Saturday, July 28, 2012

GeekDad Puzzle of the Week: Contiguous Consonants

breastroke, matchstick, corkscrew, postscriptGeekDad Puzzle of the Week: Contiguous Consonants:
Q: [GeekDad had] a lot of time to work out several phrases that incorporated words with multiple adjacent non-vowels or "contiguous consonants." For purposes of this puzzle, please consider the letter "y" strictly as a consonant. In parentheses, after each phrase is the number of words, and each word's count of contiguous consonants.
  • Ice-free, super tall buildings in Scranton (3w/6c)
  • Encoding long words in a fixed orbit (3w/6c)
  • Crazy fish-studier’s two wheeled transport (3w/5c)
  • Melodic equivalents to the "Queen of Diamonds" (Condon)(3w/6c)
  • Sufficiently valuable magic during the America’s Cup (3w/5c)
  • Where playing Beethoven on your iPhone was invented (3w/5c)
  • Artificial disk-flip game (2w/5c)
  • Rotational energy "battery," 10-10 meters across (2w/5c)
For those that have struggled with the math problems, this might be more up your alley. The hardest one, in my opinion, is the 4th one; I'm not completely happy with my answer. Which ones do you find tricky? Remember don't give anything away since this is a contest with a prize. Feel free to read the full puzzle details on the GeekDad site and submit your answers by Friday for a chance at the $50 prize.

Edit: The deadline has passed and I've posted our answers in the comments. I'm still waiting to see the intended answers, especially for #4.

Thursday, July 26, 2012

NPR Sunday Puzzle (Jul 22, 2012): Sports Section

Sports ballsNPR Sunday Puzzle (Jul 22, 2012): Sports Section:
Q: Name a sport in two words — nine letters in the first word, six letters in the last — in which all six vowels (A, E, I, O, U, and Y) are used once each. What is it?
Grr!!! I spent a good amount of time trying to find a 9-5 answer before finding out that there was a mistake in the posted puzzle... I bet a few of you were similarly confused. I was working on "organized rugby" as a potential answer until I saw the correction.

Edit: The hints were "Grr" which are the starting consonants of the answer, "9-5" as in odds of 9 to 5, "posted" as in race results being posted and "bet" as what you might do at one of these races.

Saturday, July 21, 2012

GeekDad Puzzle of the Week: Dog Siblings

Black Labs, mrpattersonsir@flickrGeekDad Puzzle of the Week: Dog Siblings:
Q: A guy with a black lab said that his dog, Selkie, has five brothers and sisters in town. "But I’ve never run into one of them," he said. "I wonder what are the chances of that?"

Imagine that each of the six dogs goes out somewhere an average of once every three days. And imagine that between trails and parks and fields there are 200 places a dog can go, all (let's say...) with equal probability.

If it's been exactly two years — 730 days — since Selkie's owner picked her up from the litter, what are the chances that during this time Selkie would NOT see a doggie sibling?
For extra credit, what are the chances over the same time that any sibling will meet any other sibling?
I've got my answers which I will reveal after the deadline. In the meantime, feel free to solve it and submit your answer to GeekDad.

Edit:  The deadline was Friday, so here is how I went about solving the puzzle.

The key to this puzzle is figure out the chances of dogs not meeting on one day.  From there it is easy to figure out the chance of them not meeting for 730 days. And then if necessary, you can figure out the probability of the opposite case (meeting) by subtracting from 100%.

Part 1 - Selkie doesn't meet a doggie sibling

In order for a dog to be at a specific location, they must be out (with 1/3 probability) and at that specific spot (1/200 probability). That means there is a 1/600 chance of a specific dog being out at a specific location.  Thinking of the negative probability, that means there is a 599/600 chance that a dog is *not* at a specific location.

Selkie will *not* meet another dog on a specific day if,
1) Selkie is at home (2/3)
2) Selkie is out at any location (1/3) and dog 1 is not there (599/600) and dog 2 is not there (599/600) and dog 3 is not there...

