Archive for December, 2015

Probability, the Playoffs, and the Indianapolis Colts

With one week left in the NFL season, Dan Graziano had this to say about the Indianpolis Colts’ chances of making the playoffs:

Indianapolis can still win a third straight AFC South title. Really, it can. All it needs is to win and then have the Texans, Bengals, Chargers, Jets, Saints, Chiefs, Patriots and Browns all lose. The league will throw in the partridge in a pear tree.

Colts HelmetIf the Colts win and the Houston Texans lose, both would be 8-8, and the first four tiebreakers for deciding which team makes the playoffs – record against one another, record against divisional opponents, record against common opponents, and record within the conference – would not be enough to decide who makes the cut.

It then comes down to strength of victory and strength of schedule. And for things to play out in the Colts’ favor, a lot of things have to go their way.

Fox Sports referred to this as long shot, comparing it to a recent win by a horse who was 200-to-1:

Sure, going 9 for 9 here looks dim, but long shots come in every once in a while.

Are the Colts’ odds as good as that horse’s? Seems not.

Using a simplistic model, assume that each of the nine necessary outcomes are equally likely. That alone would put the Colts’ odds at 511-to-1. (Since 29 = 512.)

But it’s not that simple. The following chart from 538.com gives the probability of each team winning their game this weekend:

NFL Week 17 - Game Probabilitiess

The good news is that the Colts have an 81% chance of winning their game against Tennessee. The bad news is that it seems unlikely that any of the Texans, Bengals, Chiefs, or Patriots will lose, let alone all four of them. So putting all those numbers together, the Colts’ chances of making the playoffs are:

0.81 × 0.20 × 0.22 × 0.84 × 0.54 × 0.69 × 0.15 × 0.18 × 0.77 = 0.0002 = 0.02%

or, more precisely, about 4,311-to-1. That’s more than a long shot; that’s an extended-to-an-unfathomable-distance shot.

The Colts are in the unenviable position of Lloyd Christmas in Dumb and Dumber, “So, you’re telling me there’s a chance…”

December 31, 2015 at 5:05 pm 1 comment

Math Problems for 2016

“What homework do you have to do tonight?”

I ask my sons this question daily, when I’m trying to determine if they’ll need to spend the evening doing word study or completing a math worksheet, or if we’ll instead be able to waste our time watching The Muppets or, perhaps, pulling up the animated version of Bob and Doug Mackenzie’s 12 Days of Christmas on Dailymotion.

When I asked this question last night, though, the answer was surprising:

We have to do our reading, but we already completed your math problem.

My problem? I had no idea what this meant. So they explained:

It’s not a problem you gave us. It’s one we got from [our teacher], and it says, “This problem was written by Patrick Vennebush.”

I was puzzled, but then it dawned on me. I asked, “Does it have a monkey at the top with the word BrainTEASERS?”

“Yes!”

“Which problem?”

“It’s about the word CAT.”

I knew the problem immediately. It’s the Product Value 60 brainteaser from Illuminations:

Assign each letter a value equal to its position in the alphabet (A = 1, B = 2, C = 3, …). Then find the product value of a word by multiplying the values together. For example, CAT has a product value of 60, because C = 3, A = 1, T = 20, and 3 × 1 × 20 = 60.

How many other words can you find with a product value of 60?

CAT

As it turns out, there are 14 other words with a product value of 60. Don’t feel bad if you can’t find them all; while they’re all allowed in Scrabble™, the average person won’t recognize half of them.

You can see the full list and some definitions in this problem and solution PDF.

This problem resurfaced at the perfect time.

With 2016 just around the corner, no doubt many math teachers will present the following problem to students after winter break:

Find a mathematical expression for every whole number from 0 to 100, using only common mathematical symbols and the digits 2, 0, 1, and 6. (No other digits are allowed.)

And that’s not a bad problem. It gets even better if you require the digits to be used in order. For instance, you could make:

  • 2 = 20 + 16
  • 9 = 2 + 0 + 1 + 6
  • 36 = (2 + 0 + 1)! × 6

But that problem is a bit played out. I’ve seen it used in classrooms every year since… well, since I used it in my classroom in 1995.

