You are asked a question and you don't have a clue about it. But, luckily you are given $K$ ($K \ge 4$) possible answers and only one is the right one. Because you choose your answer randomly you would like to increase your chance of answering the question right. You have two options:

2. Choose an answer randomly, exclude it and then choose another answer randomly from the other $K-1$ given answers.

So the question, would you have better chance if you choose option 2 than option 1? And if it's better, how many times would be the best option, and stopping before it become a bad idea (excluding large part of the answers).

Let's say. $K=4$. Then you'll have $25$% odds choosing the right answer with option 1.
If you exclude that answer you have $33$% or $0$% odds of choosing the right answer.

• If you can exclude one that you know has less than $25\%$ chance of being right, you are better off to do so. If you exclude one that has the same chance of being right as the others, it doesn't help (nor does it hurt). – Ross Millikan May 30 '13 at 18:00

These are two different procedures, whose outcome is selecting one of the $K$ answers. All $K$ outcomes are equally likely, so the procedures are identical in outcome. Procedure 1 is simpler and quicker, so is to be preferred.

Clarifying edit: Think of these procedures as "black boxes" that produce a guess. It doesn't matter how the procedures work, they produce a guess. By symmetry, none of the answers are any more likely than any of the others to be chosen, because the black box doesn't know the answer to the question. Hence every such mechanism is identical to every other such mechanism.

• Let's say with my example. Wouldn't be better to exclude one answer randomly? You have just 25% of choosing the right answer. Would be better to risk? – Stefan4024 May 30 '13 at 17:57
• It seems that option 1 is better. Because if you choose option 2 you'll end up with the same chance, but you may exclude the right answer and then you'll have 0% chance of guessing the right answer – Stefan4024 May 30 '13 at 18:05
• @Stefan4024, let me ask you this. If $K=4$, what is the probability that $A,B,C,D$ are chosen via either option 1 or option 2? Can they be anything other than 25% each? – vadim123 May 30 '13 at 18:09
• Every answer have the same chance of being the true one. And I even test this using randomizer and after 50 tries a get 13 answer right after the first try and 12 after excluding all. And that means that no matter which option do you choose you'll have the same chance of guessing it right – Stefan4024 May 30 '13 at 18:17

To see why the second strategy is identical in outcome to that of the first strategy, note that after selecting randomly one answer, say randomly having selected $A$, out of four choices, so you the probability A is correct is $\dfrac 14.\;$

Alternatively, skip randomly chosen A (which has a $25$% chance of being correct, and move to the second answer, so if K = 4, and you exclude A, you have a $(1 - \frac 14) =\frac 34$ probability that one of the remaining choices (B, C, D) are correct, and $\;\frac 13\;$ probability that the choice randomly selected from those three remaining choices (B, C, D) is correct.

In all, with the second method, you have a $\dfrac 34 \times \dfrac 13 = \dfrac 14$ probability of selecting the correct answer using the second method...and so on:

If you exclude randomly selected A, then randomly selected C, then you'll have a $100 - 25 - 25 = 50$% chance that $B$ or $D$ is correct. Randomly selecting one of those will give you gives you a $50$% chance that if one of the two are correct, selecting it. But that means you have a $\frac 12 \times \frac 12 = \frac 14$ (25%) chance of selecting the correct answer if exluding two options randomly, and randomly selecting among the final two options.

And so too with excluding 3 gives you a 25% chance that the one not excluded, hence must be selected, is correct.

So with either method, you have a $25\%$ chance of selecting the correct answer for any given question.

Of course, if you study even just just a little bit before the exam, and attend class regularly, you can surely improve those odds, and not have to resort to "random selection." ;-)

• I think I need more studying! :-) +1 – Amzoti May 31 '13 at 0:38

Intuitively, it should be clear that picking one, throwing it away, and picking another will give the same probability as just picking one.

We can explain it a little more using the notion of conditional probability:

Suppose you follow the procedure of picking one answer at random, throwing it away, and picking one of the remaining ones as your final answer.

The probability that you pick the correct answer on the first try is $1/4$ and the probability that you do not pick the correct answer on the first try is $3/4$.

The probability that you pick the correct answer on the second try assuming that you picked the correct answer on the first try is $0$. Thus the probability that you pick the correct answer on the first try and again on the second try is $0$.

The probability that you pick the correct answer on the second try assuming that you did not pick the correct answer on the first try is $1/3$.

Thus the probability that you do not pick the correct answer on the first try and you pick the correct answer on the second try is $\frac 3 4 \frac 1 3 = 1/4$.

Thus the total probability that you pick the correct answer on the second try is $0 + 1/4 = 1/4$, which is the same you'd have gotten otherwise.