Probability of throwing two dice and getting one 3 / 5.

Question

Imagine playing the game Yahtzee and you are going for a large straight ( every dice fits in, example: $1, 2, 3, 4, 5$). On your first roll you get $1, 2$ and $3$ already, but you still need the $4$ & $5$. You have two rolls left. Now I want to know how big the probability is, that you get one $4$ and one $5$ in the next $2$ rolls.

I can't wrap my head around it. Even worse: I want to calculate the same thing for the case you get only $1$ and $2$ on your first throw and now you need a $3$, $4$ and $5$ in your next two throws...

If anybody has an idea let me know :)

Answer 1

Hint on first case:

There are $3$ mutually exclusive routes:

• in the first of the two rolls you throw a $4$ and a $5$
• in the first of the two rolls you throw exactly one of the numbers $4$ and $5$ and in the second of the two rolls you throw the other number.
• in the first of the two rolls you throw not a $4$ and not a $5$ and in the second you throw both.

Find the three corresponding probabilities. Then - because the routes are mutually exclusive - the summation of these probabilities is the answer.

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Edit concerning the second problem (needed are $3$, $4$ and $5$ in at most two throws)

If $\hat{S}$ is a random set then for a finite set $S$ it can be proved that: $$P\left(\hat{S}=S\right)=\sum_{T\subseteq S}P\left(\hat{S}\subseteq T\right)\left(-1\right)^{\left|S\right|-\left|T\right|}\tag1$$

This result shows up if we start with the observation that: $$P\left(\hat{S}=S\right)=P\left(\hat{S}\subseteq S\right)-P\left(\bigcup_{s\in S}\left\{ \hat{S}\subseteq S-\left\{ s\right\} \right\} \right)$$ and apply inclusion/exclusion on the second term on RHS.

Further if there is symmetry in the sense that probabilities like $P\left(\hat{S}\subseteq T\right)$ only depend on the cardinality of $T$ then $(1)$ can further be worked out, resulting in: $$P\left(\hat{S}=S\right)=\sum_{k=0}^{\left|S\right|}\binom{\left|S\right|}{k}\left(-1\right)^{\left|S\right|-k}p_{k}\tag2$$ where $p_{k}$ stands for $P\left(\hat{S}\subseteq T\right)$ whenever $\left|T\right|=k$.

If moreover $\hat{S}$ takes values in $\wp X$ where $\left|X\right|=m$ then we can deduce from $(2)$ easily that: $$P\left(\left|\hat{S}\right|=n\right)=\binom{m}{n}\sum_{k=0}^{n}\binom{n}{k}\left(-1\right)^{n-k}p_{k}\tag3$$

Now let $D_{1},D_{2},D_{3}$ denote the outcomes of $3$ dice and let: $$\hat{S}:=\left\{ D_{1},D_{2},D_{3}\right\} \cap\left\{ 3,4,5\right\} $$ Then we have the following expression for the probability that you are seeking for:

$$P\left(\left|\hat{S}\right|=3\right)+P\left(\left|\hat{S}\right|=2\right)\frac{1}{6}+P\left(\left|\hat{S}\right|=1\right)\frac{2}{36}+P\left(\left|\hat{S}\right|=0\right)P\left(\left|\hat{S}\right|=3\right)\tag4$$ In this context we have: $$p_{k}=\left(\frac{3+k}{6}\right)^{3}\text{ for }k=0,1,2,3$$and based on $(3)$ we find:

• $P\left(\left|\hat{S}\right|=0\right)=\binom{3}{0}p_{0}=\frac{27}{216}$
• $P\left(\left|\hat{S}\right|=1\right)=\binom{3}{1}\left(p_{1}-p_{0}\right)=3\left(\frac{64}{216}-\frac{27}{216}\right)=\frac{111}{216}$
• $P\left(\left|\hat{S}\right|=2\right)=\binom{3}{2}\left(p_{2}-2p_{1}+p_{0}\right)=3\left(\frac{125}{216}-\frac{128}{216}+\frac{27}{216}\right)=\frac{72}{216}$
• $P\left(\left|\hat{S}\right|=3\right)=\binom{3}{3}\left(p_{3}-3p_{2}+3p_{1}-p_{0}\right)=\frac{216}{216}-\frac{375}{216}+\frac{192}{216}-\frac{27}{216}=\frac{6}{216}$

Substituting these values in $(4)$ we find:

$$\frac{6}{216}+\frac{72}{216}\times\frac{1}{6}+\frac{111}{216}\times\frac{2}{36}+\frac{27}{216}\times\frac{6}{216}=\frac{5382}{46656}=\frac{299}{2592}$$