So here are two versions of a problem — the first one being for younger folks — using the year and based on the Product Value 60 problem above:

How many words can you find with a product value of 16?

How many words can you find with a product value of 2016?

There are 5 words that have a product value of 16 and 12 words that have a product value of 2016 (spoiler: those links will take you to images of the answers). As above, you may not recognize all of the words on those lists, but some will definitely be familiar.

December 19, 2015 at 8:33 am Leave a comment

Rebus Redux

I recently received an email from Akram in Lebanon who saw the Rebus Quiz on the MJ4MF website and suggested the following rebus:

ENTURY

The answer is below, but I’ll give you two hints.

The first is that you might say the same thing to a calculus student who spent 4 hours completing an integration problem only to forget the constant.

The second hint is that Akram’s suggested rebus reminds me of this one, which you may have seen before and is very appropriate for the holidays:

ABCDEFGHIJKMNOPQRSTUVWXYZ

The answer to this one is below, also.

But Akram’s email convinced me that it was time to create a few more, so here are five all-new rebuses (or is that rebi?) for you to enjoy.

1. Rebus 1
2. Rebus 2
3. Rebus 3
4. Rebus 4
5. Rebus 5

To put a little space between the rebuses and the answers below, here’s one more rebus from Sandy Wood:

Really, Really Bad Rebus #32

solution at Brain Games from Mental Floss


SPOILER ALERT… Answers Below

ENTURY: Long time, no C (see)

ABCDEFGHIJKMNOPQRSTUVWXYZ: No L (noel)

  1. A + 6 backwards = AXIS
  2. SINE + double U = SINEW
  3. ONION + PI in the middle = OPINION
  4. Dr. OZ + ONE = OZONE
  5. STUN + TWO + MAN = STUNTWOMAN

December 17, 2015 at 7:16 am 2 comments

NPR Puzzle Combinations

During yesterday’s NPR Sunday Puzzle, puzzlemaster Will Shortz presented the following challenge:

I’m going to give you some five-letter words. For each one, change the middle letter to two new letters to get a familiar six-letter word. For example, if I said FROND, F-R-O-N-D, you’d say FRIEND, because you’d change the O in the middle to I-E.

He then presented these nine words:

  • EARLY
  • TULIP
  • MOURN
  • JUROR
  • FUTON
  • DEITY
  • PANDA
  • SLOTH
  • VISOR

You can figure out the answers for yourself. For those that give you real trouble, you can either listen to the broadcast or search for the answer at More Words.

For those of you who don’t know who Will Shortz is, you have something in common with detective Jake Peralta from Brooklyn Nine-Nine:

The puzzle was fun. But what was more fun was the conversation that our family had about it. After the third word, Alex announced, “This shouldn’t be that hard. There are only 676 possible combinations.”

What he meant is that there are 26 × 26 = 676 possible two-letter combinations, which is true.

He continued, “But you can probably stop at 675, because Z-Z is pretty unlikely.”

I smiled. He had chosen to exclude Z-Z but not Q-K or J-X or V-P.

Yet his statement struck me as a challenge. Is there a five-letter word where the middle letter could be replaced by Z-Z to make a six-letter word? Indeed, there are several:

  • BUSED or BUSES
  • CONED or CONES
  • FETED
  • FUMED or FUMES
  • GUILE
  • MEMOS
  • NOBLE
  • PITAS
  • RAVED or RAVES
  • ROVED or ROVER or ROVES
  • TAPAS
  • WIDEN
  • WINED or WINES

None of them are perfect, though, because Z-Z is not a unique answer. For instance, ROVER could become ROBBER, ROCKER, ROMPER, ROSTER, or ROUTER, and most puzzle solvers would surely think of one of those words before arriving at ROZZER (British slang for a police officer).

From the list above, the best option is probably GUILE, for two reasons. First, stumbling upon GUZZLE as the answer seems at least as likely as the alternatives GUGGLE, GURGLE, and GUTTLE. Second, the five-letter hint has only one syllable, but the answer has two, and such a shift makes the puzzle just a little more difficult.

