Category Archives: Asked and Answered

Asked & Answered 6.0

Blame it on a simple twist of Hickenlooper fate.

I confess: it is only because former Colorado Gov. John Hickenlooper decided to run for president that I decided to write this post.  His candidacy intrigues me — if Hickenlooper were to win, he would would join Dwight Eisenhower as our only four-syllable presidents and he would be the first president with a 12-letter surname.

Hickenlooper himself talks about his “funny name” but how unusual is it, relative to other U.S. Presidents?  How does it affect his chances?  What is a presidential name, anyway? Since no one is addressing these important questions, I am happy to do so — not only for Hickenlooper’s sake but for all of the 2020 presidential candidates.

Here is how we will proceed.  First, we will assign weights to the names of our presidents based on their positive or negative influence.  We will then use those weights to tabulate the desirability of three properties of presidential names — length, consonant-vowel ratio and last letter.  Finally, we will calculate a Sounds Like A President (SLAP) score for each candidate’s name based on its properties and their historical desirability.

• • • 

Let’s preface the discussion by comparing the length of the surnames of U.S. presidents to those of the population at large.  The chart below shows the prevalence of last names of various lengths for both groups.[1]  Note how the presidents with 8 to 10 letters in their last names are over-represented with respect to the population:

We should not conclude from this chart that long names confer an electoral advantage.  Perhaps the performance of those presidents was so poor that a longer name now carries a negative connotation.  Or it could be that no one even remembers certain presidents and therefore the characteristics of their names are irrelevant.  This is why we must begin the analysis by calculating Presidential Name Weights (PNW):

PNW (for President x) = Reputation Score  x  Memorability Factor

Using Presidential Name Weights (ranging from -1.0 to 1.0) is more realistic than giving equal weight to each name without regard to how or whether a president is remembered.

The Reputation Score of each president is derived from the 2019 Survey of U.S. Presidents conducted by Siena College Research Institute.[2]  In this survey, 157 presidential scholars and historians rated the presidents on their abilities and accomplishments, and the results were ranked from 1 (best) to 44 (worst).  I re-scaled the rankings so that Reputation Score ranges from 1.0 (best) to -1.0 (worst) and 0.0 represents an average (mediocre) president.

The Memorability Factor for each president comes from a study by Roediger and Desoto, in which participants were asked to name as many presidents as they could remember.  This factor ranges from 0.0 to 1.0, reflecting the fraction of participants who were able to name a given president.[3]

We can now calculate the Presidential Name Weight for each president as the product of his reputation and his memorability.  The results (below) are sorted from most positive to most negative:

This chart tells us that the names Washington, Lincoln, Jefferson, Roosevelt and Kennedy carry the most positive weights, whereas Nixon, Hoover, Johnson, Bush and Trump suffer the most negative weights.  Candidates whose names remind us of Washington or Lincoln would score highly relative to candidates with names like Bush and Trump.

• • • 

Our next step is to construct a Desirability Table for each property of a presidential name.  For instance, how desirable is having 9 letters in one’s last name?  How desirable is having a name that ends in n?  And so on.

This example shows how a Desirability Table is built.  Say we want to know the desirability of having a surname with x letters.  1) We list all the presidents with x-letter surnames and add up their Presidential Name Weights.  2) We divide that sum by the total PNW for all presidents to yield the desirability factor D for an x-letter surname.  Repeat these steps for every possible value of x to complete the Letter Count Desirability Table.  And so on.

Here then are the Desirability Tables for the presidential name properties in our analysis.  Each D-factor is a value between -1.0 and 1.0, and the highest D-factor for each property is shown in green.

The tables reveal that the highest-scoring presidential name would be 9 letters long, have a consonant-to-vowel ratio between 2.1 and 2.6, and end in the letter n.  Interestingly, none of our presidents’ names have all three of these features.[4]

The last item we need to address before calculating Sounds Like A President (SLAP) scores for the candidates is the weight we should assign to each property.  Here, we will elect to base a property’s weight on the amount of variation in its D-factors.[5]  Sparing the reader my lengthy justification, the weights we will assign to letter count, consonant-vowel ratio and last letter are w(LC) = 0.265, w(CV) = 0.275 and w(LL) = 0.460 respectively. 

