Probability of dice sum simply more than 100

It is a wonderful trouble. The opportunity of striking $101$ first is remarkably huge. Allow $a(x)$ be the opportunity of striking $101$ first, beginning at $x$.

We address recursively by establishing $a(106)=a(105)=a(104)=a(103)=a(102)=0$ and also $a(101)=1$. After that, for $x$ from $100$ to $0$, placed $a(x)={1\over 6}\sum_{j=1}^6 a(x+j)$. By the revival theory, $a(x)$ merges to $1/\mu$, where $\mu=(1+2+3+4+5+6)/6=7/2$. The value $a(0)$ is really near this restriction.

Actually, the entire striking circulation is about $[6/21,5/21,4/21,3/21,2/21,1/21]$.

Included : Let $a^\prime(x)$ be the opportunity of striking $102$ first, beginning at $x$. After that $a^\prime(x)$ pleases the very same reappearance as $a(x)$ for $x\leq 100$, yet with various border problems $a^\prime(106)=a^\prime(105)=a^\prime(104)=a^\prime(103)=a^\prime(101)=0$ and also $a^\prime(102)=1$. Consequently
$$a^\prime(x)=a(x-1)-a(100)a(x)$$ and also $$a^\prime(0)\approx {1\over \mu}-{1\over 6}{1\over \mu}={5\over 21}. $$ The remainder of the striking circulation can be assessed in a comparable means.

My effort : a mix of my comment and also Shai is comment.

A dividing suggests a dividing of n making use of just 1,2,3,4,5,6.

Variety of means to reach 101 : Number of dividings of 101.

Variety of means to reach 102 : dividings of 96+dividings of 97 ...+dividings of 100. Given that we can have 96+6, 97+5, ...100+2.

...

Variety of means to reach 106 : Partitions of 100. Given that we can reach 106 just by including in 100 after that rolling 6.

A dividing of n will certainly be the umpteenth coefficient x in $\prod_{k=0}^6 \frac{1}{1-x^k}$ making use of the geometric collection. Yet a less complicated monitoring is that $a_n = a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}+a_{n-6}$. Utilizing this

Number of means to reach 101 is $a_{101}=a_{100}+a_{99}+a_{98}+a_{97}+a_{96}+a_{95}$ Number of means to reach 102 is $a_{100}+a_{99}+a_{98}+a_{97}+a_{96}$

...

Number of means to reach 106 is $a_{100}$

Since $a_n > 0$ we see that 101 will certainly take place one of the most regularly.

Maybe I really did not recognize the inquiry, yet it appears clear, taking into consideration $\lbrace 95 + j + k:j = 0, \ldots ,5;k = 1, \ldots 6 \rbrace$, that you need to pick $101$. Below, $95+j$ represents the amount prior to going beyond $100$ for the very first time, and also $95+j+k$ to the amount when going beyond $100$ for the very first time.

EDIT :

Let me specify a little on my argument over. First, you can conveniently show by induction on $j$ that if the existing amount is $100 - j$, $j=0,\ldots,4$, after that there is no purely far better selection than $101$ (the base instance $j=0$ is unimportant). After that take into consideration the instance where the existing amount is $95$, in conclusion that $101$ is purely the most effective selection.

101 is one of the most potential stop state.

The procedure "replace the last throw of the dice by the one that offers an amount of 101" is a probability - preserving makeover on the set of feasible video games. It does not cover the entire set (i.e. the photo of the makeover is a part) of feasible video games quiting at 101, such as the ones with a change from 95 to 101 by a roll of 6. This reveals that all the terminal states greater than 101 have a smaller sized probability than that of ending up at 101.