But while Alex had reduced the field of possibilities to 675, the truth is that the number was even lower. The puzzle states that one letter should be “changed to two new letters,” which implies that there are only 25 × 25 = 625 possibilities. Although that cuts the number by 7.5%, it doesn’t help much… no one wants to check all of them one-by-one to find the answer.

When Will Shortz presented DEITY, the on-air contestant was stumped. So Will provided some help:

I’ll give you a tiny, tiny hint. The two letters are consonant, vowel.

Alex scrunched up his brow. “That’s not much of a hint,” he declared.

Ah, but it is — if you’re using brute force. To check every possibility, this reduces the number from 625 to just 21 × 5 = 105, which is an 80% reduction.

Still, Alex is correct. The heuristic for solving this type of puzzle is not to check every possibility. Rather, it’s to think of the word as DE _ _ TY, and then check your mental dictionary for words that fit the pattern. It may help to know that the answer isn’t two consonants, but most puzzle solvers would have suspected as much from the outset. In the English language, only SOVEREIGNTY, THIRSTY, and BLOODTHIRSTY end with two consonants followed by TY.

Below are five-letter math words for which the middle letter can be changed to two new letters to form a six-letter word. (Note that the answers aren’t necessarily mathy.)

DIGIT :: DI _ _ IT (unique)

POINT :: PO _ _ NT

FOCUS :: FO _ _ US

MODEL :: MO _ _ EL (unique)

POWER :: PO _ _ ER

RANGE :: RA _ _ GE (unique)

SOLID :: SO _ _ ID (unique)

SPEED :: SP _ _ ED

And below, your challenge is reversed: Find the five-letter word that was changed to form a six-letter math word.

CO _ EX :: CONVEX (unique)

LI _ AR :: LINEAR (unique)

OR _ IN :: ORIGIN

RA _ AN :: RADIAN (unique)

SE _ ES :: SERIES

SP _ RE :: SPHERE

Enjoy!

December 8, 2015 at 6:20 am Leave a comment

Dreidel is Not Fair

Dreidel

play an online version of dreidel at the Jewish Outreach Institute (but turn the sound off)

Is dreidel fair?

The rules of dreidel are straightforward. At the beginning of each round, players put one coin into the pot. (For young kids, the “coins” are actually chocolate pieces in the shape of a coin and wrapped in gold foil. This is known as geld, and as far as I’m concerned, chocolate is a currency to kids.) Players then take turns spinning the dreidel, and a reward is earned based on which of the four Hebrew letters appears on top when the dreidel stops spinning:

  • Nun: nothing.
  • Hey: half the pot.
  • Gimel: all of the pot.
  • Shin: put one in.

Play continues clockwise, with each person spinning the dreidel until Gimel occurs and all coins are removed from the pot. At that point, everyone antes another coin, and a new round starts with the next player.

Officially, a player is out of the game when she or he has no coins left to contribute to the pot, and the game ends when one person has all the coins. But practically speaking, the game often ends much earlier, because players get bored and quit or, in the case of very young kids, the game lasts beyond bedtime and the children are pulled away by their parents.

No matter how the game ends, though, it’s not fair.

The following table is courtesy of Paul J. Nahin (Will You Be Alive 10 Years from Now?, Princeton University Press, 2014, p. 81). It shows the amount, over the long run, that each player will win during a dreidel game.

Player
1 2 3 4 5
Number
of
Players
2 1.143 0.857
3 1.361 0.956 0.680
4 1.617 1.102 0.757 0.524
5 1.900 1.267 0.855 0.580 0.398

In other words, the first player has a significant advantage over the others. In a game of five players who start with 10 coins each, the first player will finish the game with 19 coins, on average, whereas the fifth player will finish with just 4 coins. That’s if the game ends early. If played until one person gets all the coins, then the first player is five times more likely to win than the fifth player.

This disparity in odds is likely the reason that an unofficial rule of dreidel is that the youngest player goes first, the second-youngest player goes second, and so on.

The word dreidel is Yiddish and means “to turn around.” Because the dreidel is, after all, a top.

Explicit LyricsThis fact is not lost on comedian Lewis Black, who has some thoughts on the matter.

Happy Chanukah!

December 4, 2015 at 6:47 am Leave a comment


About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.

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