Suppose you toss a coin and will win $1 if it comes up heads. If it comes up tails, you toss again. This time you will receive $2 if it comes up heads. If it comes up tails, toss again. This time you will receive $4 if it comes up heads. Continue in this fashion for a total of 10 flips of the coin, after which you receive nothing if it comes up tails. What is the mathematical expectation for this game?
How did you come up with .5 as the probability?
Well, if you're looking for a mathematical expectation, I'd say that the only thing you can really count on in this game is disappointment! But let's crunch the numbers.
First flip: You have a 1/2 chance of getting heads and winning $1, and a 1/2 chance of getting tails and moving on to the next flip.
Second flip: Assuming you got tails in the first flip, you now have a 1/2 * 1/2 = 1/4 chance of getting heads and winning $2, and a 1/2 * 1/2 = 1/4 chance of getting tails and moving on to the next flip.
Continuing this pattern, for each subsequent flip the probability of getting heads diminishes by half.
So, if we look at the expected value for each flip, we have the following:
First flip: (1/2 * $1) + (1/2 * 0) = $0.50
Second flip: (1/4 * $2) + (1/4 * 0) = $0.50 * 2 = $1.00
Third, Fourth, Fifth, Sixth, Seventh, Eighth, Ninth, and Tenth flip: Since the expected value for each flip is halved, the expected value for each of these flips is $1.00 * (1/2)^n, where n is the flip number. So, we have:
Third flip: $1.00 * (1/2)^3 = $0.125
Fourth flip: $1.00 * (1/2)^4 = $0.0625
Fifth flip: $1.00 * (1/2)^5 = $0.03125
Sixth flip: $1.00 * (1/2)^6 = $0.015625
Seventh flip: $1.00 * (1/2)^7 = $0.0078125
Eighth flip: $1.00 * (1/2)^8 = $0.00390625
Ninth flip: $1.00 * (1/2)^9 = $0.001953125
Tenth flip: $1.00 * (1/2)^10 = $0.0009765625
Now, let's sum up all these values to get the mathematical expectation:
$0.50 + $1.00 + $0.125 + $0.0625 + $0.03125 + $0.015625 + $0.0078125 + $0.00390625 + $0.001953125 + $0.0009765625 = $1.6865234375
So, the mathematical expectation for this game is approximately $1.69. But remember, that's just the expected value - the actual outcome may be very different!
To determine the mathematical expectation for this game, we need to calculate the expected value for each individual coin flip and then sum them up.
For the first coin flip, the probability of getting heads is 1/2, and the reward is $1. So the expected value for the first flip is (1/2) * $1 = $0.50.
For the second coin flip, if it comes up tails, you toss again. The probability of getting heads on the second flip is (1/2)^2 = 1/4, and the reward is $2. So the expected value for the second flip is (1/4) * $2 = $0.50.
Similarly, for the third coin flip, the probability of getting heads is (1/2)^3 = 1/8, and the reward is $4. So the expected value for the third flip is (1/8) * $4 = $0.50.
Following the same pattern, we can calculate the expected value for each subsequent flip, which will also be $0.50.
Since there are 10 coin flips in total, the mathematical expectation for this game is:
$0.50 + $0.50 + $0.50 + ... + $0.50 = 10 * $0.50 = $5.
Therefore, the expected value for this game is $5.
To find the mathematical expectation for this game, we can calculate the expected value for each possible outcome and then sum them up.
Let's analyze the possible outcomes:
1. The coin comes up heads on the first toss, and you receive $1.
2. The coin comes up tails on the first toss, but heads on the second toss, and you receive $2.
3. The coin comes up tails on the first two tosses, but heads on the third toss, and you receive $4.
4. ...
We can see a pattern here: the amount you receive doubles with each consecutive flip until the 10th flip. So, the amounts you receive can be represented as powers of 2: $1, $2, $4, $8, $16, ..., $512.
To calculate the probability of each outcome, we need to consider the probability of getting heads or tails on each toss. Since we are assuming a fair coin, the probability of getting heads or tails is both 0.5.
Now, let's calculate the expected value for each outcome:
1. Probability of 1st outcome: P(heads on first toss) = 0.5
Amount received: $1
Expected value: 0.5 * $1 = $0.5
2. Probability of 2nd outcome: P(tails on first toss, then heads on second toss) = 0.5 * 0.5 = 0.25
Amount received: $2
Expected value: 0.25 * $2 = $0.5
3. Probability of 3rd outcome: P(tails on first two tosses, then heads on third toss) = 0.5 * 0.5 * 0.5 = 0.125
Amount received: $4
Expected value: 0.125 * $4 = $0.5
We can see that the expected value is the same ($0.5) for each outcome.
Since there are a total of 10 possible outcomes (10 tosses of the coin), the total expected value is:
Total expected value = $0.5 * 10 = $5
Therefore, the mathematical expectation for this game is $5. In the long run, you can expect to win an average of $5 per game.
Ah, a variation on the good old St Petersburg paradox.
Prob of winning $1 is .5
Prob of winning $2 is .5*.5
Prob of winning $4 is .5*.5*.5
Prob of winning $8 is .5*.5*.5*.5
and so on,
Expected value is sum over all possible outcomes, the probability times the value of the prize.
E = .5*1 + .25*2 + .125*4 + ...
= .5 + .5 + .5
The expected value from each flip is $0.5 So, after 10 flips, the expected value is 5. (after 100 it would be 50)