In other words, the chance that Selkie doesn't meet any other dog on a specific day is:
2/3 + 1/3 x (599/600)^5 ≈ 99.7231466062%

And the chance that Selkie doesn't meet any dogs for 2 years (730 days) is:
[ 2/3 + 1/3 x (599/600)^5 ]^730 ≈ 13.2148023616%

A: (Part 1) The chance that Selkie does NOT see a doggie sibling is around 13.2148%

Part 2 - Chances that any sibling meets another sibling

This is the much tougher question.  First let's make a table of probabilities of having 0 dogs out, 1 dog out, 2 dogs out, etc.

The chance that 'n' dogs are out on any one day is C(6,n) x (1/3)^n x (2/3)^(6-n).
And given 'n' dogs are out, the chance that they do NOT meet is 200/200 x 199/200 x 198/200 ... (using the number of dogs that are out.)

Thus, the chance that no dogs meet on a single day is around 99.1717389885%
The chance that no dogs meet for 730 days is (99.1717389885%)^730 ≈ 0.2307745729%
Subtracting from 100%, you get the probability that at least two siblings will meet (100% - 0.230774573%) = 99.769225427%

A: (Part 2) The chance that any of the siblings meet during those 2 years is around 99.7692%

Thursday, July 19, 2012

NPR Sunday Puzzle (Jul 15, 2012): Is There a Doctor in the House?

DoctorsNPR Sunday Puzzle (Jul 15, 2012): Is There a Doctor in the House?:
Q: The name of something that you might see your doctor about is a two-word phrase. Three letters in each word. When these six letters are written without a space, a three-letter word can be removed from inside, and the remaining three letters in order also form a word. What's interesting is that the four three-letter words — the two in the original phrase, the one that was removed, and the one that remains — all rhyme. What is the original phrase?
The picture this time is from Halloween 2005 when we all went as various doctors. Speaking of doctors, is there ever one in the house?

Edit: Perhaps too obviously, a moving, emotional theatrical performance can result in there not being a dry eye in the house.

Wednesday, July 18, 2012

GeekDad Puzzle of the Week: Math Trolls

Troll Bridge SignGeekDad Puzzle of the Week: Math Trolls:
Q: Nora is taking a trip to visit her Grandmother in northernmost New York State this week, to bring her some freshly picked berries. On the way there, she has to cross a total of 30 bridges, and under each of the these bridges lives a troll. Each troll is aware of their bridge number, and either demands or gives berries based upon the rarest or most applicable description of their bridge. They demand or give berries according to the following schedule:
  • Trolls under odd numbered bridges demand half of your berries.
  • Trolls under even numbered bridges demand 20 berries.
  • Trolls under prime numbered bridges give you half again the number of berries you are carrying.
  • Trolls under perfect square numbered bridges demand a quarter of your berries.
  • Trolls under perfect cube numbered bridges give you the number of berries you are carrying, doubling your number of berries.
If trolls round up in their demands (i.e., if you have 57 berries at the foot of a bridge best described as odd numbered, you will cross it with 28 berries), what is the minimum number of berries Nora must start with so that she ends up with 1,000 berries when she arrives at her Grandmother’s house?
The solution has already been posted on the GeekDad website. If you want to solve it yourself, read no further. A detailed breakdown of my solution is given in the comments.

Thursday, July 12, 2012

NPR Sunday Puzzle (Jul 8, 2012): And the Oscar Goes To...

Oscar statuettesNPR Sunday Puzzle (Jul 8, 2012): And the Oscar Goes To...:
Q: Think of a well-known actor, three letters in the first name, seven letters in the last. One of the letters is an "S." Change the "S" to a "K" and rearrange the result, and you'll name a well-known fictional character. Who is it?
Add an "F" to the fictional character, rearrange to get part of a car.

Edit: TINKER BELL + F rearranges to LEFT BLINKER.

Thursday, July 05, 2012

NPR Sunday Puzzle (Jul 1, 2012): Retail Therapy

UPC BarcodeNPR Sunday Puzzle (Jul 1, 2012): Retail Therapy:
Q: Think of a well-known retail store chain in two words. Remove one letter from its name. The remaining letters, in order, will spell three consecutive words that are synonyms of each other. What are they? Hint: The three words are all slang.
I didn't immediately get the answer because they've been getting rid of most of these stores in our area. And the answer I have is technically one word, not two.