• • • 

We are finally ready to calculate the SLAP score for each candidate, using this formula:

SLAP (for candidate x) = w(LC) x D(LC)  +  w(CV) x D(CV)  +  w(LL) x D(LL)

For illustration, let’s consider John Hickenlooper.  Hickenlooper’s last name has 12 letters, a consonant-vowel ratio of 1.4 and ends in r, so his SLAP score is

SLAP (Hickenlooper) = (0.265 x 0) + (0.275 x 0.463) + (0.460 x -0.038) = 0.110

which means he does not have a very presidential name, indeed.

The highest possible SLAP score is 0.657, corresponding to a name which has 9 letters and 5 consonants and ends in n.  Sen. Dianne Feinstein, destiny awaits you.

• • •

The table below lists the SLAP scores for the 20 mainstream presidential candidates (plus Donald Trump).  And the name of our next president is…

Elizabeth Warren!  Warren vaults to the top of the list on the strength of her last letter n and consonant-vowel ratio of 2.0.  Warren edges out Joe Biden, whose five-letter name weighs him down, thanks to the likes of Trump, Nixon, Tyler, Hayes and Grant.

Tim Ryan of Ohio has a decent SLAP score and may be a good running mate for Warren. Bernie Sanders has another respectable showing but comes up short in the last letter race. Peter Paul Montgomery Buttigieg (his full name) is just happy his last name is not Burns.

If only Hillary Clinton were running… her Sounds Like A President score would be 0.637.

Last and least is where we find Donald Trump.  Memorably poor performance.  Negative scores in all areas.  Still doesn’t sound like a president.


[1] Data for the U.S. population is from the 2010 U.S. Census, comprising the most-common 150,000 names representing 90% of the population.

[2] As one might expect, George Washington topped the rankings and Andrew Johnson came in dead last.  Donald Trump was merely the third-worst.

[3] The Roediger and Desoto study was published in 2014, prior to the 2016 election.  As such, the president named most often was Barack Obama (100%).  I assigned Donald Trump a 100% memorability factor as well but did not adjust the figures for the other presidents.

[4] Readers may ask, why these properties and not others, such as the number of syllables?  Syllable count would be fairly redundant, as it is correlated with letter count and consonant-to-vowel ratio.  I tried to select properties that appeared to be more-or-less independent but I did not actually test their cross-correlation.

[5] Specifically, the property weights w are based on the standard deviation in D(property).  My rationale is that a factor with little variation does not differentiate the candidates as much as one with large variation.  The statisticians among my readers are sure to howl, which is OK, because I don’t know any statisticians.

More in  Asked and Answered | Read 3 comments and add yours

Asked and Answered 5.0

I recently became a bitters-and-club-soda drinker.  As I was to discover, club soda has the annoying habit of going flat before one finishes the bottle.  To reduce waste, one can buy small bottles of national brands but they are much more expensive than the one-liter or two-liter bottles of store brands.  What’s a club soda drinker to do?  Glad you asked, and here’s your answer.

If club soda did not go flat, the choice would be clear.  A six-pack of 10-ounce glass bottles of White Rock Club Soda costs $3.99 here — 33.25 cents per 5-ounce serving — while four one-liter bottles (135.3 ounces) of our store-brand club soda cost only $3.00 — 11.09 cents per serving.  This means club soda in the 10-ounce bottles is three times more expensive… at least before any drinks are poured.

I tried a few one-liter bottles and was surprised to see how much club soda went to waste.  (Soda without sparkle belongs in the drain, not the drink.)  So I sat down to figure out whether the one-liter variety was a bargain or a boondoggle.  The answer would involve Henry’s Law, the Ideal Gas Law, bottling industry data and a few simplifying assumptions. (Full details are provided in the Appendix.)

But let’s start with how club soda goes flat (Figure 1).  The icon at far left represents the bottle as you buy it — filled to the neck with liquid and then topped off with high-pressure carbon dioxide (CO2).  Then, you open the bottle, pour a serving, and re-close the bottle.  The headspace is now filled with air from your room.  Over time, as the bottle rests in your refrigerator, CO2 escapes from the soda and collects in the headspace, until the pressure reaches an equilibrium (as determined by Henry’s solubility constant for carbon dioxide).  This equilibrium CO2 pressure will be lower than the original pressure — as suggested by the lighter shades of blue in the diagram — due to the CO2 that escaped when you opened the bottle along with the CO2 that you poured into your drink.

FIGURE 1: How Club Soda Goes Flat

The club soda gets flatter each time you open the bottle and pour a serving, until hardly any gas is left.  At some point, you may as well water your plants with what remains.

