This was written Sunday October 27, 2024 and will not be revised.

1. The president is not elected by popular vote.

I remember watching Headline News on CNN a few days before the 2012 election and hearing that the election between Obama and Romney was a statistical dead heat. Each candidate was predicted to have 47% of the vote. However, the results in the Electoral College were not close Obama 332, Romney 206. People who forecasted based on a state by state basis (Nate Silver and a talented amateur at Princeton) concluded on the days before the election that the probability Obama would win was 99%.

2. Even when the difference between the candidates is less than the margin of error of the poll it still contains information.

Let’s start with the question: what is the margin of error? It is often reported, but rarely with the sample size of the poll which is needed to be able to check the calculation. Some examples with both pieces of data are Franklin-Marshall for Penn 794/4.3; Emerson College for North Carolina 1226/3.6, and for Wisconsin 800/3.4; Marist for Arizona and Georgia 1193/3.9.

If Ɵ is the estimated frequency of votes for Harris then using standard formulas from an intro states class the radius of the 95% confidence interval the margin of error is 1.96(Ɵ(1-Ɵ)/n)^0.5.  Since Ɵ is usually in (0.4,0.6) then 0.24 < Ɵ(1-Ɵ) < 0.25 and replacing 1.96 by 2 we can simplify this to 1/n^0.5 and our data points become

n,error               800,  0.0353       1200, 0.288

somewhat smaller than the reported margins of error.

3. Frequencies should be reported to one decimal place and should be recalculated to remove responses that are not one of the two major candidates to make it easier to compare predcitions.

As some readers have noticed I have not gotten to #2 yet. That is coming soon. Bloomberg, the one poll that I have seen which reports frequencies with one decimal place had on October 23:

Harris/Trump    H/(H+T)           P(H win)   538 ave.

Michigan            49.6/46.9           51.4                     79.2         H 0.6

Penn.                  50/48.2               50.9                     67.7         H 0.3

Nevada               48.8/48.3           50.3                     59.2         H 0.2

Wisconsin          48.3/48               50.2                     51.6         H 0.1

Arizona               49.1/46               50.2                     51.6         T 1.8

48. Carolina 48.5/48.8 49.4                     43.9         T 1.2

Georgia               48.4/49.9           49.3                     35.2         T 1.5

If the uncertainties look large, it is because they are based on one poll. Once one has the average of several polls then the uncertainties decrease. Of course comparing Bloomberg with 538 average shows that there are sampling biases. The Washington Post is even more discordant. They have Harris leading in all states except Arizona. I got these figures from an article in Forbes but when I went to check them the Post wanted $2 to read the article.

Combining these projections to calculate is difficult without a computer and a good programmer. 538 uses simulations to conclude that, when this was written: Trump wills with probability 0.54 and Harris with probability 0.45.

https://projects.fivethirtyeight.com/2024-election-forecast/

One final selling point for people who publish polls, if you make predictions like this you can’t be wrong.

This blog post is based on the previous one The Arithmetic of Presidential Elections. There we used data from the last three presidential elections to argue that there are Red States with 170 electoral vote, Blue States with 215, and Swing States listed below with 153.

Turning to https://projects.fivethirtyeight.com/polls/president-general/2024/

for data, we see that Biden trails in every swing state

States               Votes                 Biden                 Trump

Ohio                   18                        36.8                    46.1

Georgia             16                        38.3                    44.8

Arizona              11                        38.9                    43.9

Iowa                   6                          32                        50.4

Carolina           15                       38.0                    45.2

Florida               29                        36.5                    46.1

*Penn.               20                       41.6                    43.6

*Wisc.                10                        41.8                    42.8

Nevada              6                          38.7                    42.6

*Michigan        16                        41.2                    43.0

*New Hamp.   4                          42                        44

Since this predicts Biden will only carry the Blue States this gives the prediction above.

Flipping the four closest states marked with * would result in 273 – 265 but likely the polls do not yet fully reflect his abysmal debate performance.

Here’s a little problem to test your skills at combinatorial probability.

You have a set of six cups and saucers. Two are NC State red R, two are UNC light blue b, and two are Duke dark blue B. You place the saucers in a line on the table RRbbBB. Then a blind man comes in and puts the saucers on the cups in random order. Let M be the number of cups that match the color of the saucer they are on. Your job is to compute the distribution of M.

