# Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.

Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.

I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.

Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.

PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.

• The probability asked for in the title $P(\text{D20 beats 3 D6s})$ is the opposite of the probability asked for in the details $P(\text{3 D6s beats D20})$. – N. Shales May 11 '17 at 1:57

Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$ Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$

Given any particular value of $X,$ the probability that the twenty-sided die will roll higher is $$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$

The overall probability that the twenty-sided die will roll higher than the total on the other three dice is \begin{align} P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\ &= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n). \end{align}

The last line of that set of equations is just the expected value of $1 - \frac1{20}X.$ That is, \begin{align} P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\ &= 1 - \frac1{20} \mathbb E[X] \\ &= 1 - \frac1{20} \left(\frac{21}{2}\right) \\ &= \frac{19}{40}. \end{align}

If the question is actually the one posed in the original question body rather than in the original title, namely the probability that $X > Y,$ then we simply observe that for any given value of $X,$ $$P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$

The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$ We find that \begin{align} P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\ &= \frac1{20} \mathbb E[X] - \frac1{20}\\ &= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\ &= \frac{19}{40}. \end{align}

This should not be surprising, because it also follows from $P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$

• Please see my answer for a simple proof that the two probabilities are equal (and hence both equal to $19/40$). – Barry Cipra May 11 '17 at 13:50
• I don't know if this counts as a shortcut, but to find the expected value of X faster, you could find the expected value of one dice and just multiply it by three instead of summing all the numbers from 3-18. I feel like that just works better in my head. – Henry Weng May 11 '17 at 14:47

Whatever Bob rolls with the $6$-sided dice has a $1$ in $20$ chance of being matched, for a tie, by Bill's roll of the $20$-sided die. Whatever sum, $S=a+b+c$, Bob rolls, if you turn his dice over, the sum is $(7-a)+(7-b)+(7-c)=21-S$. Similarly, whatever number $T$ Bill rolls, if you turn the $20$-sided die over, the number is $21-T$. Thus for each outcome in which Bob wins, there is an equally likely outcome in which Bill wins, and vice versa. Hence the probability of winning for each of them is the same, namely

$${1\over2}\left(1-{1\over20}\right)={19\over40}$$

Remark: It's not literally necessary that the "complementary" number for each side of a die be the opposite face, just that there be a complemenary number somewhere. For $6$-sided dice, having opposite faces sum to $7$ is fairly standard; I believe it's also standard for $20$-sided dice to have opposite faces sum to $21$.

Also, I'd like to credit David K's answer with motivating this one. When I saw from his analysis that the two probabilities were equal, I decided there ought to be simple reason why. As luck would have it, I found one.

• The connection to my answer is even deeper than the coincidence that $P(Y>X)=P(Y<X)$: the usual easy method to calculate $E[X]$ uses the same "turn the dice over" argument (or something equivalent). But the really nice observation here is that the exact same symmetry works for both dice. – David K May 11 '17 at 13:57
• @DavidK, thanks. The symmetry principle also suffices, for example, to show that it's a fair game to roll $13$ regular dice against $7$ dodecahedral dice. In that case, however, computing the probability of a tie is not so easy. (Or is it?) – Barry Cipra May 11 '17 at 14:10

Your best bet is to keep track of two things for both sides: how likely this particular result is, and how likely anything less than this result is. Given these we can multiply them together relatively easily.

$$\begin{array}{r|rr|rr|r} x & 3\text{d}6 = x & 3\text{d}6 < x & 1\text{d}20 = x & 1\text{d}20 < x & 1\text{d}20 < x \cap 3\text{d}6 = x\\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 & 2 & 2 \\ 4 & 3 & 1 & 1 & 3 & 9 \\ 5 & 6 & 4 & 1 & 4 & 24 \\ 6 & 10 & 10 & 1 & 5 & 50 \\ 7 & 15 & 20 & 1 & 6 & 90 \\ 8 & 21 & 35 & 1 & 7 & 147 \\ 9 & 25 & 56 & 1 & 8 & 200 \\ 10 & 27 & 81 & 1 & 9 & 243 \\ 11 & 27 & 108 & 1 & 10 & 270 \\ 12 & 25 & 135 & 1 & 11 & 275 \\ 13 & 21 & 160 & 1 & 12 & 252 \\ 14 & 15 & 181 & 1 & 13 & 195 \\ 15 & 10 & 196 & 1 & 14 & 140 \\ 16 & 6 & 206 & 1 & 15 & 90 \\ 17 & 3 & 212 & 1 & 16 & 48 \\ 18 & 1 & 215 & 1 & 17 & 17 \\ 19 & 0 & 216 & 1 & 18 & 0 \\ 20 & 0 & 216 & 1 & 19 & 0 \\ \hline \text{total} & 216 & & 20 & & 2052 \end{array}$$

Total up everything in the rightmost column, divide by the totals for the $3\text{d}6 = x$ and $1\text{d}20 = x$ columns, and you win: Bob wins $\frac{2052}{4320} = \frac{19}{40} = 0.475$ of the time.