Edit: The hint was "getting rid of" which could be another slang synonym.
A: OfficeMax (remove the M) = Off, Ice, Ax as synonyms for "kill"

Wednesday, July 04, 2012

GeekDad Puzzle of the Week: Poaching Berries

Red Raspberries, photofarmer@FlickrGeekDad Puzzle of the Week: Poaching Berries:
Q: Leif and Kestrel are willing to give GeekDad a 20% cut of berries they poach for turning a blind eye. Imagine that Leif picks five berries per poach and Kestrel picks three berries per poach and that they attempt to poach once every day, with the exception of any day just after they’ve been caught. Now imagine that each time they poach berries, they have a 15% chance of getting caught. How many berries can GeekDad expect to eat each each week, averaged over time?
I'll post my thoughts on the answer after the answer is revealed (generally early next week).

Edit: The following answer was what I submitted to end up winning the puzzle:

Let's assume a few things:
1) Leif and Kestrel start in a state where they haven't ever been caught.
2) Each time they are caught, the berries that day don't count. Also, they must skip the next day until trying again.

There are essentially 3 states that each child could be in. Since the probabilities for each child are the same we can group them together, though in reality the puzzle allows for one child to be caught and the other not. The results work out the same, so let's just simplify it to a pair of children trying to poach 8 berries a day.

The children could be:
1) Caught the day before and therefore have no chance of getting berries that day (must sit out).
2) Caught today (15%, if not caught the day before)
3) Not caught today (85%, if not caught the day before)

Each day, the chance of being caught the day before is just carried forward. So on day 2, the chance is 15% they are sitting out. That leaves 85% chance they are attempting poaching. Of that 85%, there's a 15% chance they are caught (0.85 x 0.15 = 12.75%) and an 85% chance they poach successfully (0.85 x 0.85 = 72.25%).

If you repeat this, the next day (Day 3) they have a 12.75% chance of sitting out and a 87.25% chance of attempting poaching. Day 4, the chance of sitting out is 0.15 x 87.25% = 13.0875% and not sitting out is 86.9125%. Day 5, the chance of sitting out is 0.15 x 86.9125% = 13.036875% and not sitting out is 86.963125%. You can continue this progression and you will see that the chance they are sitting out approaches a value of about 13.043478%. The chance they are caught that day is equivalently about 13.043478%. That leaves a 73.913043% chance they are able to poach 8 berries with an expected return of 5.913043478 berries a day.

That equates to approximately 41.39130435 berries a week. With a 20% "commission" after awhile you will be eating approximately 8.27826087 berries each week.

Note: For a more accurate answer (rather than just a decimal approximation) we can solve this algebraically as follows:

Let p be the chance that you ARE sitting out.
Let q be the chance you are NOT sitting out.

Together these are mutually exclusive and therefore add up to 100% (or mathematically we say 1)
p + q = 1
p = 1 - q

The chance you are NOT sitting out, but CAUGHT is 0.15q
The chance you are NOT sitting out, and SUCCESSFULLY POACHED is 0.85q

We know that eventually the two values p and 0.15q end up being the same, so equate them
p = 0.15q

Substitute in 1-q:
1 - q = 0.15q

1 = 1.15q
q = 1/1.15

The chance we are caught that day is 0.15q
0.15q = 0.15(1/1.15) = 0.15/1.15 = 15/115 = 3/23
And the chance we poach some berries is 0.85q
0.85q = 0.85(1/1.15) = 0.85/1.15 = 85/115 = 17/23

For the sake of completeness that means you have:
3/23 = chance child is sitting out
3/23 = chance child was caught today
17/23 = chance child was able to poach successfully

Now multiply this last number by 8 berries attempted times 7 days and then times 1/5 (20% commission) to get the expected number of berries poached each week.
Berries per week = 17/23 x 8 x 7 x 1/5

If you reduce that to a fraction you end up with:
952/115 = 8 32/115 berries each week.

A: 8 32/115 berries each week.
8.27826086956521739130434... (underlined portion repeats indefinitely)