Comparing Expected Value of multiple iterations I'd like to ask if I am going about this problem in the right way
Say a bank has an offer that every time you deposit money, there is a 5% chance they will increase it by 150% (e.g. depositing \$4000 has a 5% chance of becoming $6000). Is is better to deposit all your money at once, or deposit it in smaller amounts to increase the chance of success?
I worked out a few iterations (some values rounded for ease of reading):

1 deposit of $4000
  95% to get 4000 (Probability*Value=3800)
  5% to get 6000 (PV=300)
  Expected Value = 3800 + 300 = 4100
2 deposits of $2000 
  90.25% to get 4000 (PV=3610) 
  4.75% to get 5000 (PV=237.5) 
  4.75% to get 5000 (PV=237.5) 
  0.25% to get 6000 (PV=15)
  Expected Value = 4100
3 deposits of $1333.33..
  85.74% to get 4000 (PV=3429.5)
  ...
  1.25% to get 6000 (PV=0.75)
  Expected Value = 4100

Even though the chance of winning the offer is increasing, the Expected Value is the same in all cases. Does this mean it makes no difference how many deposits are made?
I have no idea if I'm going about this correctly so any feedback would be appreciated, or even suggestions into areas I should be reading into!
 A: The average is the same in each case, but that doesn't mean it makes no difference.  The distribution of the final amount is different in each case.
In general, suppose an amount $M$ is deposited in $k$ equal amounts, and for each deposit, there is a probability $q$ that the deposited amount is multiplied by $(1+r)$.  Each deposit is therefore $\frac{M}{k}$, although there is a probability $q$ that it will suddenly become $\frac{M}{k}(1+r)$—an increase of $\frac{Mr}{k}$.
The average amount at the end can be determined by figuring out the average amount of each deposit.  It has a probability $1-q$ of being $\frac{M}{k}$, and a probability $q$ of being $\frac{M}{k}(1+r)$, so the average amount is
\begin{align}
E(\text{each deposited amount}) & = (1-q)\frac{M}{k} + q\frac{M}{k}(1+r) \\
                                & = \frac{M}{k}(1+qr)
\end{align}
After $k$ such deposits, the expected total deposit is therefore $k$ times that amount, or just $M(1+qr)$.  In the case of your problem, we have $M = 4000, q = 0.05, r = 0.5$, so the expected final amount is always $M(1+qr) = 4000(1+0.025) = 4100$.

Assuming that the deposits are independent (one deposit getting multiplied doesn't influence the probability of another deposit getting multiplied), the number of deposits that get multiplied follows a binomial distribution:
$$
P(\text{$j$ of the $k$ deposits get multiplied})
    = \binom{k}{j} q^j (1-q)^{k-j}
$$
Each time a deposit gets multiplied, the resulting increase is just $\frac{Mr}{k}$, so if it happens $j$ times, the increase is $\frac{Mjr}{k}$.  Since the base total deposit is $M$, we can therefore write the previous equation simply as
$$
P\left(\text{final amount is $M+\frac{Mjr}{k}$}\right)
    = \binom{k}{j} q^j (1-q)^{k-j}
$$