To get you started I will specify a probability space which is the first step in solving any problem of this type. I once thought it was good to number the cups but a student in my class this year taught me it was better to treat the two cups of a given color as indistinguishable so we have 6!/(2!2!2!) = 90 outcomes instead of 720. To help check the solution note that not only should the probabilities sum to 1, but we must have EM = 6(1/3)=2. In the next paragraph I will start to reveal the solution starting at 6 and working down, so if you want to discover it on your on you should stop scrolling.

P(M=6)=1/90. In our probability space there is only one outcome where the cups all match, which is better than the situation when the cups are numbered and there are 2 x 2 x 2.

P(M=5)=0. If say the 2R match and 2b match then the 2B must match so 5 is impossible

P(M=4)=12/90. Matching 2-2-0 is impossible by the reasoning for 5, so we must have 2-1-1. There are 3 ways to pick the color with two matches, and for each color with only one match 2 choices of where the matching cup is. The rest of the outcome is now forced, e.g., RRbBBb.

P(M=3)=16/90. Matching 2-1-0 is impossible so we must have 1-1-1. We can pick the locations of the matching cups in 2 x 2 x 2 ways. The other three nonmatching cups must be either BRb or bBR

P(M=2)=27/90. We can have 2-0-0. Once we pick the double matching color in 3 ways the rest is forced, e.g. RRBBbb. We can have 1-1-0. We can pick the color with no match in 3 ways and 2 x 2 ways for the location of the matching cups. Suppose dark blue has no match. Then the two B cups must be on R and b, but there are 2 ways to put R and b on the B sauces for a total of 24 + 3 = 27.

P(M=1) =24/90. We can pick the location of the matching cup in 6 ways. Suppose it is the first R saucer. The second Red saucer can be B or b (2 ways). If it is B then we have bb on the Blue saucers, and we can have RB or BR on the b sauces (x2). If it is B then we have BB on the blue saucers and we have two possibilities on the B saucers, but autocorrect in Word will not let me type them.

P(M=0) =10/90. We can have BB on the Red saucers and then must have RR on b and bb on B. The situation is similar for bb on Red. This gives 2 outcomes. If we have {Bb} on the red saucers then we must have {Rb} on b and {Rb} on B where the set braces indicate we have not specified the order, so there are 2 x 2 x 2 = 8 outcomes.

1+12+16+27+24+10=90, 6 x 1+12 x 4+3 x16 +2 x 27+1 x 24 = 180 (so the mean is 2).

Tyro is a bit of crosswordese that means beginner or novice. Writing this reminds me of my first WORDLE in which I failed to guess TACIT in six tries. A tweet related to this puzzle which found its way into Rex Parker’s NYTimes Xword blog said something like the following: The answer reminds me of why I don’t do crosswords they are done by old people writing old words into the grid.

Turning to the main subject, as most of you probably know in WORDLE you get six tries to guess a five-letter word. On each turn you guess a five-letter word, a rule which prevents you from guessing say AEIOU to find out what vowels are present. If a letter is in the correct location  it shows green. If it is in the puzzle but not in the right place then it is white. If it is not in the answer it is gray. (Colors may vary) A copy of a computer key board on the screen allows you to enter you guesses and shows the status of each letter you have guessed.

As I start to give my advice I must admit I am still a novice but that never stopped TRUMP from pontificating on how to be president. In thinking about how to play WORDLE it is useful to know how frequently letters are used in the English language.

When Samuel Morse wanted to figure this out in the 1800s, he looked at the frequency of letters in sets of printers type which he found to be (numbers in thousands) E (12), T (9), A, E, I, O, S (8), H (6.4), R (6.2), D(4.4), L (4), U (3.4), C,M (3), etc. With computers and electronic dictionaries at our disposal we have a more precise idea (numbers are percentages).

