1

Original question link I have difficulties in understanding Prof. Christian Blatter's answers, especially for the first equation see below
$$E_X=1+{1\over 6}\cdot 0+{5\over 6}\cdot E_X\ .$$ Could anyone kindly help me to explain the equation?

Community
  • 1
luoshao23
  • 113
  • 1
    Each time you roll (or reroll) the die, it adds $1$ to the total number of times that you have rolled the die so far. You will need zero more rolls $\frac{1}{6}$ of the time. Otherwise, the remaining $\frac{5}{6}$ of the time you will still need $E_X$ many more rolls. – JMoravitz Sep 04 '19 at 15:58
  • If you prefer a more direct approach, rather than manipulating $E_X$ where it appears on both the left and the right of the equation., my answer here gives a proof as to why the expected number of independent iterations until the first success is $\frac{1}{p}$, giving the number of rolls of a die until the first $5$ as being $\frac{1}{1/6} = 6$. – JMoravitz Sep 04 '19 at 16:04
  • @JMoravitz, read through your answer to the similar question, it is also a good solution which is more direct. Now I have two alternatives. Thank you! – luoshao23 Sep 05 '19 at 01:33

1 Answers1

0

$E_x$ is the expected time to roll the first $5$. When you throw a dice, it's either going to be $5$ with the probability $\frac{1}{6}$ and the rest with the probability $\frac{5}{6}$.

Once you have thrown the dice, for the first time, you add $1$ to $E_x$ (The first term, $+1$). At the first roll, as I said, it's going to be $5$ with the probability $\frac{1}{6}$. But then you don't have to throw anymore, so there is nothing to add to $E_x$(That's $\frac{1}{6}\cdot0$ ). Lastly, If you didn't throw $5$ at the first roll with probability $\frac{5}{6}$, you then start the process from the beginning. In other words, since we have added $1$ in the equation at the beginning, which counts for the first throw, we can just think the second throw as a first throw of a new process to count $E_x$. Therefore, with the probability $\frac{5}{6}$, we start this process again. (The last term $\frac{5}{6}E_x$)

underdisplayname
  • 1,240