TV Show Probability

Question

Thesis

---

Portugal has a contest show similar to "Who wants to be a millionaire?" in which is given a question, and the contestants have to try to guess which of the 4 answers is correct (there's only one right answer).

Anyway, here's the trick: there's something called a "Joker" in which a player can choose between 2 different answers, and the "Joker" will exclude one of them. However, it can never exclude the correct answer.

The question now is: "What's the probability of the player choosing the right answer, after using the "Joker"?

First hypothesis

---

Considering that there are 4 answers, after using the "Joker," we are left with 3 possible answers:

$$4-1=3$$

Therefore, since there's only one correct answer, the probability of winning is:

$$\frac{1}{3}\approx 0.33$$

Second hypothesis

---

Having 4 answers in total (before using the "Joker"), we can group them in 2 groups of 2 answers each.

Suppose that the 4 answers are named: A; B; C; D.

We can rearrange them in 6 different combinations: AB, AC, AD, BC, BD, CD:

$${}^{4}\textrm{C}_{2}=6$$

Notice that each answer appears 3 times in all of the 6 groups, which implies that each of the two groups previously selected has a 50% of probability of containing the correct answer:

$$\frac{1}{4}=0.25$$

Each group has two possible answers:

$$2*\frac{1}{4}=0.50$$

With lack of genetality, consider the group 1 made up of answers A; B; and the group 2, made up of answers C; D:

$$\mathbf{G_{1}}=\left \{ A;B \right \}$$ $$\mathbf{G_{2}}=\left \{C;D \right \}$$

If now I select the "Joker" to be only applied to the two answers that make up group 1 ($\mathbf{G_{1}}$), the probability of group 2 ($\mathbf{G_{2}}$) containing the correct answer remains the same ($\frac{1}{2}=0.50$).

However, group 1 ($\mathbf{G_{1}}$) will lose one of the possible answers, wihtout chaning it's probability of containing the correct answer, since:

$$1-0.5=0.5$$

• 1 is the total probability;
• 0.5 is the probability of the correct answer be on group 2.

That leaves us with one answer in group 1 that has 0.5=50% of probability of being correct, and two others in group 2 each with 0.25=25% of being correct:

$$\left\{\begin{matrix} \mathbf{G_{1}}=\frac{1}{2} & \\ \mathbf{G_{2}}=\frac{1}{2} & \end{matrix}\right. \Rightarrow \left\{\begin{matrix} \mathbf{A; B}=\frac{1}{2} & \\ \mathbf{C; D}=\frac{1}{2} & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \mathbf{A}=\frac{1}{2} & \\ \mathbf{C; D}=\frac{1}{2} & \end{matrix}\right.$$ If B, was one of the wrong answers.

Which means that A has a 0.5 of probability of being correct instead of the previous 0.33.

Question

---

I know that one of these hypotheses has an error; I'm just unable to find where.

Also, I know that this question is similar to Monty Hall's problem since the "Joker" can't exclude the correct answers. However, because there are 4 answers, I'm not sure whether that would impact the probability.

Which logic is correct and "what's the probability of the player choosing the right answer, after using the "Joker"?

Answer 1

Assume that the contestant has no idea which answer is correct, chooses two of the answers randomly, and always goes with the answer that the Joker did not eliminate.

In this case, the contestant wins if and only if the correct answer was one of the two chosen. Therefore, there is a 50% chance of winning the prize.

Answer 2

There are two possible event for player.

First event = Choosing two wrong answer in the beginning

Secend event = Choosing true and a wrong answer in the beginning

First event = Lets choose two random answer for our sample space such that $C(4,2)=6$ ways . Then choose $2$ wrong answer among $3$ wrong answers such that $C(3,2)=3$ ways. Thus , the probability of choosing two wrong answer is $\frac{3}{6}$.

Now, we should eliminate one of these two wrong answers , we can select that which one will be eliminated by $C(2,1)=2$ ways. At last, find the probability of chhosing true answer among $2$ wrong and one true answer which is $\frac{1}{3}$

Result= $\frac{3}{6} \times 2 \times \frac{1}{3} = \frac{2}{6}$.

Second event = Choosing the true and one false answer.

We can select $3$ different false answer along with true answer . Hence , our probability of selecting the true ans one false is $\frac{3}{6}$. After that , we should select the true answer from a sample space consisting of $2$ false and the true such that $\frac{1}{3}$

Thus our result is $\frac{3}{6} \times \frac{1}{3} = \frac{1}{6} $

As a result , when we add those two events , we see that $\frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} = 0.5$