What's the equity of the number of highest tosses in an infinite series of cube tosses?

Question

A fair cube (values from 1 to 6) is tossed an infinite amount of times. $Y$ is the number of times a toss showed a number bigger than all the previous tosses. What's the expected value of $Y$?

I can't solve this with any of the famous distributions I know such as geometric/binomial etc...

I thought that maybe it can be solved with $\mathbb E [Y]=\sum_i^\infty P(Y\geq i)$ But I'm not sure how to calculate each one of the results.

What am I missing?

Answer 1

For $i=1,2,3,4,5,6$ let $\mu_i$ denote the expectation of the number of times a toss shows a bigger number than all previous tosses if we use a die that has $i$ distinct faces.

Then we are actually looking for $\mu_6$.

Further we find the following equalities:

• $\mu_1=1$
• $\mu_2=1+\frac12\mu_1$
• $\mu_3=1+\frac13\mu_1+\frac13\mu_2$
• $\mu_4=1+\frac14\mu_1+\frac14\mu_2+\frac14\mu_3$
• $\mu_5=1+\frac15\mu_1+\frac15\mu_2+\frac15\mu_3+\frac15\mu_4$
• $\mu_6=1+\frac16\mu_1+\frac16\mu_2+\frac16\mu_3+\frac16\mu_4+\frac16\mu_5$

Some explanation.

If the die has only one face then after throwing once we allready reached the highest possible value so no new records are to be expected. This tells us that $\mu_1=1$.

If the die has two faces then two equiprobable events are possible. The highest value is reached (and no new records are possible) or the lower value is reached. In the second case all throws that follow and do not exceed the first throw can be disregarded so actually we are in the same situation as we are if the die has only one face. That together leads to $\mu_2=\frac12\cdot1+\frac12\cdot(1+\mu_1)=1+\frac12\mu_1$.

Likewise we find the other equalities, and it is not really difficult to find $\mu_6$ on base of them:

• $\mu_1=\frac{60}{60}$
• $\mu_2=\frac{90}{60}$
• $\mu_3=\frac{110}{60}$
• $\mu_4=\frac{125}{60}$
• $\mu_5=\frac{137}{60}$
• $\mu_6=\frac{147}{60}$

Answer 2

How many times will you roll a $6$ with it being higher than all the previous rolls? Exactly once - when you roll the first $6$.

What about a $5$? Now this will happen at most once, and only if the first $5$ occurs before the first $6.$ This has probability $1/2$ (the first number $>4$ is equally likely to be a $5$ or a $6$). So the expected number of times is $1/2$.

What about a $4$? Again, it will happen at most once, and it will happen if the first $4$ is before the first $5$ and the first $6$. So the expected number of times this happens is $1/3$.

Continuing in this manner, the total expectation is $1+1/2+1/3+1/4+1/5+1/6$.