How can I express the dice probability for rolling dice vs a target number with re-rolls?
There are a number of dice rolled to get a target number of hits.
- A "hit" is a 4, 5, or 6 on 1d6.
- Every 6 can be re-rolled to get
another hit. - There is a target number of hits required for success.
For example, if you rolled 4 dice with a target of 2 hits, I would expect the chance of success to be around 66%.
However, if you wanted to calculate the odds of getting 5 hits on 4 dice, you would need to know the odds of getting a 6 on any of those 4 die and then getting a 4,5,or 6 to get that additional required hit.
I have read several articles, such as Ed Collins site, but I cannot get it to work in a Google spreadsheet.
From that, I have gathered: 1 - (3/6)^4 for 4 dice and one hit.
For 4 dice with 2 hits, I tried multiplying both probabilities together: 1 - (3/6)^4 * (3/6)^4
I also tried subtracting the total number of not 4,5,6 and then subtracting that from the total number of combinations and dividing the probability from that.
But that does not come out as expected either.
For re-rolling a 6, I looked at this post, but do not understand it:
probability involving rerolling dice
Is there a single Google formula that can represent the typical situation of rolling multiple dice vs a target number and then also the odds of re-rolling to achieve the success?
Best Answer
For a single die, let's find the expected value for the number of hits.
You have a $\dfrac{1}{2}$ chance of not getting any hits on one die.
You have a $\dfrac{2}{6} + \dfrac{1}{6}\cdot \dfrac{1}{2} = \dfrac{5}{12}$ chance of getting exactly 1 hit on one die.
You have a $\dfrac{1}{6}\cdot \dfrac{1}{3}+\dfrac{1}{6^2}\cdot \dfrac{1}{2} = \dfrac{5}{72}$ chance of getting exactly 2 hits.
In general, for $h>0$, the probability of exactly $h$ hits is $$\dfrac{5}{2\cdot 6^h}$$
I am not going to go into the proof of this, but I will demonstrate that the total probability is preserved.
$$\sum_{h\ge 1} \dfrac{5}{2\cdot 6^h} = \dfrac{5}{12}\sum_{h\ge 0} \left(\dfrac{1}{6}\right)^h = \dfrac{5}{12}\cdot \dfrac{6}{5} = \dfrac{1}{2}$$
Plus the probability of $h=0$, which is $\dfrac{1}{2}$.
So, this gives the expected value:
$$\dfrac{5}{12}\sum_{k\ge 0}(k+1)\left(\dfrac{1}{6}\right)^k = \dfrac{3}{5}$$
So, that is the expected value for the number of hits.
Continuing, let's create some notation.
Let $P(n,h)$ be the probability of rolling $n$ dice and getting exactly $h$ hits. Let $P(n,h^+)$ be the probability of rolling $n$ dice and getting at least $h$ hits.
$$P(n,h) = \sum_{\sum_{i=1}^n a_i = h}\prod_{i=1}^n\begin{cases}\tfrac{1}{2}, & a_i = 0 \\ \tfrac{5}{2\cdot 6^{a_i}}, & a_i>0\end{cases}$$
Basically, you are summing over all nonnegative integer solutions to the Diophantine equation: $$a_1+\cdots + a_n = h$$
This tells you which dice rolled hits, and the total number of hits is $h$. Then, multiply the probabilities that the chosen die scores that many hits.
This is an impractical approach. Instead, we can try to break down the probabilities.
Example:
$$P(4,2^+) = 1-P(4,0)-P(4,1) = 1-\left(\dfrac{1}{2}\right)^4-\dbinom{4}{1}\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^3 \approx 72\%$$
What I am doing here is for $P(4,0)$, this is the probability that not a single die rolled a hit. For $P(4,1)$, this is the probability that exactly one die rolled exactly one hit.
$$P(4,5^+) = 1-P(4,0)-P(4,1)-P(4,2)-P(4,3)-P(4,4)$$
This is a bit trickier to calculate.
$$\begin{align*}P(4,4) & = \dbinom{4}{1}P(1,4)P(3,0)+\dbinom{4}{2}\dbinom{2}{1}P(1,3)P(1,1)P(2,0)+\dbinom{4}{2}P(1,2)^2P(2,0)+\dbinom{4}{3}\dbinom{3}{1}P(1,2)P(1,1)^2P(1,0)+P(1,1)^4 \\ & = 4\left(\dfrac{5}{2\cdot 6^4}\right)\left(\dfrac{1}{2}\right)^3+12\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+6\left(\dfrac{5}{72}\right)^2\left(\dfrac{1}{2}\right)^2+12\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)+\left(\dfrac{5}{12}\right)^4 \\ & = \dfrac{865}{6912}\end{align*}$$
Essentially, what I am calculating is the probability of the following:
Only one die hits, but four times
2 dice have hits, one rolls 3 hits one rolls 1 hit
2 dice hit, both roll 2 hits
3 dice hit, one rolls 2 hits, two others roll 1 hit each
4 dice hit, one hit each
$$P(4,3) = \dbinom{4}{1}\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{1}{2}\right)^3 + \dbinom{4}{2}\dbinom{2}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+\dbinom{4}{3}\left(\dfrac{5}{12}\right)^3\left(\dfrac{1}{2}\right) = \dfrac{205}{864}$$
$$P(4,2) = \dbinom{4}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{1}{2}\right)^3+\dbinom{4}{2}\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)^2 = \dfrac{85}{288}$$
$$P(4,5^+) = \dfrac{55}{768}$$
Edit: How to build an Excel spreadsheet to perform these calculations: