Buddy's Blog

Just a note to Torbjörn and Divad. You two  guys are too smart for me. But I don’t believe Torbjörn. There must have been a flaw in your program. Maybe in your random number generator. And Divad mentioned Marilyn Vos Savant. If you don’t recognize the name, she was supposedly the worlds smartest genius. I had correspondence with her years ago because I mentioned her in the strip. So I started following her advice column. One day a guy wrote in about a social problem that he was having. And she answered it as such. She didn’t recognize the statement of his problem was an example of Russell’s Paradox.

Words of wisdom from a couple philosophers among you…

If you’re still wondering about the birthday question, here is the answer:

https://www.bing.com/search?form=MOZLBR&pc=MOZI&

q=How+many+people+in+a+room+have+same+birthday

I just got back from one of my entertaining gigs. This one went well. It was in a memory care facility. That’s good. Nobody will remember.

I hope you had a nice picnic. Meanwhile Arnold has discovered his true identity…

Gary posed a math question in comments.

Since we are currently discussing interesting shortcuts to solve problems, here is one for you math enthusiasts:
How can one easily compute the sum of all of the numbers from X to Y in literally seconds (versus adding them all up sequentially)? For instance, what is the sum of all the numbers between 1 and 10. Or between 1 and 20? It takes no longer than a few seconds to find the answer.

Here’s another  problem : I play the Queen of Hearts at the Elks Lodge. If they pick your ticket which has a number from 1 to 52 on the first drawing, and the number on your ticket corresponds to the number associated with the Queen of Hearts you win the jackpot. Tickets cost $2.00. Of course at the first drawing there are 52 cards in the deck , so someone’s chance of winning is 1/52. The next drawing assuming someone didn’t win at the first drawing is 1/51, etc. I want to know before the first drawing, how many cards must be drawn before the odds are 50/50. The answer is n where ∑1/52 + 1/51 +…1/n = 1/2. Of course the odds before any particular drawing is 1/n where n is the number of cards left in the deck. The Elks cap the jackpot at $15,000 because the money (so they tell me) goes to any of their several charities. Three months ago they drew my ticket and my card was the JACK OF HEARTS! Geeze! The Piranha Club should do the Queen of Hearts.

On the island, Anna Maria Island where I grew up, at the Moose Lodge the jackpot is over $250,000! That’s because it’s the biggest membership in all the Mooses in the country. And it’s the biggest because it’s smack on the beach. When I was a very young man what is now the Moose Lodge was an Elks Lodge. I spent the best New Years Eve in my life there once.

I’ll be at the Elks this Wednesday. There are 4 cards left.

Yeah, right. Frau G is going to a high school football game with one of her lady friends. Manatee High, my ol’ alma mater. I’m sitting home. I did my time watching boys run around when I was a soccer dad and Loinfruit was 8 years old.

This ad for TikTok just popped up on my IPad.

Heck, that’s not so clever. Here’s one that I just made up:

Speaking of my Wordle solution, last night I was playing. I put down my four special words:

twang, doers, filch, bumpy

And with two lines left, this is what I had:

         _ a e r S

The S was the only letter in the right place. The possible solutions are:

raves  rakes  rares  razes

So there were two guesses left and four possible solutions. So you see, my method is not entirely fool proof.  BTW the answer was razes.

I just noticed that this post didn’t get published yesterday. Sorry. I have another post that will go online at 18:00 EDT

Does anybody play Wordle? If you do, I’m going to show you how to beat it almost every time. The secret to beating the game is not by guessing the letters. The best strategy is to eliminate letters. You eliminate as many letters as possible so at the end there are only a few possibilities. Now I used to use these five letter combinations: rates, filch, mound and gawky.  But after you enter these four words there would still be seven letters that you haven’t tried. Lately I’ve used these words: doers, filch, bumpy and twang. It took me an hour or so to come up with them. These words all together use twenty distinct letters, so there are only six letters that I haven’t tried. Not only that, those six letters are v, k, j, x, q and z, the least frequently used in our language. Usually you will get all five letters, barring duplicates, in your first four words. Here’s a chart that I put together showing the relative frequency of letters as they appear in the English language, and the frequency of the of the letters in my trial words. The bottom entries are the sums of the frequencies:

To enhance your chances of winning your first entry would be the word doers (highest frequency of appearance) followed by twang, filch and bumpy. Notice that the remaining six letters have the lowest probability of appearing in the solution.

Now, of course, if you have five correct letters before using all four words, you stop there and work out the solution. But if you use all four trial words you still have two more trial words left use.  Suppose you have two trial words left and there are three possible solutions. What you have to do is find words that use two of the possible letters. If one of them is correct, good. If neither is correct then the remaining solution is the one you want. Occasionally you might find a situation where there are, say, three possible solutions involving two letters. That’s happened to me, but I can’t recall the specific circumstances. You just have to be clever in the trial words that you choose.

In the worse case, and maybe, the only case wherein you might lose, is if you are left with four possible solutions with only two trial words left.