In a game you flip a coin twice, and record the number of heads that occur. You get 10 points for 2 heads, zero points for 1 head, and 5 points for no heads. What is the expected value for the number of points you’ll win per turn?
I'd have to say
10(1/4) + 0(1/2) + 5(1/4) = 3.75
To find the expected value, we need to determine the probability of each outcome (in this case, the number of heads) and multiply it by the corresponding number of points. Then, we sum up these products to get the overall expected value.
Let's break down the possibilities:
1. Two heads: The probability of getting heads on a coin flip is 1/2. Since we flip the coin twice, we multiply 1/2 by 1/2 to get 1/4. So, the probability of getting two heads is 1/4, and the corresponding points are 10.
2. One head: Again, the probability of getting heads on a coin flip is 1/2. Now we need to consider the arrangement of the two coin flips. There are two possibilities: heads followed by tails or tails followed by heads. Each of these combinations has a probability of 1/2 * 1/2 = 1/4. Therefore, the total probability of getting one head is 1/4 + 1/4 = 1/2, and the corresponding points are 0.
3. No heads: This time, we are considering the probability of getting tails on both coin flips. Just like before, each coin flip has a probability of 1/2, so the probability of getting tails on two flips is 1/2 * 1/2 = 1/4. Therefore, the probability of getting no heads is 1/4, and the corresponding points are 5.
Now, we can calculate the expected value by multiplying each outcome's probability by its respective points and summing them up:
(1/4 * 10) + (1/2 * 0) + (1/4 * 5) = (10/4) + 0 + (5/4) = 2.5 + 0 + 1.25 = 3.75
Therefore, the expected value for the number of points you'll win per turn in this game is 3.75.