E: 11.16                             A: 8.50                R: 7.58                I: 7.55                  O: 7.16                     41.95

T: 6.95                 N: 6.65                S: 5.74                L: 5.49                 C: 4.54                + 29.73 = 71.68

U: 3.63                D: 3.38                P: 3.17                 M: 3.01               H: 3.00                + 16.19 = 87.87

G: 2.47                B: 2.07                F: 1.81                 Y: 1.78                 W: 1.29               9.42

K: 1.102              V: 1.007              X: 0.290              Z: 0.272               J,Q: 0.196              2.93

Here the numbers in the last column are the sum of the numbers on the row and we have made 26 divisible by 5 by putting J and Q which have the same frequency to 3 significant figures into the same entry. This table become somewhat irrelevant once you visit

https://leancrew.com/all-this/2022/01/wordle-letters/

to find the letter frequencies in five letter words.

A: 10.5                E: 10.0                 R: 7.2                   O: 6.6                  I: 6.1                    40.4

S: 5.6                   T: 5.6                   L: 5.6                   N: 5.2                  U: 4.4                  + 26.4   = 66.8

Y: 3.6                   C: 3.6                   D: 3.3                  H: 3.1                  M: 3.1                 + 16.7   = 83.5

P: 3.0                   B: 2.7                   G: 2.6                  K: 2.1                   W: 1.6                 12.0

F: 1.6                   V: 1.1                   Z: 0.6                   X,J: 0.4                Q: 0.2                  4.3

Here E has fallen from the #1 spot. However, with the exception of Y climbing from 19^th to 11^th and P dropping from 13^th to 16^th it doesn’t seriously change the rankings, so I am not going to change my blog post due to this late breaking information.

The next thing to decide about WORDLE is what is your definition of success. I think of the game as being like a par-5 in golf. To take the analogy to a ridiculous extreme you can think of the game as par-5 in a tournament which uses the modified Stableford scoring system (like the Barracuda Open played at a course next to Lake Tahoe). Double bogey or worse (= not solving the puzzle) is -3, bogey (six guesses) -1, par (five) 0, birdie (four) 2, eagle (three) 5, and double eagle (two) 8 points.

I am not one who is good at brilliant guesses, so my personal metric is to maximize the probability of solving the puzzle. Hence I follow the approach of Zach Johnson who won the 2007 Masters by “laying up” on each par five. Most of these holes are reachable in two (for the pros) but 13 and 15 have water nearby so trying to hit the green in two and putting your ball in th water can lead to a bogey or worse. Zach hit his second shots to within 80-100 yards of the green so he could use his wedge to hit the ball close and make old school birdie.

My implementation of his strategy is to start with TRAIL, NODES, and CHUMP which covers all five traditional vowels and has 15 most frequent letters. The expected number of letters in the word this uncovers is (to use the five letter word frequencies) is  0.835 x 5 =  4.175 if all five letters in the word are different. (Recall from elementary probability that if X_i is the indicator of the event that the letter appear among the first 15 in frequency then E(X₁ + … + X₅) = 5EX₁  Dividing by 5 shows that the expected number of letters in the right position is 0.835 (assuming again all letters are different), so on the average we expect a green and three yellos..

Of course the answer can have repeated letters and can be chosen by the puzzle creator to be unusual, e.g., EPOXY or FORAY which were recent answers. (It is now April 8). In several cases my first three guesses have produced only 2 letters in the word, which makes the birdie putt very difficult. Even when one has four letters, as in  _OUND, possibilities are bound, found, mound, pound, round, sound, wound, even though some of these are eliminated if they are in the first 15 guessed.

If there are three (or more) possibilities for the one unknown letter, then it can be sensible to use a turn to see which of these are possible in order to get the answer in two more guesses rather than three. Or you can be like Tiger one year at Augusta and “go for it all.” give your birdie putt on the 15^th hole a good hard rap and watch it roll off the green into the creek. Fortunately for him, the rules of golf allowed him to play his next shot from the previous position.

These rules I have described are just to give you a start at finding a better strategy. You should choose your own three words not only to feel good about having done it yourself, but because the order of the letters can influence the probability of success. Of course you can also choose only to guess two (or only one) and then make your guess based on the result.  When I get several letters on the first two guesses, I have often substituted another word for CHUMP to get to the solution faster but I have often regretted that. On the otherhand sometimes when I play CHUMP I am disappointed to get no new positive information about what is in the word