Routine Probability Question, Computational Issue, Alternate Way to Solve? I feel like my approach and reasoning for the question provided below are good. Considering it is from a multiple choice exam with a strict time limit, I am not sure what other approach one might take. However, once I derive my final expression (which I believe to be accurate), I am unable to compute its decimal value for computation reasons. Please help me to see if either:

1) I am wrong

2) There is an alternative approach that will avoid the computational complexity

3) There is a trick to computing the expression I am stuck on.

The question is:


  An investor buys \$100 of worth of a stock. Each month, the investment
  has probability .5 of increasing \$1.10 and probability .5 of
  decreasing \$0.90. Changes in price each month are mutually
  independent. What is that probability that after 100 months the value
  of their stock is worth more than \$91?


How I reasoned through this was to picture the different possibilities each month as a tree - each node has the two options: increase 1.10 or decrease .90. For month 100 there are $2^{100}$ nodes. Since mutually independent, the probabilities along these  nodes are all $(.5)^{100}$. Now we just need to count all the branches that give a final value greater than \$91. If there are $k$ of them, the final probability we seek is $k \cdot (.5)^{100}$. 

From here I proceed by considering $$100 +m \cdot (1.10) - (100 - m) \cdot (.9) &gt; 91$$ where $m$ represents the number of months the stock increased, and $(100-m)$ is the number of months it decreased. Solving yields $m = 40.5$ but we want integers so $41 \leq m \leq 100$. This gives 60 different combinations $(\text{increase, decrease}) = (41,59), (42, 58), \dots , (100,0)$ that result in a success branch.

However, for each combination we need to account for all the permutations. I.e, for say $(41,59)$ there would be ${100\choose 41}$ arrangements in which this success branch occurs. This gave me my final expression which should evaluate to the desired probability, $$P[X \geq 91] = (.5)^{100} \sum_{j=41}^{100} {100 \choose k}.$$ 

This is not easily computable, even with a calculator. So now I'm not sure what to do. I also don't see a very clean way to apply the Central Limit Theorem, which would be my usual go to in situations like this. 

The options were:

a) .63

b) .75

c) .82

d) .94

e) .97

I dare you to ask me ‘what have you tried’ :) 

As mentioned, this question comes from a multiple choice exam where time is crucial. You are expected to spend 4 - 8 minutes per problem, so there must be some reasonably efficient approach to this.
After leaving this question unanswered for quite some time, I stumbled upon it today and decided to see what I could do. I have found a way to compute the expression I derived above, but I am still not satisfied with this method because I know they wouldn't expect this solution on a timed exam, so I will also present an approximation.

Exact Computation:



As I mentioned we have 61 ordered pairs (decrease, increase) that result in a success branch. However, we can exploit the symmetry of the binomial coefficient. For the first 41 ordered pairs (0,100), (1, 99),$ \dots $, (40,60) that result in a success branch, there are corresponding ordered pairs that give a fail branch (100,0), (99,1),$ \dots $, (60,40). By the symmetry of the binomial coefficient, the corresponding ordered pairs give the same number of arrangements, so the probabilities that corresponds to each of these two sets of outcomes is equal, call it $a$. By definition $a$ is the probability of the compliment event P[X $\leq$ 91], but we have found a portion of the original event that shares this probability.   

For the event we are interested in, the total probability is then $$P[X &gt; 91] = a + (1/2)^{100} \cdot \sum_{k = 41}^{59} {100 \choose k}.$$ Call the term on the right $c$ so that P[X > 91] = $a + c$. Then from above, we also know that $2a + c = 1$. Since we have complimentary events. Thus if we can get a value for $c$ we can determine $a$ which gives the desired probability. 

For computing $c$, we exploit the symmetry further since 100 - 41 = 59, 100 - 42 = 58, etc. With this, we can compute $c$ as $$c = (1/2)^{99} \sum_{k = 41}^{49} {100 \choose k} + (1/2)^{100} {100 \choose 50}.$$

Although this is still not exactly "nice" or "easy" to compute, it is computable. I was able to plug this into my simple TI-30 calculator in just a minute or two and get a value of $c \approx .943111$. With this once can determine $a$, then we have $$c + a = P[X &gt; 91] \approx .9715$$ which is indeed one of the options for answers.

But, all of this was a lot of work for a question that should be done in 6 - 8 minutes. So, there is also this;

Approximation via C.L.T



Let $X_i$ represent the change in value at the end of any particular month, so $$X_i:= \begin{cases} 1.1, \hspace{2mm} P = .5\\ -.9, \hspace{2mm} P = .5 \end{cases} $$
Then the total change at the end of the hundred month is $$X = \sum_{i = 1}^{100} X_i$$ and we have $Z = 100 + X$ represents the value of the stock after the 100th month. Then $$P[Z &gt; 91] = P[100 + X &gt; 91] = P[X &gt; -9].$$

The expected value and variance of $X_i$ can easily be computed so that by the Central Limit Theorem $X \sim N(10,100)$. Employing the standard normalization we have $$P[X &gt; -9] = P[Z &gt; -1.9] = P[Z &lt; 1.9] \approx .9713 \hspace{2mm} (\text{via a Z-table}).$$

This answer is consistent with what we found previously.

I suppose this method is perhaps what joriki was suggesting in his answer, though a lot more specific, and I wasn't sure what values he was suggesting I take logarithms.
Take logarithms and approximate the resulting sum by a normal distribution with mean and variance given by the sums of the corresponding quantities of the individual changes.