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.


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.


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.