Author: Puzzle Prime

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Seeing Theory is a beautifully designed website, which aims to educate people about probability theory via series of visual and interactive lessons. If anyone is struggling to grasp some of the basic concepts in this field of mathematics or is just getting into it, the website can be a very useful learning tool. Seeing Theory was designed by Daniel Kunin as an undergraduate project in Brown University and has won numerous awards. To visit the website, click the banner below.


Two moms, Sarah and Courtney, are talking to each other.

Sarah: I have two children.
What is the probability that both of Sarah’s children are boys?

Courtney: Me too! Do you have any boys?
What is the probability that both of Courtney’s children are boys?

Sarah: Yes, I do! What is your younger child?
What is the probability that both of Sarah’s children are boys?

Courtney: It is a boy. He is so mischievous!
What is the probability that both of Courtney’s children are boys?

Sarah: Is he Sagittarius? Sagittarius boys are known to drive their mothers crazy. I can testify from personal experience.
What is the probability that both of Sarah’s children are boys?

Courtney: No, but actually I have the opposite personal experience to yours.
What is the probability that both of Courtney’s children are boys?

Sarah: Well, I guess astrology does not always get it right.

Courtney: I assume it does about half of the time.

The answers are: ~1/4, ~1/4, ~1/3, ~1/2, ~23/47, 1.


Initially, we do not have any information about the children and therefore the chance that both of them boys is 1/2 × 1/2. This applies to the first and the second question.

After Sarah says that she has at least one boy, there are equal possibilities that she has Boy + Boy, Boy + Girl, or Girl + Boy. Therefore, the chance that both children are boys is 1/3.

After Courtney says that her younger child is a boy, the only remaining question is what is the gender of her older child, and therefore the chance is 1/2.

The fifth exchange implies that Sarah has a Sagittarius boy. There are 23 combinations such that both children are boys and at least one of them is Sagittarius. There are 47 combinations such that at least one of the children is a Sagittarius boy. Therefore, the chance that both children are boys is 23/47.

Finally, Courtney says that her younger child, which is a boy, is not Sagittarius, but her personal experience with Sagittarius boys is positive. Therefore, her older child is a Sagittarius boy and the chance is 1.


Creative expression, learning, and focusing are some of the most important activities which children should be encouraged to do from an early age. Driven by this idea, Elizabeth Carpenter has published several oversized books, which give kids the opportunity to solve beautifully drawn line mazes, color them, and learn interesting trivia all at once. The Mummy Mazes Monumental Book contains 28 poster-size mazes based on Ancient Egypt themes, along with explanations about each of the included objects. The Dino Mazes Colossal Fossil Book contains 31 poster-size mazes, depicting various dinosaurs, accompanied by descriptions and quick facts about them. Recently, Elizabeth also published a Mandala Mazes book, which can be popular among older people looking for fun, relaxing activities as well. In terms of difficulty, the Dino and the Mummy mazes seem to fall on the easier side, while the Mandala mazes are a bit more challenging. After being completed, the mazes can be detached and used as posters, even though we think they look best organized together. All three books offer great quality and we would highly recommend them to any maze enthusiast.

  • about 30 over-sized mazes per book
  • beautiful line mazes, suitable for coloring
  • books include interesting trivia
  • highly recommended

by Elizabeth Carpenter

Alex and Bob are playing a game. They are taking turns drawing arrows over the segments of an infinite grid. Alex wins if he manages to create a closed loop, Bob wins if Alex does not win within the first 1000 moves. Who has a winning strategy if:

a) Alex starts first (easy)
b) Bob starts first (hard)

Remark: The loop can include arrows drawn both by Alex and Bob.

In both cases, Bob wins. An easy strategy for part a) is the following:

Every time Alex draws an arrow, Bob draws an arrow in such a way, that the two arrows form an L-shaped piece and either point towards or away from each other. Since every closed loop must contain a bottom left corner, Alex cannot win.

For part b), Bob should use a modification of his strategy in part a). First, he draws a horizontal arrow. Then, he splits the remaining edges into pairs, as shown on the image below. If Alex draws one arrow on the grid, then Bob draws its paired arrow, such that the two arrows point either towards or away from each other. The only place where a loop can have a bottom left corner is where Bob drew the first arrow. However, if a loop has a bottom left corner in this positio, then it must have at least one more bottom left corner, which is impossible. 


Bob and Jane are taking turns, placing knights and coins respectively on a chessboard. If Bob is allowed to place a knight only on an empty square which is not attacked by another knight, how many pieces at most can he place before running out of moves? Assume that Jane starts second and plays optimally, trying to prevent Bob from placing knights on the board.

Bob can place at most 16 knights. One way to do this is to keep placing knights only on the 32 white squares. In order to see that Jane can prevent Bob from placing more than 16 knights, split the board in four 4×4 grids. Then, group the squares in each grid in pairs, as shown on the image below. If Bob places a knight on any square, then Jane will place a coin on its paired square. This way Bob can place at most one knight on each of the four red squares, one knight on each of the four green squares, one knight on each of the four brown squares, and one knight on each of the four blue squares. Therefore, he can not place more than 64/4 = 16 knights on the board.


For our July giveaway we are partnering with our friends at, a great website for games, toys, and fun Halloween costumes. We have prepared for you a colorful picture with lots of hidden stars in it. Answer correctly what their number is and on 15th of July you could be the lucky winner who gets more than $100 worth of puzzles and games, including “Perplexus”, “Codenames”, “Manifold”, and many more!