How many times does it take to look at the six sides of a dice

How many times does it take to look at the six sides of a dice – Penden’s Notes on Probability Theory

source

When I was watching the video a few days ago, I found such a question

answer

Reduced to the coin problem

general practice

• Assuming that you can see the front and back of the coin twice, then the situation may be "front and back" or "anyway" (the other two are "front and back", "reverse and negative"), and the probability is 1 2 \frac{ $\frac{1}{2}$；
• Assuming that you can only see the front and back of the coin three times, then the situation may be "heads and tails" or "heads and tails" (the other two are "heads and heads" and "reversal and back"), and the probability is 1 4 \ $\frac{1}{4}$(Because of "positive positive", the probability of "anti-negative" is $\frac{1}{2}$)；
• And so on...

n	2	3	$\cdots$	k
p	$\frac{1}{2}$	$\frac{1}{4}$	$\cdots$	$\frac{1}{2^{k-1}}$

$$\begin{array}{rl}in\_& =2\ast \frac{1}{2}+3\ast \frac{1}{4}+\cdots +k\ast \frac{1}{2^{k-1}}\\ 2En& =2+3\ast \frac{1}{2}+\cdots +k\ast \frac{1}{2^{k-2}}\\ downminusupin& =2+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{k-2}}-k\ast \frac{1}{2^{k-1}}=3\end{array}$$

recursive method

General $E_2$Recorded as the average number of times used to see two people, $E_1$Recorded as the average number of times E 2 used to see one side
$$\begin{array}{rl}{E}_2& =\frac{1}{2}(1+E_1)+\frac{1}{2}(1+E_1)\end{array}$$
Where the previous $\frac{1}{2}(1+E_1)$ represents the average number of times required to cast heads for the first time (this$E_1$Indicates the average number of times required to cast to the tail), the latter $\frac{1}{2}(1+E_1)$ represents the average number of times required to cast tails for the first time (this$E_1$represents the average number of tosses required to land heads);

E $E_1$If it means the average number of times
$$E_1=\frac{1}{2}+\frac{1}{2}(1+E_1)$$
where the previous$\frac{1}{2}$Indicates that the first cast is negative, and the next one is $\frac{1}{2}{E}_1$Indicates the first time it was voted heads;

E 1 = 2 E 2 = 3 can be solved from E_1 = 2 \\ E_2 = 3
$$\begin{gathered} E_1=2 \\ {E}_2=3 \end{gathered}$$

back to the dice problem

If the general solution is still used for dice, it will be very complicated; so take the recursive method
$$\begin{array}{rl}{E}_6& =\frac{1}{6}(1+E_5)+\cdots +\frac{1}{6}(1+E_5)\\ & =(1+E_5)\\ E_5& =\frac{1}{6}(1+E_5)+\frac{5}{6}(1+E_4)\\ & =\frac{6}{5}+E_4\\ E_4& =\frac{2}{6}(1+E_4)+\frac{4}{6}(1+E_3)\\ & =\frac{6}{4}+E_3\\ E_3& =\frac{3}{6}(1+E_3)+\frac{3}{6}(1+E_2)\\ & =\frac{6}{3}+E_2\\ E_2& =\frac{4}{6}(1+E_2)+\frac{3}{6}(1+E_1)\\ & =\frac{6}{2}+E_1\\ E_1& =\frac{5}{6}(1+E_1)+\frac{1}{6}\end{array}$$

解得
$$\begin{gathered} E_1=6 \\ {E}_6=1+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+6 \end{gathered}$$