The question is, for a given serving size and bottle size, how many servings can one pour before the contents become undrinkable?  In my case, a serving is 5 oz (148 ml), I drink one serving a day, and I consider club soda too flat to drink if it has less than half of the original fizz (your mileage may vary).  This is the case I will address first, comparing the 10-ounce bottles to the one-liter bottles.

Let’s assume we refrigerate the bottle at 4C (39F).  Once a day, we take it out of the fridge, open it, pour the 5-oz serving, then close the bottle and return it to the fridge, all done as quickly as possible so that the contents do not get a chance to warm up.  We repeat these steps every day until the bottle is empty or the soda is flat.  The following chart (Figure 2) shows the resulting carbonation level for each serving, for both sizes of bottles:

FIGURE 2: Carbonation Level vs Serving Number (One 5-Oz Serving per Day)

We see that the 10-ounce bottle delivers two acceptable servings — the carbonation level of the second serving (57%) satisfies my 50% specification.  But the real surprise is that the one-liter bottle yields just one more drinkable (61%) serving — the fourth serving has only 36% of the original fizz and the fifth 17%.

So, by my criterion, almost 56% of the one-liter bottle is undrinkable.  This increases the effective cost-per-serving of a one-liter bottle from 11 cents to 25 cents.  Even so, it is still less expensive than the 33 cents-per-serving cost of a 10-ounce bottle.

Pouring two servings at a time (1o ounces) makes the one-liter bottle an even better value, but not by much.  The first two servings would be full strength of course, and the next two would have 72% of the original fizz.  The final two servings, however, would drop to 34%. The effective cost-per-serving in this scenario would be 19 cents, and 41% of the contents would still be undrinkable.

To completely eliminate waste from a one-liter bottle, you would have to either (a) lower your fizz-level standards, or (b) belt down half the bottle (3-plus servings) each time that you open it.  Only then will it cost you 11 cents per serving.

The bottom line is, a one-liter bottle of club soda at $0.75 is a better buy than a six-pack of glass bottles at $3.99, even if you have to share half of it with your houseplants.  But if the one-liter bottle costs $1.00, it is a break-even proposition.

Keep the Fizz Alive

Want to get the most fizz for your club-soda buck?  Research by Nestlé [1] suggests that soda in one-liter plastic (PET) bottles loses about 10% of its carbonation every 60 days when stored at room temperature.  This is because CO2 gas diffuses through the walls of the bottle.  (Bottlers compensate for the loss in shelf-life by adding extra carbonation to PET bottles.)  There are three takeaways from this:

(1) Buy your club soda from a busy supermarket.  This increases your chance of drinking  recently-bottled higher-carbonation product.

(2) Only buy as many bottles as you will use this week.  For the same reason.

(3) Refrigerate your club soda.  This reduces the internal pressure — and thus the rate of CO2 loss due to diffusion — by about one-third compared to room-temperature storage.  Not only that, cold liquid dissolves more CO2 so less gas is lost when you open the bottle.

Follow this advice and you will earn an A for effervescence.  And don’t forget to recycle.


To estimate the carbonation in a bottle of club soda at each step of its product life, I made a number of reasonable simplifying assumptions:

  • The bottle is completely rigid.
  • The gases and liquids follow Henry’s Law and the Ideal Gas Law.
  • When the bottle is filled, the headspace gas is assumed to be pure CO2 .
  • The carbonic acid formed by the reaction of CO2 and water is insignificant.
  • The diffusion of CO2 through the walls of the PET bottle is ignored.
  • Evaporation of the soda is insignificant.
  • The liquid and gas are at equilibrium before the bottle is opened.
  • Bubbles that escape from the soda just before it is poured are ignored.
  • The contents of the bottle remain at 4C (39F) while the soda is poured.
  • When the soda is poured, the gas in the headspace is replaced entirely by air.
  • The amount of CO2 and water vapor in the air can be ignored.

We begin by writing down the initial conditions in each bottle (Figure 3).  I will use liters as the unit of volume, atmospheres as the unit of pressure, and moles for the mass of CO2. (One atmosphere is essentially sea-level pressure, and one mole of CO2 is about 44 grams.)

FIGURE 3: Initial Conditions

A few notes on these facts and figures.  I measured the total volume of each bottle in my kitchen by filling them to the brim with water.  I assumed that the actual amount of soda in each bottle was exactly what the label claimed (bottling equipment is very accurate).  The initial headspace is the difference between those numbers.

