If you play roulette, are you going to break-even? Maybe sometimes, but not if you play forever, because the expected value is negative. The expected value finds the average outcome of random events.

A square with a unit length has two arbitrary points chosen inside it. What is the average distance between the two points?

If this can be represented in the form \(\dfrac {a + \sqrt{b} + c \ln (d+\sqrt{e})}{f}\), where \(a, b, c, d, e, f\) are positive integers, and the fraction is simplified as far as possible, find \(a+b+c+d+e+f\).

There are 3 kegs of water. The first keg has 1 liter, the second keg has 2 liters, and the third keg has 3 liters of water in it.

Each day our traveler picks one keg at random with each keg having the same chance of being picked, and drinks one liter of water from it. When he empties a keg, he throws it away. When he is left with one keg, he records the amount of water in it.

What is the expected amount of water (in liters) remaining in that last keg?

If this expected value can be expressed as \( \dfrac ab\), where \(a\) and \(b\) are coprime positive integers, enter \(a+b\).

\(\)

**Note:** You may use a calculator if you want.

**Bonus:** Solve this problem for 5 kegs.

A row of street parking has spaces numbered 1 through 100, in that order. Each person trying to park drives by the spaces one-by-one starting at number 1.

At each open space, they are told how many open spaces, \(S,\) they haven't passed (including the current space). They park in that space with probability \(\frac{1}{S};\) otherwise, they move on to the next open space.

What is the expected value for the parking spot of the \(100^\text{th}\) person to arrive (assuming no one has left and only one car goes through the parking process at a time)?

**How long does it take you to roll a Yahtzee?**

The dice game Yahtzee is a classic. Each player takes turns rolling 5 dice, 3 times each. After each roll, the player can set aside some dice that they want to "keep" and reroll the rest in order to get a better score. The best roll that someone can have is called a "Yahtzee," or 5 dice of the same number, e.g. five 1s, five 2s, etc.

It's easy to determine the probabilities of getting a Yahtzee in 1 roll. You have 5 dice, and a dice has 6 sides, so the chance of getting all 5 the same is \(\left(\frac{1}{6}\right)^5\), but since there are 6 ways to get a Yahtzee, the probability is really \(\left(\frac{1}{6}\right)^4 = \frac{1}{1296}\). From this, we can determine that one should expect to roll 5 dice 1296 times before getting a Yahtzee, right?

If you've ever played, you'll realize that the chances of getting a Yahtzee in a game are higher than that. This is because of the rule that you can keep some dice aside after each roll. If after each roll you decide to keep the dice aside that have the number that shows up the most, given an infinite number of rolls (not limited to 3 rolls as in real games), what is the expected number of turns it should take you to get a Yahtzee?

Round your answer to the nearest whole number.

**Example sequence of rolls:**

Roll 1: 1, 2, 4, 4, 5. Keep the two 4s.

Roll 2: 4, 4, 6, 3, 3. Keep the two 4s.

Roll 3: 4, 4, 1, 1, 1. Keep the three 1s.

Roll 4: 1, 1, 1, 5, 1. Keep the four 1s.

Roll 5: 1, 1, 1, 1, 1. Done after 5 rolls.

Suppose the Cookie Monster has two (distinguishable) jars of cookies, each of which initially contains \(20\) cookies. The Monster then starts to eat the cookies one at a time such that each time he takes a cookie from a jar chosen uniformly at random.

Now the Cookie Monster never looks into the jars, so he never knows how many cookies are left in either jar until he reaches in to one and finds that that jar is empty. When he first discovers that one of the jars is in fact empty, the probability that there are (strictly) fewer than \(3\) cookies in the other jar is \(S.\)

Find \(\lfloor 1000S \rfloor\).

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