How to find the probability of $P(Y=6)$

Question

Anna writes down a random sequence created by the following process. She repeatedly rolls a fair 6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if any), then she writes the new number down and then continues rolling. Otherwise, she does not write the new number down and the process ends.

Let X be the length of Anna’s sequence, and Y be the last number in her sequence.

For example, if Anna rolled 1 then 4 then 5 then 4, her sequence would be 1,4,5 and the random variables X, Y would take values X = 3 and Y = 5.

Question:

what is $P(Y=6)$

My method

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I know I can write down all the possible ways that the last number of the sequence could be 6, and then I add up all the probabilities.

1. $1,2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})(1)$
2. $2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})$
3. $3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$
4. $1,2,4,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$

etc.

But this method is way too time-consuming as there are too many possible sequences.

I also tried to do it a different way by using $1 - $(The possible sequences that the last number is not 6), but there are even more sequences to take into account this way.

Can anyone show me a much more efficient way of finding the solution to this problem?

Answer 1

$Y = 6$ means we have maximum of six rolls. If none of the six rolls is $6$, we would have at least one of the numbers between $1$-$5$ appear twice breaking the sequence and the process would stop.

Probability of getting $6$ in the first roll is $\frac{1}{6}$

Probability of getting $6$ in the second roll is $ \frac{5}{6} \cdot \frac{1}{6}$.

Now what is the probability of getting $6$ in the third roll? The first two rolls have to one of the remaining $5$ numbers and in the increasing order, which is simply ${5 \choose 2}$ out of $6^2$ possibilities as the order is pre-determined. The third roll has to be $6$.

Similarly for getting $6$ in the fourth, fifth or the sixth roll. So,

$ \displaystyle \small P(Y = 6) = \frac{1}{6} \cdot \left[1 + {5 \choose 1} / 6 + {5 \choose 2} / 6^2 + {5 \choose 3} / 6^3 + {5 \choose 4} / 6^4 + {5 \choose 5} / 6^5\right]$

Answer 2

Question:

what is P(Y=6)

observe that the sequence's length can only be $X \in\{1,2,3,4,5,6\}$

$X=1$ when you roll immediately 6

$X=5$ when you roll 4 numbers in increasing order, and then 6. The only favourable sequences are, in this case

$1234$

$1235$

$1245$

$1345$

$2345$

...and 6 on the fifth roll. Observe that these cases are exactly $\binom{5}{4}$

thus here the probability to get $Y=6$ is $\frac{5}{6^4}\cdot\frac{1}{6}$

Using this procedure, you easy get

$$\mathbb{P}[Y=6]=\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{10}{6^2}\cdot\frac{1}{6}+\frac{10}{6^3}\cdot\frac{1}{6}+\frac{5}{6^4}\cdot\frac{1}{6}+\frac{1}{6^6}=\approx 0.36$$

Answer 3

Note you do not need to count each single combinations, as any combinations of the same length, as any of these have the same probability of happening. So fixing that the last position must be $6$, there is $0,...,5$ free places. If we try to assign the numbers $1,...,5$ to these places we get ${5\choose k}$ possibilities. This gives us a probability of $$ \sum_{k=0}^5 {5 \choose k} 6^{-(k+1)}$$