I was recently reviewing my notes on the Binomial Trees and MonteCarlo (MC) methods for option pricing. 
I noticed that while in the Binomial Trees, we use the probability weighted average, we use the simple average in the MC. I've taken this for granted and just used the MC method....but I started questioning why we take a simple average for calculating the option value. I understand that we're drawing independent random samples from the normal distribution....but isn't the prob of drawing say 0.65 different from drawing 1.465 (etc.)? If so, why don't we do a probability weighted expectation (like in the Binomial Tree) instead of the simple/equally weighted expectation in MC? I referred to Hull and Glasserman....but couldn't find a convincing enough explanation of the topic.

Would appreciate some guidance. Thanks!

The two methods are mathematically equivalent. In the MC, for each trial you will simulate the terminal stock price, [$]S_T[$] and calculate the option value. It's true that you will then take a simple average of the option values over all MC trials. However, each terminal stock price will occur (in the limit of a very large number of trials) with its correct risk-neutral probability. 

If you don't believe it, just plot a histogram of the [$]S_T[$] values after, say, 100,000 MC trials. You will see the familiar log-normal distribution.  

Thanks Alan - I'm good with the shape of the final distribution......no concerns there. But are you also implying that since the terminal stock price occurs with its true probability (as shown by the histogram of terminal values), it's then ok to use a simple average? I'm trying to understand why we use the simple average. Thanks!!

cdsharm75, if I understand your confusion correctly, I think it comes down to how the risk premium is implemented.

Both the binomial tree and the Monte Carlo methods really describe categories of models -- or maybe it would be better to say "processes of estimation" -- rather than very particular models. So depending upon what you learned as the binomial tree model and the Monte Carlo method, what I write here might not line up exactly, but probably you'll be able to figure out exactly what the distinction is.

Basically in option pricing you have the "real world" probabilities and the "risk-neutral" -- or to be more general, "risk-adjusted" -- probabilities. In what is almost certainly the most common Monte Carlo set-up, all you're working with is the risk-neutral probability, which includes the risk premium. So you just take the simple averages of your results to produce prices. It would probably be messy to try to produce prices using a Monte Carlo simulation based on real world probabilities and then somehow weight the ending values or even the average of the ending values for the risk premium.

Binomial tree models, on the other hand, can pretty easily be done in two different ways: either you use a real world drift and risk-neutral volatility -- giving weights other than 0.5 to the "up" and "down" possibilities -- or you use a risk-neutral drift and equally weighted "up" and "down" probabilities. They should converge to the same results as the number of steps you use increases.

If that's not what confuses you, then maybe it's as Paul alludes, and you're misunderstanding how the Monte Carlo method works: it doesn't matter what the probability of the random outcome you actually get is; what matters is that you got it, and you therefore include it in a simple average to get your results.

And this is why you want to include a lot of outcomes in a Monte Carlo simulation: the more outcomes you use, the less likely it is that your results will be skewed by just a few statistical outliers.

Thanks Alan - I'm good with the shape of the final distribution......no concerns there. But are you also implying that since the terminal stock price occurs with its true probability (as shown by the histogram of terminal values), it's then ok to use a simple average? I'm trying to understand why we use the simple average. Thanks!!

Yes -- if the terminal stock price occurs with its true probability (as shown by the histogram of terminal values), then the MC simple average is equivalent to the probability-weighted average of the option payoff. Here is the proof.

Proof.
For simplicity and definiteness, suppose zero interest rates and the terminal stock price can take on only three values [$]S(k),  (k = 1,2,3)[$], with risk neutral probabilities [$]p_k[$], where [$]p_1 = 0.3, p_2 = 0.5, p_3 = 0.2[$]. The option payoff function is some [$]w(S(k))[$]. The initial option value [$]V[$], computed by probability weighting the payoffs is:

[$] V = \sum_{k=1}^3 p_k \, w(S(k))[$].

Now, compute V by a MC with equal weighting of  MC outcomes; each MC draw is indexed by [$]n = 1, 2, \cdots, N[$] and suppose there are [$]N = 100,000[$] draws for definiteness:

[$] V_{MC} = \frac{1}{N} \sum_{n=1}^N w(S_n)[$]

But there are only three distinct possible values for [$]S_n[$]. Each [$]S_n[$] will equal either [$]S(1), S(2)[$] or [$]S(3)[$]. By assumption, the MC frequencies match the risk-neutral probabilities. This means [$]S_n = S(1)[$] occurs 30,000 times, [$]S_n = S(2)[$] occurs 50,000 times, and [$]S_n = S(3)[$] occurs 20,000 times. So,

[$] V_{MC} = \frac{30,000}{100,000} w(S(1))  + \frac{50,000}{100,000} w(S(2)) + \frac{20,000}{100,000} w(S(3)) =   \sum_{k=1}^3 p_k \, w(S(k)) = V [$]

@Alan - really appreciate the detail.....it's super clear now....thanks a ton!

@Paul and @Marsden - thank you both for the detailed explanations - it's all very clear now - thanks!
@Marsden  yes - it was a misunderstanding how the MC works