Your friend tosses three coins and you roll a single die. If the number on the die you roll is less than or equal to the number of heads that your friend tosses, you win $X. If not, you lose $1. How large should X be in order for this to be a fair game?

I'm not sure how this is calculus... it sounds more like plain probability. So you can figure that there are 2^3*6=48 total possibilities for the outcomes of your tosses (6 for the die and 2 for each coin which are then multiplied together). If your friend only tosses 1 or no heads at all, you lose no matter what you roll. If your friend gets 2 heads, there are 3 ways in which he/she can get it (HHT, HTH, THH) and you have to roll a one. Thus there are 3 ways here in which you can win. If your friends flips all three as heads, there are 2 ways you can win (rolling a 1 or a 2) and thus there is a total of 5 ways for you to win. Now, there are also 48-5=43 ways for you to lose, so the ideal ratio for wins loses (because of the odds) is 43:5 = 8.6:1. Therefore X should be 8.6.

To determine the value of X for this game to be fair, we need to calculate the probabilities of winning and losing.

There are 3 coins, so each coin has 2 possible outcomes (heads or tails). This means there are a total of 2^3 = 8 possible outcomes for the coin tosses.

Let's first calculate the probability of getting zero heads:

Since each coin toss has 2 possible outcomes, the probability of getting tails on a single toss is 1/2. Therefore, the probability of getting tails on all three coin tosses is (1/2) * (1/2) * (1/2) = 1/8.

Now, let's calculate the probabilities of getting one, two, or three heads:

Getting exactly one head can occur in three different ways: HTT, THT, or TTH. Each of these outcomes has a probability of (1/2) * (1/2) * (1/2) = 1/8. Since there are three possible outcomes, the probability of getting exactly one head is 3 * (1/8) = 3/8.

Getting exactly two heads can occur in three different ways: HHT, HTH, or THH. Each of these outcomes has a probability of (1/2) * (1/2) * (1/2) = 1/8. Since there are three possible outcomes, the probability of getting exactly two heads is 3 * (1/8) = 3/8.

Getting exactly three heads can only occur in one way: HHH. This outcome has a probability of (1/2) * (1/2) * (1/2) = 1/8.

Now that we have calculated the probabilities, let's determine the expected value of the game.

If the number on the die is 1, the probability of winning is 1/8 (because there is only one outcome with zero heads).

If the number on the die is 2, the probability of winning is 3/8 (because there are three outcomes with exactly one head).

If the number on the die is 3, the probability of winning is 3/8 (because there are three outcomes with exactly two heads).

If the number on the die is 4, 5, or 6, the probability of winning is 1/8 (because there is only one outcome with three heads).

Therefore, the expected value of the game can be calculated as:

(1/8) * X + (3/8) * X + (3/8) * X + (1/8) * (-1) = 0.

Simplifying this equation:

(8/8) * X + (8/8) * X + (8/8) * X + (1/8) * (-1) = 0.

8X + 8X + 8X - 1/8 = 0.

24X - 1/8 = 0.

24X = 1/8.

X = (1/8) / 24.

Simplifying this further, X = 1/192.

Therefore, for this game to be fair, X should be equal to 1/192 (approximately $0.0052).