GATE DS&AI 2024 | Question: 26

Question

​​​​​A fair six-sided die (with faces numbered $1,2,3,4,5,6$ ) is repeatedly thrown independently.

What is the expected number of times the die is thrown until two consecutive throws of even numbers are seen?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Let $x$ be the expected number of throws until two consecutive throws show an even number for a fair six-sided die and $y$ be the expected number of die throws to get the $\{2,4,6\}$ second time provided we get the $\{2,4,6\}$ first time

So, we can write the recurrence as:   

$$x=\frac{3}{6}(1+y)+\frac{3}{6}(1+x)\;\;\;\;\; (1)$$  

$$y=\frac{3}{6} \times 1 + \frac{3}{6} \times (1+x)\;\;\;\;\; (2)$$   

After Solving these two equations, you will get $x=6$

So, Answer is 6.

(You could also write it in one equation by including the interpretation of $y$ in equation (1) but I have written two equations so that it can be understandable easily)

-----------------------------------------------------------------------------------------------------------------------------------------------------

Now, let's try to understand it.

Here, $\mathbb{P}(\{2,4,6\})=\mathbb{P}(\{1,3,5\})=\frac{3}{6}$

Now, In equation (1),

first $\frac{3}{6}$ shows the probability of getting an even number i.e. $\mathbb{P}(\{2,4,6\})$ and second $\frac{3}{6}$ shows the probability of not getting an even number i.e. $\mathbb{P}(\{1,3,5\})$

$(1+y)$ represents that we have consumed one die throw to get an even number and then we again throw the die to get another even number.

$(1+x)$ represents that we have consumed one die throw to get an odd number and then we have to repeat the procedure to get the two consecutive even numbers.

In equation (2),

First $\frac{3}{6}$ shows the probability of getting an even number in second time i.e. $\mathbb{P}(\{2,4,6\})$ and second $\frac{3}{6}$ shows the probability of not getting an even number second i.e. $\mathbb{P}(\{1,3,5\})$

$1$ represents that we have consumed one dice throw to get an even number second time and we are done.

$(1+x)$ represents that we have consumed one die throw to get an odd number and then we have to repeat the procedure again to get the two consecutive even numbers.

----------------------------------------------------------------------------------------------------------------------------------------------------

If you are not understanding it then you can check this comment first and then come back to this question so that it will be easy for you to know how the second variable comes into the picture.

Comments

Doubt : What is wrong in this ?
To solve this problem, let's define a random variable \( X \) as the number of times the die is thrown until two consecutive throws of even numbers are seen.
Let's denote the expected number of throws until two consecutive even throws are seen by \( E(X) \).
Now, let's analyze the possible scenarios:
1. If the first throw results in an odd number, the situation doesn't change, and we're still looking for two consecutive even throws. The probability of this happening is \( \frac{1}{2} \).
2. If the first throw results in an even number:
  - If the second throw also results in an even number, we've found our two consecutive even throws, and the process stops. The probability of this happening is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
   - If the second throw results in an odd number, we're back to the initial situation, and we have to start over. The probability of this happening is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
So, we can set up the following equation for \( E(X) \):
\[ E(X) = \frac{1}{2} \times (1 + E(X)) + \frac{1}{4} \times (2) + \frac{1}{4} \times (1 + E(X)) \]

Solving this equation gives us the expected value of \( X \), the number of throws until two consecutive even throws are seen. Let's solve it:

\[ E(X) = \frac{1}{2} + \frac{1}{4} \times 2 + \frac{1}{4} \times E(X) \]

\[ E(X) = \frac{1}{2} + \frac{1}{2} + \frac{1}{4} \times E(X) \]

\[ \frac{3}{4} \times E(X) = \frac{3}{2} \]

\[ E(X) = \frac{3}{2} \times \frac{4}{3} = 2 \]

So, the expected number of times the die is thrown until two consecutive throws of even numbers are seen is 2.

---

@NarutoUzumaki

your equation would be:

$ E(X) = \frac{1}{2} \times (1 + E(X)) + \frac{1}{4} \times (2) + \frac{1}{4} \times (2 + E(X)) $

because you have consumed 2 dice throws for even and odd numbers.

After solving this equation, you would get 6.

---

@ankitgupta.1729 got it thankyou  sir !