He could be right

Probability Level 3

Able and Brainy are stumped by a multiple choice problem with three options and each of them takes a wild guess at it. When the proctor is distracted, they look at what answer the other has put down. Each of them has a fifty-fifty chance of either keeping his answer or copying his friend's. They get two more opportunities to crosscheck before they turn the test in. On both occasions, they behave exactly the same way, with the same probabilities.

What is the probability that they both have the correct answer at the end?

If the probability is in the form $\frac mn$ , where $m$ and $n$ are coprime positive integers, submit your answer as $m+n$ .

The answer is 47.

2 solutions

Manuel Kahayon
Dec 15, 2016

There's a $\frac{1}{3}$ chance they got the same answer at the start (up to the end). Given that they have different answers to begin with, the probability that they still have different answers at the end of the test is $\frac{1}{8}$ , since at any given cross-check we have a $\frac{1}{2}$ chance that they do not get the same answers (either both copy each other or they both don't, with $\frac{1}{4}$ probability each case) and there are 3 cross-checks, thus $\frac{1}{2}^3 = \frac{1}{8}$ is the probability they don't have the same answer at the end, given that they started off with different answers.

The probability that they then get the same answers at the end even if they started off with different answers is $1-\frac{1}{8} = \frac{7}{8}$ . The probability that they then get the same answer in the end all in all is $\frac{1}{3} + \frac{2}{3} (\frac{7}{8}) = \frac{11}{12}$ .

The probability that this answer they agreed upon is correct is $\frac{1}{3}$ . Our desired answer then becomes

(prob. that they agree)(prob. that they are right, given that they agree) = $\frac{11}{12}(\frac{1}{3} = \frac{11}{36}$ ,

And $11+36 = \boxed{47}$ .

Marta Reece
Dec 1, 2016

With three options, all equally likely, the probability that they both guess correctly to start with is 1/9. The probability that they are both wrong is 4/9. It's the case when they start with one right and one wrong answer that is interesting. Probability of this happening is 4/9.

With probabilities 1/2 of either changing or not, the chances that they are still in disagreement after one round are reduced by a half, halved again in the second round, and again in the third. So the probability of them still disagreeing after 3 rounds of cross-pollination is 1/8. This gives them a 7/8 chance of agreeing with each other and a 7/16 chance of agreeing on the right answer.

There was a 1/9 probability that they were right to start with (in which case they didn't change anything) and a 4/9 probability that they started with one of them right, in which case they had a 7/16 probability of both getting it right in the end. So the final calculation is

$\frac{1}{9}+\frac{4}{9}\times\frac{7}{16}=\frac{1}{9}+\frac{7}{9\times4}=\frac{4+7}{36}=\frac{11}{36}$

I did it slightly differently, but got the same result.

$P(\text{same answer in the end}) = \frac{1}{3} + \frac{2}{3} \cdot \frac{7}{8}$

The first $\frac{1}{3}$ is if the initial guesses are the same. The $\frac{2}{3}$ is if the initial guesses are different. And, $\frac{7}{8}$ is the probability they will eventually have the same answer if they didn't originally, as Marta describes above.

And, of course this needs to be multiplied by $\frac{1}{3}$ , the probability that the same answer is in fact the correct answer.

So,

$P = \frac{1}{3}(\frac{1}{3} + \frac{2}{3} \cdot \frac{7}{8}) = \frac{11}{36}$ $11+36=\boxed{47}$

Geoff Pilling - 4 years, 6 months ago

This is the same approach I took. I like using the symmetries of "same answer", since all of the answers are symmetric up until deciding on the "right" one.

Eli Ross Staff - 4 years, 6 months ago

If one of them always stayed while the other always copied, wouldn't they end up with a 100% chance of agreeing and a 1/3 chance of getting it right? That would give a 12/36 chance of both being right.

Kevin Boyd - 4 years, 6 months ago

Kevin is right, if the probabilities of them copying or keeping their own answers were also unknown, with the provision that the final probability would be maximized, his would be the correct (and trivial) solution. To avoid this, I have rephrased the problem. I apologize for the mistake, and thank you, Kevin, for your comment.

Marta Reece - 4 years, 6 months ago

Wonderful problem.If one of them has guessed the correct answer and the other one has guessed incorrect, then finally the probability that they both get it right comes out to be function in a and b.This function gets maximized at a=b=1/2.How did you know it beforehand itself?

Indraneel Mukhopadhyaya - 4 years, 6 months ago

What about the chance if they both started out the same and correct, then after round 1, the chance if neither of them changes the answer is 1/4 and round two the chance of both holding onto the same correct answer is 1/4 1/4 now. So wouldn’t it be 1/16 1/9 for the first scenario ?

Tony Yin - 3 years, 5 months ago

The probability of changing the answer is not 1/4 if both have the same answer. They only change their answer if they copy the other guy's different answer.

Marta Reece - 3 years, 5 months ago

He could be right

The answer is 47.

2 solutions

0 pending reports