# Probability of dice sum simply more than 100

Can a person please overview me to a means through which I can address the adhering to trouble. There is a die and also 2 gamers. Moving quits as quickly as some goes beyond 100 (not consisting of 100 itself). Therefore you have the adhering to selections : 101, 102, 103, 104, 105, 106. Which need to I pick offered front runner. I'm assuming Markov chains, yet exists a less complex means?

Many thanks.

MODIFY : I created dice as opposed to die. There is simply one die being rolled

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2019-12-02 02:48:26
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Answers: 4

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.

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2019-12-03 05:11:48
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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.

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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.

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2019-12-03 04:23:35
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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.

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2019-12-03 04:21:53
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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.

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2019-12-03 04:15:54
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