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I thought of this strategy to answer multiple choice questions that seems to beat the odds.

Assume that there are 4 choices.


  1. Pick one at random
  2. Eliminate one of the other 3 (assume that you can eliminate one of the options because you know it cannot be right)
  3. Switch to one of the other 2 options that remain.

So, it would now seem that the probability that you get the question correct is $3/4 * 1/2 =3/8$ which is higher than what would be expected if you eliminated the option and then picked at random ($1/3$).

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well if you can always spot one false answer from any set of 3 answers with 100% accuracy, it means that you can always spot at least two false answers with 100% accuracy. So a better strategy would be to select randomly between the two answers left. Of course an even better strategy is to detect all 3 false answers with 100% accuracy and always go with the correct one. – mercio Apr 5 '12 at 12:35
up vote 2 down vote accepted

This is a variation of the Monty Hall problem. The issue is that if you know one answer is wrong, you don't have an initial probability of $1/4$ to be right, but $1/3$ (assuming you didn't choose the known wrong answer). And in that case, you get the correct $2/3*1/2=1/3$ as expected. Now, if you initially chose the known wrong answer, then you know the probability after switching is $1*1/3$ - also as expected.

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What if I do it in the order presented. I first pick an answer, then of the remaining 3, I eliminate one and then I switch. The first step is done with probabiliy 1/4. – picakhu Apr 5 '12 at 14:57
The same argument applies: Since there are only 3 valid choices left, your random choice is correct with probability $1/3$, not $1/4$. – Desiato Apr 5 '12 at 15:13

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