Author: Eugene Fischer

I can’t believe it actually happened! Go Spurs Go!

I became a Dad eight days ago.

My son’s name is Ezra Owen Fischer-MacQueen. At birth he weighed 8 lbs. 1 oz. and was 21.75 in. long. He has a full head of dark hair and bright, alert eyes.

I won’t be posting any pictures of him here, or writing about him here much at all for now. But if you want to see pictures of my kid and read whatever myself or my wife have to say about him, you can do so at EzraOwen.net.

I recently watched Matt Parker’s video “The Unexpected Logic Behind Rolling Multiple Dice and Picking the Highest” (which—for two dice—players of D&D 5e might know as “rolling with advantage”). In this video he calculates the expectation value for rolling 2, 3, and 4 dice and picking the highest value. True to the title, I found the answer quite unexpected!

Once I’d seen the result, though, it seemed to me that there was a geometric way of arriving at it that’s more straightforward than the one Matt used. It also points toward a proof of the general case, which is something Matt leaves as an exercise to the viewer.

UPDATE: My friend Will has pointed out an error in this work. He’s writing up a post with the correct derivation, which I’ll link to here when he posts it. For now, let me explain my mistake.

My basic argument relied on the fact that, for any n-sided die, the expectation value is the midpoint of the number line interval from 1 to n. This is true.

I then said that, when rolling two n-sided dice, you could fix the higher value v_{high} in place and treat the lower value as the roll of a v_{high}-sided die. Then the expectation value for v_{low} is halfway between 1 and v_{high}. This is also true.

I further argued that you could reverse this, holding v_{low} fixed, and determining that the expectation value for v_{high} is halfway between v_{low} and n. This, too, is true.

I then said that you could use these facts together to conclude that the expectation values must partition the interval from 1 to n exactly into thirds. This step is false, or at least doesn’t follow from the previous two facts. The expectation values are not the same as individual rolls of the dice, and I can’t necessarily treat them as rolled values. This last step swaps out the specific values I was holding constant for the expectation values of those dice. There’s no reason to believe I can legitimately do that, so the whole of the rest of the argument falls apart.

(There was also a trivial condition with 2-sided dice not fitting one of the equations from Matt’s video that I was working off of, which I’m sure Will will include in his writeup.)

UPDATE 2: Here’s Will’s post, wherein he derives the general case formula for the expected value of rolling m dice with n sides each with advantage to be n - \sum\limits_{i=1}^{n-1}(\frac{i}{n})^m. More commentary on this, and how expectation values partition the interval from 1 to n, when I have time to play around with it more.

Grant Sanderson, who runs the excellent mathematics YouTube channel 3Blue1Brown, recently posted a video premised around introducing generating functions, and how they can be used to solve problems. At the end of the video he gives four puzzles for further exploring generating functions. One of them was this:

Compute the sum \sum\limits_{k=0}^n2^k\binom{n}{k}.

(Hint, expand the expression (1+x)^n.)

First, let’s look at that sum.

\sum\limits_{k=0}^n2^k\binom{n}{k} = 2^0\binom{n}{0} + 2^1\binom{n}{1} + \dots + 2^{k-1}\binom{n}{k-1} + 2^k\binom{n}{k}.

Doing as the hint says,

(1+x)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \dots + \binom{n}{k-1}x + \binom{n}{k}.

This gives us a polynomial with coefficients close to that of the desired sum. We’re just missing the powers of 2, which can be easily added in by changing the binomial we’re expanding.

(2 + x)^n = 2^0\binom{n}{0}x^n + 2^1\binom{n}{1}x^{n-1} + \dots + 2^{k-1}\binom{n}{k-1}x + 2^k\binom{n}{k}.

The coefficients of f(x) = (2 + x)^n are exactly those of the desired sum. So the sum be found by evaluating f(1) = (2 + 1)^n = 3^n. Thus, we get the answer that \sum\limits_{k=0}^n2^k\binom{n}{k} = 3^n.

That was a pretty surprising result. When I saw it, I thought, “3^n is the number of ways to make n choices when there are 3 options for every choice. I bet there’s an elegant combinatorial proof of this identity.”

Sure enough, there is, and it didn’t take too long to find.

Given a set of n objects, consider the number of different ways that Alice can choose k of those objects, 0\le k \le n, and then give a subset of the k objects she has chosen to Bob.

For any given value of k, Alice has \binom{n}{k} ways to choose k objects. She can then select any of the 2^k subsets of k to give to Bob. So, for a given k, the number of ways for Alice to choose k objects and then give some of them to Bob is 2^k\binom{n}{k}. Summing over all possible values Alice can choose for k gives \sum\limits_{k=0}^n2^k\binom{n}{k}.

But instead of first choosing k objects and then giving a subset of those to Bob, Alice could instead consider each of the original n objects and make one of three choices: leave it alone, take it for herself, or give it to Bob. Since there are three possible choices for each of n objects, the number of ways this can be done is 3^n.

