# Possible Error in the First Example of 'Sum Of Expectations' (Week 2, Lesson 1 - Describing Distributions)

Hi,

There seems to be a mistake in the video explanation regarding the expected value of the game.

The video outlines the expected value (E(X)) as the sum of the expected value from the coin toss (X_coin) and the die roll (X_dice), resulting in \$4. However, this doesn’t account for the fact that the die roll only matters if you win the coin toss (get heads).

Here’s a breakdown of the correct approach:

1. Coin Flip:
• Expected value (X_coin) = (Probability of heads) * (Amount won on heads) + (Probability of tails) * (Amount won on tails)
• Expected value (X_coin) = (1/2) * (\$1) + (1/2) * (\$0) = \$0.50
1. Die Roll (only if Heads):
• Since the die roll only happens after winning the coin toss (heads), we need to consider the conditional probability.
• Expected value (X_dice) = (Probability of winning the coin toss) * (Average win from die roll)
• Expected value (X_dice) = (1/2) * ((1 + 2 + 3 + 4 + 5 + 6) / 6) = \$1.75
1. Overall Expected Value (E(X))
• This is the average of the winnings from both independent events (coin toss and die roll after heads).
• E(X) = X_coin + X_dice = \$0.50 + \$1.75 = \$2.25

Therefore, the expected amount you win per game (E(X)) is actually \$2.25, not \$4. The conditional nature of the die roll seems to be missed.

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