I'm analyzing the game Yahtzee in which a player can roll five dice three times and at each interval can choose to roll/not roll certain dice or stop rolling altogether. The goal is to get different combinations like four-of-a-kind, a straight (1-2-3-4-5 or 2-3-4-5-6), etc.

I found that binomial probability was exactly what I needed. Using the formula `P(X=r) = nCr * pʳ * (1-p)ⁿ⁻ʳ`

I could basically say "if I roll five dice, what are the chances 3 of them match" and that formula gives me a percentage chance. That works great!

However, referencing the research done here, your chances of getting a Yahtzee (five of a kind) is about 4% if you use all three rolls. Your chances of rolling a Yahtzee in your first roll is 0.08% but the probability of doing it over three rolls is much higher because you could set aside some matches and only roll the remaining dice you need to match.

**But how can you calculate this at each interval?**

**Example:** Say I want a four-of-a-kind and I roll a `4 5 3 4 2`

. A smart player would re-roll the `5 3 2`

in hopes of getting more `4`

's. Binomial probability tells us you have a 6.94% chance of getting two `4`

's in the next roll. So how can I calculate the chance I get two `4`

's in either the next roll or the one after that?

Here's a representation of how you can accomplish this feat. If you need 2 `4`

's, are rolling three dice, and can roll twice, you have the following win possibilities:

- (1) 4 X X – (2) 4 X
- (1) X X X – (2) 4 4 X
- (1) 4 4 X

I can use binomial probability to calculate the chances of getting each of these. Should I average them to get what I need? I feel like I'm so close, hopefully one of you guys knows a formula or can help me out.

## Best Answer

Some formality and structure should help you tackle the problem you mention in your example.

Take $E$ as the event of getting at least two more fours in the next roll or the one after that, $X$ as the number of fours in the first roll, and $Y$ as the number of fours in the second roll.

Using total law of probability, $$\begin{eqnarray*}\mathbb{P}(E)=\mathbb{P}(E|X=0)\mathbb{P}(X=0)+\mathbb{P}(E|X=1)\mathbb{P}(X=1)+\mathbb{P}(E|X\geq 2)\mathbb{P}(X\geq 2)\end{eqnarray*}$$ Since $\mathbb{P}(E|X\geq 2)=1$ and $X\sim \text{Binomial}(3,1/6)$ the aforementioned reduces to $$\mathbb{P}(E)=\frac{125}{216}\mathbb{P}(E|X=0)+\frac{25}{72}\mathbb{P}(E|X=1)+\frac{2}{27}$$ Now for $x=0,1$ $$\mathbb{P}(E|X=x)=\sum_{y=0}^{3-x}\mathbb{P}(E|X=x,Y=y)\mathbb{P}(Y=y|X=x)$$ Because $\mathbb{P}(E|X=x,Y=y)=1_{\{x+y\geq 2\}}$ and $Y|X\sim \text{Binomial}(3-X,1/6)$ we have $$\mathbb{P}(E|X=0)=\mathbb{P}(Y\geq 2|X=0)=\frac{2}{27}$$ Furthermore, $$\mathbb{P}(E|X=1)=\mathbb{P}(Y\geq 1|X=1)=\frac{11}{36}$$ Putting all this together, $$\mathbb{P}(E)=\frac{5203}{23328}\approx 0.22$$