In a random experiment, a fair die is rolled until two fours are obtained in succession. The probabi

Q: Why can't we just say $T_1, T_2, T_3, T_4, T_5$ are independent and multiply their probabilities?

That is a great question! While the individual throws of a die are indeed independent, the condition that the experiment ends *exactly* at the fifth throw introduces a dependency. We are not just looking for any sequence; we are looking for a specific subset of sequences that satisfy the 'stopping time' condition. We must account for the constraints that prevent the game from ending earlier.

Q: Why do we subtract 1 from the total outcomes of $(T_1, T_2)$?

We subtract 1 because the pair $(4, 4)$ is a 'forbidden' state for the first two throws. If the first two throws were $(4, 4)$, the game would have ended at $T_2$, which violates our requirement that the game ends at $T_5$. By subtracting this one case, we ensure that the game continues past the second throw.

Q: What if $T_3$ was 4? Why does that matter?

If $T_3$ were 4, then the sequence would look like $(T_1, T_2, 4, 4, 4)$. The pair $(T_3, T_4)$ would be $(4, 4)$, meaning the game would have ended at the fourth throw. Since we are calculating the probability of the game ending *at* the fifth throw, we must exclude any scenario where the game ends at the fourth throw. Thus, $T_3$ cannot be 4.

Question

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

Options:

(A) $150/6^5$

(B) $175/6^5$

(C) $200/6^5$

(D) $225/6^5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the threshold of a probability experiment. You have a fair, six-sided die in your hand, and the rule is simple: you roll it repeatedly until you see two fours in a row.

We want to know the probability that this game ends exactly on the fifth throw. Let us break this down into a sequence of five slots: $T_1, T_2, T_3, T_4, T_5$.

### The Ending Constraint

For the game to end at the fifth throw, the condition of 'two consecutive fours' must be met for the first time at $T_5$. This means the fourth and fifth throws must be fours.

We lock these in: $T_4 = 4$ and $T_5 = 4$. There is only one way for this to happen for each slot.

To ensure the game ends *exactly* at the fifth throw, we must ensure it did not end at the fourth. This implies that the pair $(T_3, T_4)$ cannot be $(4, 4)$. Since $T_4$ is already $4$, this forces $T_3 
eq 4$.

### The $T_3$ Trap

Now, let us look at $T_3$. We have established that $T_3$ cannot be $4$.

Since a die has six faces $\{1, 2, 3, 4, 5, 6\}$, and $T_3$ cannot be $4$, there are exactly $6 - 1 = 5$ possible values for $T_3$. This is a crucial step to avoid counting sequences where the game ended early.

### The Beginning Constraint

Finally, we look at the first two throws, $T_1$ and $T_2$. The game must not have ended at the second throw, meaning the pair $(T_1, T_2)$ cannot be $(4, 4)$.

The total number of outcomes for two throws is $6 	imes 6 = 36$. We subtract the one forbidden outcome, $(4, 4)$, leaving us with $36 - 1 = 35$ valid pairs for the first two throws.

We do not need to worry about the game ending at the third throw because we have already constrained $T_3$ to not be $4$. This automatically prevents the pair $(T_2, T_3)$ from being $(4, 4)$.

### Synthesis and Calculation

The number of favorable outcomes is the product of the number of ways to fill each slot: $35$ ways for $(T_1, T_2)$, $5$ ways for $T_3$, and $1$ way each for $T_4$ and $T_5$.

Multiplying these gives us:
$$35 	imes 5 	imes 1 	imes 1 = 175$$

The total number of possible outcomes for five throws is $6^5 = 7776$. Therefore, the probability is:
$$\frac{175}{6^5} = \frac{175}{7776}$$

This elegant result shows how constraints in probability act like filters, narrowing down the vast space of possibilities to the specific events we care about. The final probability is **$\frac{175}{7776}$**.

Animated Solution for Mathematics - Probability: In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e E x p er im e n t$

$F i x in g t h e E n d in g C o n d i t i o n$

$T h e C o n s t r ain t o n T_{3}$

$C h o i ces f or T_{3}$

$C o n s t r ain t so n T_{1} an d T_{2}$

$C o u n t in g V a l i d P ai r s f or (T_{1}, T_{2})$

$T o t a l F a v or ab l e O u t co m es$

$F ina l P r o babi l i t y C a l c u l a t i o n$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Ending Constraint

The $T_{3}$ Trap

The Beginning Constraint

Synthesis and Calculation

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