# Probability law

1. Jun 18, 2014

### darrin016

Feel free to move if this is in the Erin section.
I'm no mathematician. Second year of college going to my astrophysics degree.
Something I don't understand is this
Probability law. The following is copied from Wikipedia
"For example, if the chance of rain in Moscow on a particular day in the future is 0.4 and the chance of an earthquake in San Francisco on that same day is 0.00003, then the chance of both happening on that day is 0.4 × 0.00003 = 0.000012, assuming that they are indeed independent."
They also have a equation for a monkey typing the word banana. 1/50 for 50 keys on a board. Banana has 6 letters so it was explained as
1/50 *1/50 six times. And that was the solution to the chances of the monkey randomly typing the word banana. I disagree. Yes that is the chance it hits the keys. But this equation doesn't take into consideration the variable of it actually happening in the correct order. I feel that 1/50 X 6 just gives us the answer for the probability of the monkey randomly hitting the letters b a n a n a but not in a specific order. I believe spelling the word banana would be even less of a chance that 1/50 6 times added together. Maybe something like
1/50 x 1/50 x 1/50 x 1/50 x1/50 x1/50 x 1/48 for all the possible ways the letters can fall.. Does this make sense to anyone?

2. Jun 18, 2014

### Chronos

You are confusing permutation with straight probability. Coming up with a 'b', 3 'a' s and 2 'n' s without regard for the order, while still relatively small, is much more probable than 'banana' in that exact sequence. On the first keystroke you have 3 chances in 50 [b,a,n], and you always have at least 2 chances in fifty for the next keystroke, and in the worst case you have 1 chance in fifty for the rest. For 'banana' in that exact order, the odds are 1 in 50 for each keystroke. Clearly, the permutation sequence is more than 6 times more probable than the exact order.

3. Jun 18, 2014

### darrin016

Okay so the 1/50 are the odds of spelling banana out correctly in order.. So even with the 1/50 chance, there isn't another variable that comes into play? As far as getting all the 1/50 letters to happen exactly in the precise order?

4. Jun 18, 2014

### Staff: Mentor

Imagine that we have a very large number of monkeys, each sitting in front of it's own typewriter. They all strike a key... One in fifty of them will hit the "b" key. We shoot all the other monkeys (or release them to go play in the jungle if you prefer).

When our remaining monkeys all strike a key again, one in fifty of them will hit the "a" key. That's one-fiftieth of one-fiftieth of our original collection of monkeys.

We keep these around and give them their one chance in fifty of hitting the "n" key, and so forth. Each time, one in fifty monkeys score the right next letter and the other forty-nine out of fifty drop out. That's where the 1/50 x 1/50 x1/50... six times comes from.

If you want the probability of getting the six letters right in any order, that's a bit trickier. If every monkey types six letters, there are 50x50x50x50x50x50 different sequences they can type. How many of these sequences contain one "b", two "n"s, and three "a"s?

5. Jun 18, 2014

### TumblingDice

The calculation is all good as it is. The parts that represent getting the letters in the correct order are the "ones" in all numerators. That's because, for each letter in "banana", there is only 1 correct key (out of 50) that will continue to produce the desired sequence. Ergo the product of all six 1/50 values represents the entire probability.

6. Jun 18, 2014

### Delta²

A simpler way to see it is that the probability for the monkey to hit the word banana is the same with the probability to hit any other word (one specific word not all together) that has 6 letters. All the words with 6 letters are 50^6 . So the probability to hit one specific word is 1/50^6.

The probability to hit any word that is an anagram of the word banana is (6!/(2!x3!))/50^6. That is so because all the anagrams of the word banana are not exactly 6! as one might thought at first glance but because "n" appears 2 times and "a" appears 3 times we have to do a division by 2!x3!.

7. Jun 19, 2014

### D H

Staff Emeritus
No, no, no!

The probability that the monkey will correctly press the "b" is 1/50. Note that 49 out of 50 times, the monkey doesn't even do that. If the monkey does that first task correctly, the probability of getting the next letter ("a") correct is 1/50. If the monkey does both the first and second task correctly, the probability of getting the next letter ("n") correct is 1/50. And so on.

However, the probability that the monkey will hit the "b" key and then the "a" key, in that order, is 1/2500. The probability the monkey will hit the "b" key, and then the "a" key, and then the "n", in that order, is 1/125000. By the time you get to "banana", the odds are ridiculously against the monkey typing that by random choice.

8. Jun 19, 2014

### darrin016

So the probability is smaller than 1/50^6 that it types the correct order?
I feel like 1/50^6 is like resetting after each key. And doesn't take into consideration the small chance they do it in order. Even though the shooting the monkey does make sense.

9. Jun 19, 2014

### Staff: Mentor

$1/50^6$ is the probability of getting six letters correct in the correct order. And yes, it is like resetting each after key. Suppose you walked in on the experiment after five trials; you are looking at a whole bunch of monkeys who have successfully typed "banan" and are about to type another letter. One in fifty of them will hit an "a". That's the exact same odds that any monkey in our (then much larger) ensemble of monkeys had of hitting a "b" on the very first trial.