The Die Game

You pick a number between 1 and 6 and keep throwing a die until you get it. Does it matter which number you pick for maximizing the total sum of the numbers in the resulting sequence?

In the example below, the picked number is 6 and the total sum of the numbers in the resulting sequence is 35.

No matter what number you pick, the expected value of each throw is the average of the numbers from 1 to 6 which is 3.5. The choice of the number also does not affect the odds for the number of throws until the game ends, which is 6. Therefore, the total sum is always 3.5 × 6 = 21 on average, regardless of the chosen number.

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Population

In certain society all parents stop having children right after they get their first boy. After 1000 years, approximately what will be the percentage of the women in the society?

The answer is 50%. The reason is that no matter what birth plan the society comes up with, the chance for having a boy or a girl during every birth is 50/50.

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The Monty Hall Show

You are in Monty Hall’s TV show where in the final round the host gives you the option to open one of three boxes and to receive the reward inside. Two of the boxes contain just a penny, while the third box contains $1.000.000. In order to make the game more exciting, after you pick your choice, the rules require the host to open one of the two remaining boxes, such that it contains a penny inside. After that he asks you whether you want to keep your chosen box or to switch it with the third remaining one. What should you do? This is the so called “Monty Hall” problem. The answer is that in order to maximize your chances of winning$1.000.000, you should switch your box. The reason is that if initially you picked a box with a penny, then after switching you will get a box with $1.000.000. If initially you picked a box with$1.000.000, then after switching you will get a box with a penny. Since in the beginning the chance to get a penny is 2/3, then after switching your chance to get $1.000.000 is also 2/3. If you stay with your current box, then your chance to get$1.000.000 will be just 1/3.

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Gun Duel

Mick, Nick, and Rick arrange a three-person gun duel. Mick hits his target 1 out of every 3 times, Nick hits his target 2 out of every 3 times, and Rick hits his target every time. If the three are taking turns shooting at each other, with Mick starting first and Nick second, what should be Mick’s strategy?

Clearly, Mick should not aim for Nick, because if he kills him, then he will be killed by Rick. Similarly, Nick should not aim for Mick, because if he kills him, then he also will be killed by Rick. Therefore, if Rick ends up against alive Mick and Nick, he will aim at Nick, because he would prefer to face off a weaker opponent afterward. This means that if Nick is alive after Mick shoots, he will shoot at Rick.

Thus, if Mick shoots at Rick and kills him, then he will have to face off Nick with chance of survival less than 1/3. Instead, if he decides to shoot in the air, then he will face off Nick or Rick with chance of survival at least 1/3. Therefore, Mick’s strategy is to keep shooting in the air, until he ends up alone against one of his opponents.

Envelopes with Numbers

You are given 2 sealed envelopes with numbers inside. You are told that one of the numbers is twice as much as the other one. You grab one of the envelopes and right before you open it, you make the following calculation:

“If this envelope contains X inside, then the other envelope contains either X/2 or 2X inside. Since the chance that the other envelope contains a larger number is exactly 50%, the expected money I will get after switching is X/4 + X = 1.25X > X. Therefore, I should switch!”

Clearly, this reasoning is wrong, since you can’t possibly deduce which envelope of the two contains a larger number. Where is the mistake?

The trick is that conditionally on the fact that your envelope contains X, it is not true that the other envelope has 50% chance of containing either X/2 or 2X. The reason is that it is impossible that all amounts of dollars appear in the envelopes with the same probabilities (densities). Thus, for example, if it is very unlikely that an envelope contains more than 1000, and you open an envelope with 800 inside, you will not think that the other envelope has 50% chance of containing 1600.

Fish in a Pond

There are 5 fish in a pond. What is the probability that you can split the pond into 2 halves using a diameter, so that all fish end up in one half?

Let us generalize the problem to N fish in a pond. We can assume that all fish are on the boundary of the pond, which is a circle, and we need to find the probability that all of them are contained within a semi-circle.

For every fish Fᵢ, consider the semi-circle Cᵢ whose left end-point is at Fᵢ. The probability that all fish belong to Cᵢ is equal to 1/2ᴺ⁻¹. Since it is impossible to have 2 fish Fᵢ and Fⱼ, such that the semi-sircles Cᵢ and Cⱼ contain all fish, we see that the probability that all fish belong to Cᵢ for some i is equal to N/2ᴺ⁻¹.

When N = 5, we get that the answer is 5/16.

Moms’ Talk

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.

Explanation:

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.

Discuss this puzzle in the forum.

Royal Wedding

A prince decides to get married to the prettiest girl in his kingdom. All 100 available ladies go to the palace and show themselves to the prince one by one. He can either decide to marry the girl in front of him or ask her to leave forever and call the next one in line. Can you find a strategy which will give the prince a chance of 25% to get married to the prettiest girl? Can you find the best strategy?

Remark: Assume that the prince can objectively compare every two girls he has seen.

A strategy which ensures a chance of 25% is the following:
The prince banishes the first 50 girls which enter the palace and then gets married to the first one which is prettier than all of them (if such one arrives). If the prettiest girl in the kingdom is in the second 50, and the second prettiest girl is in the first 50, he will succeed. The chance for this is exactly 25%.

The best strategy is to wait until ~1/e of all girls pass, and then choose the first one which is more beautiful than all of them. This yields a chance of ~37% for succeeding. The proof is coming soon.

Islands and Bridges

You need to cross a river, from the north shore to the south shore, via a series of 13 bridges and six islands, which you can see in the diagram below. However, as you approach the water, a hurricane passes and destroys some (possibly none/all) of the bridges. If the probability that each bridge gets destroyed is 50%, independently of the others, what is the chance that you will be able to cross the river after all?

Imagine there is a captain on a ship, who wants to sail through the river from West to East. You can see that he will be able to do this if and only if you are not able to cross the river. However, if you rotate the diagram by 90 degrees, you can also see that the probability that you cross North-South is equal to the probability that he sails West-East, and therefore both probabilities are equal to 50%.

Lost Boarding Pass

There are 100 passengers which are trying to get on a plane. The first passenger, however, has lost his boarding pass, so decides to sit on an arbitrary seat. Each successive passenger either sits on his own seat if it is empty or on an arbitrary other if it is taken. What is the chance that the last person will sit in his own seat?

The chance is 50%. Indeed, the last passenger will either have to sit in his own seat or the one which belongs to the first passenger. Since there hasn’t been a preference made by anyone at any time towards any of these two seats, the probability that either of them is left last is 1/2.