According to Steen and Ashurst [2], bottle-filling is usually performed at 14C (about 57F) and the carbonation pressure for club soda is about 60 psi, or just over 4 atmospheres [3].  Assuming that all the gas in the headspace is carbon dioxide, we can use the Ideal Gas Law and Henry’s constant to calculate the amount of CO2 in the gas and liquid as bottled.

Initial moles CO2 in headspace (n0) from the Ideal Gas Law [4]:

\text{(1)}\hspace{5em}n_0 = p_0V_0 / RT_0

where p0 is the CO2 pressure (4.083 atm), V0 is the initial headspace volume (0.010 liters or 0.85 liters depending on bottle size), R is the gas constant (0.08206 liter-atm/mol-deg) and T0 is the filling temperature in degrees Kelvin (14C + 273.15 ≈ 287K).

Initial concentration of CO2 in liquid (c0) from Henry’s Law [5]:

\text{(2)}\hspace{5em}c_0 = p_0H_{14\text{C}}

where H14c is Henry’s constant at 14C (refer to above table). 

We now know the total moles (m0) of CO2 in the bottle:

\text{(3)}\hspace{5em} m_0 = n_0 + c_0L_0

where L0 is the initial liquid volume (0.296 or 1.000 liters depending on bottle size).

Chilling the bottle to 4C does not change the moles of CO2 in the bottle but it does affect the CO2 liquid-to-gas ratio.  We find the new liquid concentration (c1) at 4C by combining the Ideal Gas Law and Henry’s Law, then substituting for pressure p1 and rearranging:

\text{(4)}\hspace{5em} n_1 = p_1V_1/RT_1 = m_1 - c_1L_1\;,

\text{(5)}\hspace{5em} p_1 = c_1/H_{4\text{C}}\;,\hspace{.75em}\text{and so...}

\text{(6)}\hspace{5em} c_1 = m_1H_{4\text{C}}/(L_1H_{4\text{C}}+V_1/RT_1)

where m1m0 L1 = L0 and V1 = V0 (since nothing has been removed from the bottle).  Knowing ci lets us back-calculate n1 and p1 from Equations (4) and (5).

We now pour the first serving of soda, whose volume is LS .  The pour reduces the liquid in the bottle to L2 = L1Land the total moles of CO2 to m2 = m1n1c1 LS , as we assume all CO2 in the headspace is lost.  The headspace volume is now V2 = V1 + LS .  We quickly close the bottle and return it to the refrigerator, so that T2T1 .  While the bottle rests, the CO2 concentration of the liquid gradually decreases from c1 to cand the CO2 gas pressure in the headspace rises to p2 .  The values of c2 and p2 are unknown, but we can solve for them using Equations (5) and (6) with new subscripts:

\text{(5')}\hspace{5em} p_2 = c_2/H_{4\text{C}}\;,\hspace{.75em}\text{and...}

\text{(6')}\hspace{5em} c_2 = m_2H_{4\text{C}}/(L_2H_{4\text{C}}+V_2/RT_2)

This is the procedure I used to generate the CO2 concentration results in this article.


[1] Profaizer, Mauro. “Shelf life of PET bottles estimated via a finite elements method simulation of carbon dioxide and oxygen permeability.” Italian Food and Beverage Technology, vol. 48, 2007.

[2] Steen, David P., and Ashurst, Philip R. (editors).  Carbonated soft drinks: formulation and manufacture.  Oxford Ames, Iowa: Blackwell Publishers, 2006.

[3] Spangenberg, Craig. “Exploding Bottles.”  Ohio State Law Journal, vol. 24, 1963, p. 513.

[4] Khan Academy ( is one of thousands of sites with info on the Ideal Gas Law.  The Ideal Gas Law (pV=nRT) is to chemistry what Newton’s Second Law (F=ma) is to physics.

[5] Choose your own reference: Henry’s Law on Wikipedia or Henry’s Sparkling Water on ice.

More in  Asked and Answered | Read 5 comments and add yours

Asked and Answered 4.0

I am one of those annoying persons who is kind and cooperative in everyday interactions but whose competitive bloodlust rises to the fore when Scrabble is involved.  Ask my wife.  She stopped playing Scrabble with me years ago, even after I began to spot her 100 points. That was little help, as far as she was concerned, because it only made me dig my heels in that much harder to overtake her.

I play online now, as I assume most players do, because who stops by your place to play Scrabble these days?  I am a good casual player but not tournament-level by any means — not that I care to be.  I have only recently begun to learn the two-letter-words and I am constantly surprised by the odd three-letter-words my opponents play.

