Never before have I read something on the internet and felt compelled to post a comment. you fuckin suck. seriously.

 Please watch the meanness and the language on this blog.  My parents read this. 

I feel sorry for them.

I ususally post under "Anonymous." Obviously that wasn't me. How rude!

Not enough information...although I'm sure if you asked a bunch of monkeys, one of them would answer you correctly

Pretty sure I gave enough info...what else do you need?

How smart are the monkeys?

 average monkey smart

Is it 4 to the 50th power monkeys? The problem with monkey analysis is that it's not a complete random number generator. The monkeys might influence one another, after all monkey see, monkey do. Monkeys tend to pick "B" repeatedly because monkeys think that B means banana.

 Okay, counselor Pirates, I will hit up my calculator and get back to you.  My gut tells me your guess is too high, but we will see...

I bet it's .35*(4^50) monkeys. If you're only trying for 35% accuracy. But, I don't think it's entirely right because monkeys aren't random number generators. They're not dependent variables. If monkey-see monkey-do holds true, then they could very well all hit the same button each time (see supra re bananas)

What's up Ricky. You lawyers may be smart - most probably smarter than monkeys even ... but it looks like you need the help of a Financial Engineer with this problem. ;)

You'd need ~13 monkeys taking the test such that there is a greater than 50% chance that at least one of them scores 35% or better.

A single monkey has a score that follows a binomial distribution with 50 trials and a success of each trial of 25% (ie, 50 questions, 4 choices each question). This binomial distribution can be approximated with a normal distribution with mean of [50 * .25] and standard deviation of [sqrt(50 * .25 * .75)]. To get at least 35% correct on the test, the monkey would need at least 17.5 questions correct, which is 1.63 standard deviations from the mean [(17.5 - mean) / std dev]. A z-score of 1.63 or better happens 5.12% of the time [1 - (standard normal cumulative distribution function evaluated at 1.63)] meaning that a single monkey would score 35% or better with probability 5.12%.

Now we want to know how many monkeys it takes to get a 50% probability or better that at least one of them scores at least 35%. With a batch of n monkeys, the probability that NONE of them score 35% or better is [(1 - 5.12%) ^ n]. So, the probably that AT LEAST ONE of them scores 35% or better is [1 - (1 - 5.12%) ^ n]. We want this probability to be greater than or equal to 50%, so simply set this equation equal to 50%, and solve for n (consolidate terms, take log of both sides, and divide). Viola - 13.179 monkeys!

It's up to you to determine whether 13 monkeys is a comforting number or not. At any rate, I would guess there are many fewer videos of monkeys doing card tricks on YouTube. =)