Animated Solution for Mathematics - Probability: A person throws two fair dice. He wins Rs. 15 for throwing a doublet, wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

Mastering Expected Value

Imagine you are standing at a casino table, holding two fair dice. The air is thick with anticipation. You are about to roll, and the rules of the game are simple: you win big on doublets, you win a decent amount if the sum is nine, but for everything else, you pay a penalty.

This isn't just a game; it is a perfect laboratory for understanding the concept of Expected Value. In the world of JEE Advanced, probability isn't just about counting; it is about weighing the future. Let us break this down.

Mapping the Sample Space

Before we can calculate any winnings, we must understand the universe of possibilities. When you throw two dice, each die has 6 faces.

The total number of outcomes is not just 6, but $6\times 6=36$. Think of this as a $6\times 6$ grid where every cell is a unique pair $(d_1,d_2)$. This grid is our sample space, the foundation upon which all our probabilities will be built.

Identifying the Winning Conditions

Now, let us look at the events that bring us profit. First, the 'doublets'. A doublet occurs when both dice show the same number: $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$.

There are exactly 6 such outcomes. The probability of this event, which we will call $P(A)$, is $\frac{6}{36}$. The reward is a handsome gain of $X_1=+15$ rupees.

Next, we have the 'sum of 9' condition. We need to find pairs that add up to 9: $(3,6),(4,5),(5,4),(6,3)$.

That gives us 4 outcomes. The probability $P(B)$ is $\frac{4}{36}$, and the reward is $X_2=+12$ rupees. Notice how these events are mutually exclusive; you cannot roll a doublet that also sums to 9, as 9 is odd and a doublet must sum to an even number.

The 'Other' Trap

Here is where most students stumble. The problem states that for 'any other outcome', the person loses 6 rupees. We must not forget these outcomes!

We have already accounted for $6+4=10$ outcomes. Since there are 36 total, the remaining outcomes are $36-10=26$. The probability of this losing event $P(C)$ is $\frac{26}{36}$, and the value is $X_3=-6$.

The Expected Value Masterclass

Now, we bring it all together using the Expected Value formula: $E(X)=\sum P_i{X}_i$. This formula is essentially a weighted average, telling us what we can expect to happen on average over many rolls.

Let us perform the arithmetic with precision. Keeping the denominator as 36 is our secret weapon here:

Combining the numerators, we have $90+48=138$. Then, $138-156=-18$. So, our expected value is:

The Verdict

The result is $-\frac{1}{2}$. This tells us that this game is mathematically rigged against the player.

On average, every time you throw these dice, you are expected to lose half a rupee. In the cold, hard language of mathematics, the negative sign is not just a symbol; it is a warning. You have successfully navigated the sample space, identified the traps, and calculated the long-term reality of the game.