Expectation values

The second question. I appreciate any help that anyone can provide.

Spin the dreidel is a gambling game which uses a 4-sided top, each side has an equal chance of being up when the top topples.

Side 1 The player wins the whole pot
Side 2 The player wins half the pot
Side 3 Nothing happens the next player gets the top
Side 4 The player puts a coin in the pot

Whenever the pot is emptied every player puts a coin in the pot.

Assume there are six players, who have each just put a coin in the pot.

Part A What is the expectation value for the player that spins first?

Part B What is the expectation value for the player that spins second?

The following doesn't format correctly when saved. f(x) is over 0.25, and x is over 6


f(x) x
Side 1 0.25 6
Side 2 0.25 3
Side 3 0.25 0
Side 4 0.25 -1

The following doesn't format correctly when saved. f(x) is over 0.25, and then over .0625, while x and x*f(x) are for the answers

f(x) f(x) x x*f(x)
Side 1 0.25 Side 1 0.0625
Side 2 0.0625
Side 3 0.0625
Side 4 0.0625
Side 2 0.25 Side 1 0.0625
Side 2 0.0625
Side 3 0.0625
Side 4 0.0625
Side 3 0.25 Side 1 0.0625
Side 2 0.0625
Side 3 0.0625
Side 4 0.0625
Side 4 0.25 Side 1 0.0625
Side 2 0.0625
Side 3 0.0625
Side 4 0.0625



E(X) 0

Are the tables given?

In general, an expectation value is just an "average" over all possible outcomes. For instance, if we flip a coin and I pay you $1 for heads, you pay me $1 for tails, after a lot of flips I'd expect to end up basically where I started. After an insanely large number of flips, I'd almost certainly expect to end up basically where I started, relative to the number of flips (say, it might happen that after 1 billion flips, I was $100 ahead; that's not large compared to the 1 billion times it took to get there). In a specific, technical sense, the expected (expectation) value is the value almost certainly obtained after many independent repetitions of a situation. With coin flips, the expectation value is 0.5*$1 + 0.5*(-$1) = $0, where the 0.5's are the probabilities of each outcome.

In the first dreidel example, 1/4th the time you'd expect to win 6 coins, 1/4th the time you'd expect to win 3 coins, 1/4th the time you'd expect to gain (and lose) nothing, and 1/4th the time you'd expect to lose one coin. So, you'd expect to get 1/4*6 + 1/4*3 + 1/4*0 + 1/4*-1 = 8/4 = 2 coins. That is, the expected value of the first player is 2 coins.

I won't do the second one for you. (I only did the first one because it's actually a very simple example.)