To get ties, or where Bill wins, change what pairs you multiply: both $=$ ones for ties, and $3\text{d}6 < x$ and $1\text{d}20 = x$ for Bill's wins.

This particular one is interesting: because both distributions are symmetrical, and they have the same mean, Bill and Bob will both win the same proportion of the time.

It occurs to me that there's another part to this question: how do we efficiently calculate result probabilities for combinations of dice?

The answer to that is an operation called convolution, which I'll present here in discrete form.

Given two functions $f(x)$ and $g(x)$, the convolution is $$(f * g)(x) = \sum_{k = -\infty}^\infty f(k)g(x-k)$$

This can be interpreted in probability theory as the following: we have two random variables $f$ and $g$, with probability functions $f(x)$ and $g(x)$. the probability function for $f + g$ -- adding the two results together -- is equal to $(f * g)(x)$.

Obviously with those infinities in there, we have to fiddle with it a little to actually get anything done. In our case, because we're dealing with dice, our functions have what's called limited support: they're only non-zero in a small area, so we only need to cover that small area.

Let's do a specific example. Say I want to calculate the probability that I'll get a $7$ on $3\text{d}6$. $3\text{d}6$ is the same as $1\text{d}6$ + $2\text{d}6$, so I can convolve these two. I'll call their functions $f$ and $g$ respectively.

$f$ here has limited support: the only values it is non-zero for are $1$ through $6$. This allows us to change the limits of our summation to the bounds of $f$'s support.

\begin{align} (f * g)(7) &= \sum_{k=1}^6 f(k)g(7 - k)\\ &= f(1)g(6) + f(2)g(5) + f(3)g(4) + f(4)g(3) + f(5)g(2) + f(6)g(1) \\ &= \frac{1}{6}\cdot\frac{5}{36} + \frac{1}{6}\cdot\frac{4}{36} + \frac{1}{6}\cdot\frac{3}{36} + \frac{1}{6}\cdot\frac{2}{36} + \frac{1}{6}\cdot\frac{1}{36} + \frac{1}{6}\cdot\frac{0}{36} \\ &= \frac{15}{216} \end{align}

Using convolution, then, we can calculate the probabilities of the summed results of multiple dice, without necessarily considering every single simple event: to calculate, say, the probability distribution of $5\text{d}6$, we can take the distribution of $4\text{d}6$ and the distribution of $1\text{d}6$ and convolve them. And to get $4\text{d}6$'s distribution we can convolve $3\text{d}6$ and $1\text{d}6$, etc. So instead of counting out $7776$ possibilities, we instead handle $21\cdot6 + 16\cdot6 + 11\cdot6 + 6\cdot6 = 324$ total multiplications and a similar number of additions.

• I didn't read your answer or the problem, but your LateX table skills are impressive – user2879934 May 11 '17 at 0:53
• That last paragraph is an important short cut. It means that you really only need to work out the cases that result in a tie. – amd May 11 '17 at 0:55
• I literally learned how to make the table to write the answer, and beat Excel until morale impro^W^W it built most of the rows for me. – Dan Uznanski May 11 '17 at 0:56
• I was initially skeptical that this is really "Your best bet", given the other answers, but it does appear to be a more general solution that is purely computational. – 6005 May 12 '17 at 5:34

Call the probability of scoring a $k$ with the D20 $q_k=q=1/20$ and scoring a $k$ with the 3 D6'S $p_k$ then we are looking for the probability that the D20 has the greater score

$$p_3\sum_{k=4}^{20}q_k+p_4\sum_{k=5}^{20}q_k+\ldots +p_{18}\sum_{k=19}^{20}q_k$$

Or simply

$$q(17p_3+16p_4+\cdots +2p_{18})$$

but by symmetry for every way to score $k$ with 3 D6s there is a way to score $21-k$, simply by subtracting the score on each die from $7$. The probability of rolling a total of $18$ with $3$ D6s is therefore the same as rolling a total of $3$ ($p_3=p_{18}$) and rolling a total of $17$ is the same probability as rolling $4$ ($p_4=p_{17}$) etc. So

$$\text{required probability}=19q(p_3+p_4+\cdots +p_{10})=19q(p_{11}+p_{12}+\cdots +p_{18})$$

But of course

$$p_3+p_4+\cdots +p_{18}=1$$

So

$$2(p_3+p_4+\cdots +p_{10})=1=2(p_{11}+p_{12}+\cdots +p_{18})$$ $$\implies p_3+p_4+\cdots +p_{10}=p_{11}+p_{12}+\cdots +p_{18}=\frac{1}{2}$$

Giving

$$\text{required probability}=19\cdot\frac{1}{20}\cdot\frac{1}{2}=\frac{19}{40}\tag{Answer}$$

Note that, since the probability of the players getting the same score is $(1/20)\cdot 1=1/20$, then the probability that the D20 loses is the same as it winning i.e. $19/40$.