Suppose you toss a coin and it you win a dollar if it comes up heads. If it comes up tails you toss the coin again. This time you get two dollars if it is heads. If it is tails you toss it again. This time you win four dollars if it is heads, but if it is tails you toss it again. Continue in this fashion fora toatl of 10 flips of the coin, after which you get nothing if it is tails. What is the mathematical expectation for this game?
probability of winning one dollar = .5
of $2 = .5*.5* = .25
of $3 = .5^3
of $4 = .5^4
so E = .5*1 + .25*2 + .125*3 ..... .5^10 *10
To calculate the mathematical expectation for this game, we need to determine the probability of each outcome and multiply it with the corresponding payoff.
Let's break down the possible outcomes step by step:
1. On the first toss, the probability of getting heads is 1/2, and the payoff is $1.
So the expected value after the first toss is (1/2) * $1 = $0.50.
2. If the first toss is tails, we move on to the second toss.
On the second toss, the probability of getting heads is also 1/2, and the payoff is $2 (since it's the second toss and heads).
So the expected value after the second toss is (1/2) * $2 = $1.00.
3. If the second toss is tails, we move on to the third toss.
On the third toss, the probability of getting heads is still 1/2, and the payoff is $4 (since it's the third toss and heads).
So the expected value after the third toss is (1/2) * $4 = $2.00.
We continue this process for each flip up to the 10th flip. However, after the 10th flip, if it's tails, we don't get any payoff.
Hence, the expected value of the game is the sum of the expected values for each step:
Expectation = $0.50 + $1.00 + $2.00 + $4.00 + $8.00 + $16.00 + $32.00 + $64.00 + $128.00 + $256.00
If we calculate this sum, the expected value for this game would be $511.50.