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A game is played with the following rules. Three ordinary coins are flipped and count as a single "turn" in the game.
Each coin could land on heads or tails. After their turn a player is given a certain number of points depending on the
outcome.
•If the result is 3 heads, the player receives +4 points.
• If the result is 2 heads, the player receives +2 points.
• If the result is anything else, the player receives -1 point.
What is the expected value of the number of points a player will receive after a turn?

Sagot :

Answer:   0.75

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Explanation:

H = heads

T = tails

Something like HHT means we get two heads in a row at first, followed by one tail. The order matters. Since there are 2 sides per coin, and 3 coins, this leads to 2^3 = 8 different possible ways to flip three coins.

Those 8 outcomes are:

  1. HHH
  2. HHT
  3. HTH
  4. HTT
  5. THH
  6. THT
  7. TTH
  8. TTT

Again the order matters.

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Now let's define the following events

  • A = event of getting three heads
  • B = event of getting exactly two heads
  • C = event of getting 1 head, or getting 0 heads

There's only one way to get three heads, and that's HHH. So P(A) = 1/8.

There are three ways to get exactly two heads. Those three ways are: HHT, HTH, THH. This means P(B) = 3/8.

From those two probabilities, we then know that P(C) = 4/8 = 1/2 since there are 1+3 = 4 ways to get two or more heads, leaving 8-4 = 4 ways to get the other situations.

Now let's define function notation V(A) to mean "the net value of event A occurring". It's the number of points you receive if event A happens. We are given the following info:

  • V(A) = 4
  • V(B) = 2
  • V(C) = -1

From here, multiply the probability values with their corresponding net values

  • P(A)*V(A) = (1/8)*(4) = 0.5
  • P(B)*V(B) = (3/8)*(2) = 0.75
  • P(C)*V(C) = (1/2)*(-1) = -0.5

Finally, add up those results to get the expected value: 0.5+0.75+(-0.5) = 0.75

Each time a turn happens, we expect on average to get 0.75 points

Imagine a scenario where you play the game 10,000 times. If you record the result of each turn and then compute the arithmetic mean, you should get somewhere close to 0.75; it likely wont be 0.75 exactly due to the nature of theoretical probability and empirical probability not being perfectly aligned. However, due to the law of large numbers, the two should be very close when you do a large number of trials.

Because we're dealing with averages, it is possible to get a decimal value as the answer even though the point values of events A,B,C are integers.

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