Both of these methods count the number of ways the original set of n objects can be partitioned into three groups (Alice’s objects, Bob’s objects, and the leftover objects). Therefore, \sum\limits_{k=0}^n2^k\binom{n}{k} = 3^n. \blacksquare

I love this proof, and combinatorial proofs in general. While the generating function method is fascinating and no-doubt powerful, it doesn’t tell me why the identity is true. The combinatorial proof, though, tells me that the expressions are two different ways of looking at how to partition a set into three groups.

Now I’m curious if there’s a combinatorial solution to the problem that’s the main focus of the video.

I liked this recent piece from The Consilience Project. It starts with a great explication of the specific features and strategies of good faith communication, contrasted with those of bad faith. It then goes on to discuss the dangers of normalizing bad faith communication.

It also articulates well two points that I think quite important. First, that social media companies train people to engage in bad faith. This is an efficient explanation of one of my core objections to the platforms as they currently exist:

A key feature of escalating extremism is a belief that group membership requires bad faith engagements with out-groups. In these contexts, bad faith behavior is often justified to maintain in-group membership and consensus. The normalization of bad faith communication contributes to the creation of extreme in-group pressures, which can rupture identities and exacerbate mental health crises. Personal instabilities usually lead to a doubling down on the need for group membership, increasing rationalizations and amplifications of bad faith practices.^[8]

Digital media companies’ business models result in a proliferation of increasingly niche group memberships. They also incentivize public displays of conflict and bad faith communication, in order to capture attention and optimize engagement. Advertisements and propaganda dominate the social media space, driving up the total amount of bad faith communication to which people are exposed.^[9] In a very literal sense, heavy users of social media are being behaviorally entrained to engage disproportionately in bad faith communication. Politicians, public officials, and influencers of all kinds seek to exploit this environment of distrust and capitalize on the declining social value of good faith interactions. The epistemic commons is repeatedly degraded to the point of exhaustion.^[10] A “post-truth” culture is a culture of bad faith. 

Second, the need to distinguish between good-faith and naiveté. Good faith engagement is hard; it is always easier to engage naively (or in bad faith). But we need to do it anyway, and reject the easy excuses for not doing so:

Good faith communication is both a complex skill and a value commitment that shapes personal identity. In other words: doing it is sufficiently difficult that getting good at it will change the kind of person that you are. But good faith engagement is often misconceived as a simple emotional stance leading to an overzealous search for agreement. In part, this is because naive good faith interactions can often be based on an innocent orientation towards peacemaking and the avoidance of conflict. This can be seen in the good faith engagement of a child who does not know that some people are untrustworthy. Unskillfully engaging in good faith can be a danger when others are acting in bad faith. 

In some political discussions today, it is common to hear that “you can’t engage in good faith with Nazis!” A generous interpretation of this statement is that there are truly unreasonable people who must not be trusted because they have proven themselves to be dangerous and unethical. But this is not an argument against good faith communication in general. It’s an argument against engaging in naive forms of good faith communication, which would play into the hands of those actively seeking to cause harm to others. Likewise, for those who have been lied to historically and treated with disrespect by specific groups, naively engaging with those same groups again in good faith would be foolish. 

Registration for my online science fiction workshop closes at midnight Central time on April 27. If you want to participate, be sure to sign up before then. Also, use one of the discount codes I posted to save yourself some money!

My online science fiction and fantasy workshop is coming up on April 29th, and still has space available. Writefest has released some discount codes for their programming, so if you’re interested in participating, this is a perfect time to sign up.

Everyone: use code WRITEFEST22 for $35 off.

Students: use code STUDENTSROCK to get 25% off a workshop enrollment.

Teachers: use code TEACHERSROCK to get 15% off a workshop enrollment.

Writespace Houston also offers scholarships, which you can apply for here.

I’ll be leading an online science fiction and fantatsy writing workshop on April 29th. This is one of several workshops and events comprising Writefest, a hybrid writers’ workshop and festival from Writespace Houston. Here’s the description:

Science fiction and fantasy stories can go anywhere, from anti-matter galaxies at the edge of our universe to enchanted cities sparkling in universes as yet unimagined. But to bring readers along, these stories must also effectively deploy the tools of prose fiction. In this workshop, we’ll discuss your works-in-progress constructively, identifying the artistic and speculative goals visible on the page, and analyzing how elements like character, setting, and exposition are working to further those goals. Come sharpen your own narrative skills by offering useful feedback to others, and gain invaluable insight about where readers think your writing shines, and where it can be improved.

For this workshop, participants can submit short stories or excerpts (complete stories preferred). Maximum 3,000 words. The workshop will include peer critique with instructor participation.

If that sort of thing sounds like your jam, you can buy tickets for my workshop here (while they last—there is a participation cap to ensure everyone’s story has time for a full discussion).