Question

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 is 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?

Answer 1

5

Explanation:

The winnings are in G.P. : 1, 2, 4, ..... till 10 toss.

`$a_n = 1 * 2^(n-1)\ \ \ \forall \ n = 1,2,3,4,....,10$`

`$a_n$` denotes the winnings on the
`$n^(th)$` toss.

The probability of earning amount
`$a_n$` on the
`$n^(th)$` toss is =
`$\left((1)/(2)\right)^n$`

∴
`$E(X) = \sum_(n=1)^(10) \ a_n * \left((1)/(2)\right)^n $`

`$=\sum_(n=1)^(10) \ 1 * (2^(n-1))/(2^n) $`

`$=\sum_(n=1)^(10) \ (1)/(2)$`

Sum of the 1st n terms of the A.P. is :

`$=(n)/(2)[2a+(n-1)d] $`

`$=(10)/(2)[2* (1)/(2)+(10-1)* 0] $`

= 5

Therefore, E(X) = 5

Hence the expected value of the game is 5