# Binominal Distribution (urgent)

#### lkwokchu

##### New Member
kindly help (or provide hints) on below question:

Suppose that you pay \$1024 to enter a coin tossing game. A biased coin with head
probability 0.3 is tossed 10 times independently. The amount will be doubled every time a
head is obtained and halved every time a tail appears. Denote X as the number of heads
obtained and Y as the amount of money you end up with.
(a) Express Y in terms of X.
(b) Determine the expected value of Y. Is this a fair game?

#### BGM

##### TS Contributor
If you still stuck in part a), you may think of the following way first:

Let $Y_i \sim \text{Bernoulli}(0.3)$ be the indicator of the $i$-th toss having a head such that $X = \sum_{i=1}^{10} Y_i$.

Can you express the outcome after the 1st toss? If you are not sure, write it down with something like

$\begin{cases} W_H \text{ if } ~~ Y_1 = 1 \\ W_T \text{ if } ~~ Y_1 = 0 \end{cases}$

in terms of the initial wealth $W$. (assume the question have not given the number 1024 yet).

Note that while $Y_i$ is the indicator for head, we can also use $1 - Y_i$ as the indicator for tail. Two common way to combine the above written cases

1. $W_HY_i + W_T(1 - Y_i)$ in an additive model
2. $W_H^{Y_i}W_T^{1-Y_i}$ in a multiplicative model

You should be familiar with the second one if you have learned something about the Binomial.

#### lkwokchu

##### New Member
thank for the explanation. but still dont know how to incorporate "The amount will be doubled every time a head is obtained and halved every time a tail appears" in part a). could you elaborate a bit more. Thanks!

#### BGM

##### TS Contributor
Actually I have not think of a good idea to give a good hint here as those hints I have thought of will lead to the direct final answer.

Anyway, here it goes: Just following my first post, do you see why the wealth after the first toss is in the form of

$W \times W_H^{Y_1} \times W_T^{1 - Y_1}$

Please fill in the appropriate values for the factors $W_H, W_T$ and try to generalize the result.

#### lkwokchu

##### New Member
Thank for your guidance. the ans should be Y=(1024)(2)^x(.5)^(10-x), right? could you proivde hint on b) also? Thank again

#### BGM

##### TS Contributor
Yes you are correct.

The first thing in part b) is to determine the distribution of $X$

Calculating the expected value related to summing the Binomial series (Binomial Theorem).

i.e. You need to know how to expand $(a + b)^n$