# Probability of an answer being correct

Suppose that two people independently give same answer to a question (the question in not a multiple choice question and we don't know the set of possible answers). The probability of these people being correct is $p_1$ and $p_2$ respectively. What is the probability that the answer is correct ? Is it $p_1.p_2$ or $(1-(1-p_1)(1-p_2))$ ? I have been told that its the latter but I don't understand how ? Intuitively it seems right because the probability of the answer being correct should increase with more endorsements by people with high $p_i$ but what's the principled way to understand it ?

• $p_1p_2$ gives the probability that both people answer correctly. $1-(1-p_1)(1-p_2)$ gives the probability that at least one person answers correctly. they measure different outcomes. – symplectomorphic Jul 5 '16 at 7:22
• So you mean that it depends on how I define the probability of answer to be correct ? – NGInd Jul 5 '16 at 7:25
• The probability that the correct answer is given, it seems from what you have mentioned in the question, is the probability of at least one person letting out the correct answer. If at least one of the two persons tells the correct answer, the final answer is considered to be correct. Hence, the desired probability is $1-(1-p_{1})(1-p_{2})$ – PN Karthik Jul 5 '16 at 7:30
• – David Jul 5 '16 at 7:31
• But if $p_1=0$ and $p_2=\frac{1}{2}$, then the probability that the answer is correct is $p_1 p_2 = 0$ and not $1-(1-p_1)(1-p_2) =\frac{1}{2}$. – Hetebrij Jul 5 '16 at 7:31

If we would have the probability that both persons give the same wrong answer, lets say, it is $q$, then we could solve the problem as follows :

We have two possibilities to get the event : "The answers coincide " :

$1)$ Both persons are right : Probability $p_1\cdot p_2$

$2)$ Both persons give the same answer and it is wrong : Probability $q$

The probability that the answer is right woule be :

$$\frac{p_1\cdot p_2}{p_1\cdot p_2+q}$$

• See my comment above; This is correct but I think a better answer makes an unbiased or mimimum variance estimator for $q$ based on $p_1$ and $p_2$, then completes the calculation. – samerivertwice Jul 5 '16 at 12:28
• In this case, we do not have a probability, only an estimated probability. – Peter Jul 5 '16 at 12:31
• All probability is estimated probability. The only non-estimated probability of an event is always $1$ or $0$ upon observation. – samerivertwice Jul 5 '16 at 14:36
• This question seems to be about probability. The OP did not mention any estimates. – Peter Jul 5 '16 at 21:14

As given, the problem has no solution, as there's no way we can determine the probability of both people to select the same wrong answer.

An extreme case to illustrate this: Assume that the question has a billion options, and both people select the right option if they know the answer and a random option if they don't know the answer, and both have at least an 1% chance to know the answer. Under those conditions, if both selected the same answer, it's almost certain that it is the right answer, as the probability of both selecting the same answer by chance is negligible.

• It does have an answer, but the probability itself sits in a range depending on the distribution of other possible answers. Probability is always about making assessments based on incomplete information. We need to make an estimator for the probability that two incorrect answers are the same incorrect answer based on the fact that the number of alternative answers is somewhere between 0 and infinity. And $p_1$ and $p_2$ give us a means to estimate that. The probability of two samples from a Poisson distribution coinciding is probably the best model. – samerivertwice Jul 5 '16 at 12:25
• @RobertFrost One can do that but I am pretty sure that the exercise should be about Bayes' theorem. Unfortunately, it contains not enough information. – Peter Jul 5 '16 at 12:34

This is a conditional probability. What is the probability that they are both right, given that the both gave the same answer. The "SPACE" is the probability that they both gave the same answer and the "success space" is the proportion of that made up of them both giving the same correct answer.

probability both wrong = $(1-p_1)\times(1-p_2)$

Probability both right = $p_1\times p_2$

Probability both right given that one of the above the case, is:

$$\frac{p_1\times p_2}{((1-p_1)\times(1-p_2))+p_1\times p_2}$$

Which simplifies to $$p_1\times p_2$$

• It has to be the same wrong answer, just "both wrong" won't do. – true blue anil Jul 5 '16 at 9:11
• @trueblueanil ah yes, good point. – samerivertwice Jul 5 '16 at 9:17