Things were different in the analog days.  Back then, diligent players had to memorize the accepted words, because consulting a dictionary was verboten except to answer challenges. (VERBOTEN is an official Scrabble word, by the way.)  But there are no challenges in the online version — instead, players are allowed (and expected) to check whether words are valid before playing them.  And the two-letter-word list is now just a feature of the game.  So no one really needs to memorize much.

That said, there is still a place for vocabulary skills in online Scrabble, tempered as always by the luck of the draw of one’s letters.  And that, at long last, brings me to the point of this article.  I just completed a game with an anonymous online opponent named Micki.  Here is the board we played:

Scrabble Board - The 100 Billionth PersonYou can see that Micki defeated me, 339 to 308.  What you do not see is that Micki drew and played all five of the highest-scoring letters (Q, Z, J, X, K), along with seven of the ten next-highest-scoring letters (two each of F, H, V, W, Y) as well as the two blanks, each of which confer a 15 to 30 point advantage [Thomas, 2011].  Now perhaps I am biased, but this seemed to me to represent an especially good string of luck — for my opponent.

The competitive person that I am asked, what are the chances of such a lopsided draw?  The nerd that I am set about to answer it.

• • • •

At first I thought I would need to program a computer simulation of thousands of games with random word lengths, to see how likely it would be that a real game would have such a one-sided letter draw.  But then I made a few simplifying assumptions.  (That is what old engineers eat for breakfast, simplifying assumptions.  For us, they go down as easily as, say, eggs Benedict did for the last Pope.)

My first simplifying assumption was that each player will likely draw half of the stockpile of letters over the course of the game.  This assumption allows us to neglect the number of tiles drawn on a play-by-play basis.  Once we stipulate that each player will draw 50 of the 100 tiles over the course of the game, we can focus on the distribution of tiles between the two players, as if all the tiles were shuffled and dealt out to the players like a deck of cards.

To further simplify the problem, we can imagine that the special, high-scoring letters are dealt out first, followed by the low-scoring letters, using the following process.  Start with the Q.  To see which player gets the Q, the dealer flips a coin.  If the result is heads, the Q is dealt to my pile, otherwise it is dealt to my opponent.

The dealer repeats this process for the remaining high-scoring letters and the two blanks, seven tiles in all.  Since the chance of my being dealt a tile after a coin flip is 1/2, then the chance that I would be dealt all seven of the special tiles is 1/2 to the 7th power, or 1/128.  Needless to say, this is also the chance that my opponent would be dealt all seven tiles.

Though not common, landing all the best-scoring tiles is not as rare as I had first assumed. But wait — my opponent not only also played the top seven tiles but also seven of the ten four-point tiles.  So we need to do a bit more math.

In the second phase of the process, the dealer takes the ten four-point tiles and again flips a coin to determine which player is dealt each tile.  What is the chance that I will be dealt exactly three of the ten tiles?  Interestingly, I found that this specific problem was asked and answered on Mathematics Stack Exchange.  The general formula is

prob(k heads in n tosses) = C(n,k) * p k * (1-p) n-k

where C(n,k) is the number of combinations of n things taken k at a time (explained here) and p is the probability of a single event, which in our case is 1/2.  For n = 10 and k = 3, the answer is 15/128, or a little better than one chance in nine that I would be dealt three of the ten four-point tiles.*

So the overall likelihood of Micki’s lopsided draw is the product of the two probabilities we calculated, that is, 1/128 * 15/128 = 15/16384.  Expressed as odds, this would be 1091 to 1, or about the same likelihood as getting bumped from an airline flight due to overbooking.  If one looks at it this way, yes, I was a bit unlucky (or Micki lucked out, take your pick).

The big difference between one-sided Scrabble games and overbooked airline flights is that my bad luck did not result in my getting beaten up and hauled out of my recliner, kicking and screaming — instead, I wrote this post.  So there.  I’m vindicated.  But I still lost.


* Some readers with penchant for details may be asking, wait, what about the rest of the tiles?  How does the dealer make the piles come out even, when one player has been dealt most of the special tiles?  Easy.  Take the remaining tiles, shuffle them and deal them out until one player has 50 tiles, then deal whatever is left to the other player.  For good measure, make sure the unlucky player gets most of the I’s and U’s.
More in  Asked and Answered | Read 2 comments and add yours