0
$\begingroup$

I am unsure of why we sometimes subtract from the exponent in a probability formula.

For example, for this question:

A six-sided die is tossed. If it turns up a 1 or 2 the player wins. If it turns up 5 or 6 the player loses. If it turns up 3 or 4 the die is repeatedly tossed until either the same number as the first toss turns up (in which the player wins) or a 5 or 6 turns up in which case the game is lost.

Let $N$ represent the number of tosses until the game stops. Find a formula for $P(N = n)$.

The answer for $n = 1$ is $2/3$, but the answer for n greater or equal $2$ is $$\frac{1}{6}\left(\frac{1}{2}\right)^{n-2}$$

I'm guessing the numbers $1/6$ and $1/2$ are the probabilities for success and failure? I'm also confused about why you need to subtract two from $n$?

$\endgroup$
1
$\begingroup$

Suppose $n > 1$. How is it possible that $N = n$? We must get a $3$ or $4$ on the first toss (probability $1/3$), then the next $n-2$ tosses can't be $5, 6$, or the first number that was rolled (probability $(1/2)^{n-2}$), then the final toss must be either $5, 6$, or the same as the first toss (probability $1/2$). The probability of this sequence of events occurring is $(1/3) \cdot (1/2)^{n-2} \cdot (1/2)$.

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer. I still don't get the n - 2 bit, could you explain why you have the n -2 value and how you know to put it on top of the 1/2? And just another thing, how do you know to multiply all these values? Is it from a tree diagram or something? $\endgroup$ – chris Jun 3 '19 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.