proper way of calculation.
The incorrectness becomes clear when we apply the same approach to a slightly different situation: you roll a 4-sided die (with faces showing 1, 2, 3 and 4) until you roll a 4; what’s the expected number of rolls, given that you happened to roll only even numbers?
Using the same approach, you’d calculate
E[# rolls until 4|all rolls are even] = 1/4 Σ (n+1)(1/2)^n = 1
which is obviously incorrect (because how could the expected number of rolls equal the minimum possible number of rolls?). The actual correct answer is 4/3 rolls.
Another analogous example (with a normal 6-sided die) that shows that the formula doesn’t match what you meant to define:
E[# rolls until 6|all rolls are divisible by 3] = 1/6 Σ (n+1)(1/2)^n = 2/3
is obviously wrong. (The correct outcome should be 6/5 .)
The proper calculations are:
E[# rolls until 6|all rolls are even] = [ 1/6 Σ (n+1)(2/6)^n ] / [ 1/6 Σ (2/6)^n ] = [ 3/8 ] / [ 1/4 ] = 3/2
E[# rolls until 6|all rolls are divisible by 3] = [ 1/6 Σ (n+1)(1/6)^n ] / [ 1/6 Σ (1/6)^n ] = [ 6/25 ] / [ 1/5 ] = 6/5
E[# rolls until 6|all rolls are greater than 2] = [ 1/6 Σ (n+1)(3/6)^n ] / [ 1/6 Σ (3/6)^n ] = [ 2/3 ] / [ 1/3 ] = 2
E[# rolls until 6|all rolls are greater than 1] = [ 1/6 Σ (n+1)(4/6)^n ] / [ 1/6 Σ (4/6)^n ] = [ 3/2 ] / [ 1/2 ] = 3
E[# rolls until 6|all rolls are greater than 0] = [ 1/6 Σ (n+1)(5/6)^n ] / [ 1/6 Σ (5/6)^n ] = [ 6 ] / [ 1 ] = 6
and in the case of the 4-sided die:
E[# rolls until 4|all rolls are even] = [ 1/4 Σ (n+1)(1/4)^n ] / [ 1/4 Σ (1/4)^n ] = [ 4/9 ] / [ 1/3 ] = 4/3
(or you could use the elegant shortcut solution from Ben Orlin’s article, of course.)
This problem annoyed me a lot for a couple of reasons.
First, the passage, “I roll only even numbers prior to getting the 6” just kills me. I guess it’s easier if you specify that you _must_ roll at least one even number before rolling one six, but what then? The problem is actually more about the probability of rolling an even number on a six sided die that is less than six, and then rolling a six after, or an even number until you roll a six.
Furthermore, if you take the problem as stated, it’s literally, “I rolled a die ‘x’ number of times and got even numbers. What is the probability that the next roll is six?” However, I, honestly, get critics a lot for taking things ‘literally’, and have made an attempt to decipher the meaning behind what’s said. Due to the, “[something] [something] [snide remark] [the literal definition of the stated problem] [the not literal definition of the problem and I don’t even know what] is quite wrong” statement, I can only imagine that the problem is as difficult as I can make it.
[my version]
What’s the probability of obtaining a sequence when you randomly select a number between one and six, and when you must get an even number first, and if you roll an odd number, you must abscond your previous sequence and start over again. The valid sequence begins with an even number equal to or less than six, followed by any number of even numbers (even as in the number displayed on a six sided die), and terminating with a six. Given said constraints, I said, “fuck it” and moved on to the next post!
You (and I!) could calculate an answer, but the difficulty in this post is not in calculating a solution, but in the phrasing and conception of the problem. Why even bother in solving this question if the challenge is in the phrasing, and not in the solution though?
I have taken so many other math classes but for some reason 116 has been a struggle for me. Its just like something isn’t clicking. Could be I’m trying to memorize everything or that I’m trying to find things that seem to make since which helps me remember. Those things just don’t seem to be happening. This past test was so scary for me. Even thou my grade was passing I know I need to do better. I’m taking full advantage of the tutoring lab and the practice questions, those 2 things have really helped me through this first month. I’m giving all I can give and I’m praying I will pass. One thing I am very proud of when I get the wrong answer on mathlab is I have been able to go back and find my mistake which I think is huge. I believe that helps me see the equation and truly understand it. Thanks so much 🙂
116?
Just the wife and i. It felt too good to stop now; she didn t want to think, she just wanted to
feel.