Annual Puzzle and Video for Christmas 2024!

It's time for our Annual Puzzle and Video. Click the tree below to get a printable PDF of the full puzzle.

Once you have figured out the password, click the video link below to see all the fun we've had in 2024!

As always, post a comment below about your favorite part of the video to show you've solved the puzzle, but don't give away the password directly.

P.S. If you need help, here is the answer key with the password.

Annual Puzzle and Video for 2023.

Happy New Year! In Our Annual Family Video we star in our own comic book adventure as we travel back to the beginning of 2023. The password can be found by solving our annual puzzle. Click this link or the image below to get a printable PDF of the full puzzle.

Post a comment below about your favorite part of the video to show you've solved the puzzle, but don't give away the password directly.

Our Annual Puzzle and Video for 2022

Our Annual Family Video for 2022 is available now. The password can be found by solving our annual puzzle. Click this link or the image below to get a PDF of the full puzzle.

Post a comment below about your favorite part of the video to show you've solved the puzzle, but don't give away the password directly.

Our Family Puzzle and Video for 2021!

Our Family Video for 2021 is available now. The password can be found by solving the puzzle below:

Feel free to add comments about the video to show you solved the puzzle, but don't give away the password to others.

Annual Video and Puzzle for 2019

Our annual video and puzzle is available now. Simply solve the puzzle to get the password. Click on this link to the annual video and type the password to see what we have been up to in 2019.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

For those that might still be struggling, there is an answer key, but please only click on that as a last resort.

NPR Sunday Puzzle (Sep 1, 2019): You Have Less than One Week to Escape

NPR Sunday Puzzle (Sep 1, 2019): You Have Less than One Week to Escape:

Q: This is a two-week challenge. It may sound impossible, but it's not. You wake up trapped in a round room with six doors. A voice over a loudspeaker tells you that five of the doors are booby-trapped and will bring instant death if you try to open them. Only one door provides an opening that will get you out safely. The doors are evenly spaced around the room. They look exactly alike. Your only clue is that on the wall between each pair of doors is a large letter of the alphabet. Going clockwise, the letters are H, I, J, K, L and M. Which is the correct door that will get you out ... and why?

Shh! Don't give the answer away before the Wednesday 3pm ET deadline.

Edit: The room is round, so the spot where the letters wrap must be important otherwise it could have been a single wall with several doors. Also, the letters are *between* the doors. So this points to the exit door being being between M and H, but why? Reread the question and focus on opening and out; they seem to be key.

A: When marked with "OUT" the door between M and H forms a word for an opening — "MOUTH".

This is my theory as to the intended answer, but I'll have to wait to Sunday to have it officially confirmed.

NPR Sunday Puzzle (Aug 25, 2019): You Have Two Weeks to Escape

NPR Sunday Puzzle (Aug 25, 2019): You Have Two Weeks to Escape:

Q: This is a two-week challenge. It may sound impossible, but it's not. You wake up trapped in a round room with six doors. A voice over a loudspeaker tells you that five of the doors are booby-trapped and will bring instant death if you try to open them. Only one door provides an opening that will get you out safely. The doors are evenly spaced around the room. They look exactly alike. Your only clue is that on the wall between each pair of doors is a large letter of the alphabet. Going clockwise, the letters are H, I, J, K, L and M. Which is the correct door that will get you out ... and why?

Read carefully. Don't choose a booby trapped door or you can kiss your *** goodbye!

Christmas Puzzle for 2018

Our annual Christmas puzzle is available now. Follow the simple directions, eliminate those words and see what's left...

As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by solving the puzzle.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

If you need help with the puzzle or if you missed the reference in our Christmas family photo, here is the answer to the puzzle. But try to solve it yourself before looking there.

Christmas Puzzle for 2017

Our annual Christmas puzzle is available now.
As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by solving the puzzle first.

Note: If you need some help, the full answer is posted here, but try solving it without help first... it's more fun that way.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

Christmas Snowflake Matching Puzzle for 2016

Our annual Christmas puzzle is available now.
As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by solving the puzzle first.

Note: If you need some help, the full answer is posted here, but try solving it without help first... it's more fun that way.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

Christmas Puzzle for 2015

Our annual Christmas puzzle is available now.
As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by solving the puzzle first.

Note: If you need some help, the full answer is posted here, but try solving it without help first... it's more fun that way.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

Christmas Puzzle 2014 - Power of TEN

Our annual Christmas puzzle "Power of TEN" is available now. Every answer is a word containing the letters TEN. For example, if the clue were "These keep your hands warm when playing in the snow", the answer would be mittens.

As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by using the first letter of each word. As a hint, the password could be used to describe an occasion such as the new year.

Note: If you need some help, the full answer is posted here, but try solving it without help first... it's more fun that way.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

Christmas Puzzle 2012 - Snowflake Maze

Our annual Christmas puzzle is available now. It's a fun maze in the shape of a snowflake. As in prior years, the reward for solving is a video Christmas card, but you'll need to figure out the password by solving the puzzle.

Note: If you need some help, the full answer is posted here, but try solving it without help first... it's more fun that way.

Feel free to add a comment below to let us know that you successfully figured it out (without giving away the answer to others). We are always looking for new ideas for next year's Christmas puzzle, so submit those too.

GeekDad Puzzle of the Week: Contiguous Consonants

GeekDad 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.

GeekDad Puzzle of the Week: Dog Siblings

GeekDad 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%

GeekDad Puzzle of the Week: Math Trolls

GeekDad 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.

GeekDad Puzzle of the Week: Poaching Berries

GeekDad 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

Rearrange:
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)

GeekDad Puzzle of the Week: When Are the Odds Even?

GeekDad Puzzle of the Week: When Are the Odds Even?:

Q: If we have a bag containing equal numbers of black and white marbles, and we pull out one marble, the odds of it being black are even. If we have a bag containing 120 marbles, 85 of which are black, the odds of us pulling out two marbles and them both being black is also even — (85/120)x(84/119) = 0.5 or 50%.

If the largest bag we have can hold 1,000,000 marbles, for how many sets of marbles (i.e., the 120 marbles described above are one set) can we pull two marbles and have a 50% chance of them being the same designated color? Are there any sets of marbles for which we can pull three marbles and have a 50% chance of them being the same designated color? If so, how many?

After the solution is revealed, I'll post the details of my answer.

Edit: GeekDad Puzzle Solution:
In the first case you are essentially looking for integer solutions to:
a(a-1) = 2b(b-1)

There are EIGHT sets under 1 million that will result in even odds when 2 balls are drawn.

4 marbles (3 black) --> 4 x 3 = 2(3 x 2)
21 marbles (15 black) --> 21 x 20 = 2(15 x 14)
120 marbles (85 black) --> 120 x 119 = 2(85 x 84)
697 marbles (493 black) --> 697 x 696 = 2(493 x 492)
4,060 marbles (2,871 black) --> 4,060 x 4,059 = 2(2,871 x 2,870)
23,661 marbles (16,731 black) --> 23,661 x 23,660 = 2(16,731 x 16,730)
137,904 marbles (97,513 black) --> 137,904 x 137,903 = 2(97,513 x 97,512)
803,761 marbles (568,345 black) --> 803,761 x 803,760 = 2(568,345 x 568,344)

Interestingly, the next number in each sequence can be computed as follows:
a(n) = 6a(n-1) - a(n-2) - 2

So for example, the next numbers in the sequence would be:
Total balls: 6 x 803,761 - 137,904 - 2 = 4,684,660 marbles
Black balls: 6 x 568,345 - 97,513 - 2 = 3,312,555 black

Integer sequences: A011900 and A046090

In the second case you are looking for integer solutions to:
a(a-1)(a-2) = 2b(b-1)(b-2)

There is only ONE set under 1 million that will result in even odds when 3 balls are drawn.

6 marbles (5 black) --> 6 x 5 x 4 = 2(5 x 4 x 3)

NPR Sunday Puzzle (Jun 12, 2011): Sam Loyd's Hat Rack Puzzle

NPR Sunday Puzzle (Jun 12, 2011): Sam Loyd's Hat Rack Puzzle:

This Hat Rack Puzzle by Sam Loyd was published 100 years ago in Woman's Home Companion:

Q: A hat room contains a wall with 49 pegs, arranged in a 7-by-7 square. The hat clerk has 20 hats that are to be hung on 20 different pegs. How many lines, containing four hats in a straight line, is it possible to produce? A line can go in any direction: horizontally, vertically or obliquely. To explain your answer, number the pegs in order, from 1 in the upper left corner to 49 in the lower right corner; list which pegs you put the 20 hats on, and give the total number of lines containing four hats in a row.

Liane has left, but it also seems like the NPR website editors are gone. Last week they had "goose" as a two word phrase (instead of "roast goose") and this week they misspelled Sam Loyd (as Sam Lloyd). Anyway, back to the puzzle; not counting rotations and reflections, I have 3 ways to get the answer.

Edit: If you re-read my post you'll see the phrase "are gone" at the end of the first sentence. This is a homophone of Argon with atomic number 18, a clue to there being 18 lines in the solution(s).

A: I found 3 main solutions (not counting reflections and rotations). Click each one to see a larger view with any rotated/reflected variants.

Write the Alphabet Backwards, Quicker than Forwards!

I was reminded recently of a way to impress your friends and perhaps win a bet too. It involves writing the alphabet backwards faster than they can forwards.

The key is to learn the backwards alphabet as "words" rather than individual letters. If you break it into chunks of 4 letters (with 2 left over) you have:

ZYXW VUTS RQPO NMLK JIHG FEDC BA
Phonetically think of this as the phrase:
"Zixwa Vuts Irqpo Nimlick Jig Fedic Bah"

Practice saying this as you write each set of letters one after the other. With a little practice you'll be able to write this very quickly.

Now you are ready to challenge your friends to a race. You can even bet them that you'll write the alphabet backwards faster than they write it forwards. The reason it works is you won't need to stop, think, sing that alphabet song, go back a few letters, etc. You simply write down your 7 "words" as quickly as possible and you are sure to beat them.

Enjoy!