# Two dice with the same probability for each sum? [duplicate]

A friend invites you to play a game. The game is using two standard six-sided dice with faces numbered 1, 2, 3, 4, 5, 6 each.
As usual, the dice are considered distinguishable, i.e. throwing a 1 with dice 1 and a 2 with dice 2 is different than throwing a 1 with dice 2 and a 2 with dice 1 even in both cases the sum is 3. You friend claims that each sum from 2-12 on both dice appears with the same probability

How can you prove if this is true or not?

• What is the relevant definition of "two standard six-sided dice" here? I'm asking with regards to rot13(n ybnqrq qvr cbffvoyl abg orvat fgnaqneq naq bccbfvgr ahzoref abeznyyl univat fhz frira) here – Braegh Dec 28 '19 at 23:59
• Is quantum entanglement between the cubes allowed? – A. P. Dec 30 '19 at 0:43
• Is the lateral-thinking tag really appropriate for this question? It seems to be a relatively straightforward maths/probability question...? – Chris Dec 30 '19 at 9:34
• @Chris, I removed the lateral-thinking tag. – ThomasL Dec 30 '19 at 15:44
• @A.P. I never thought about quantum entanglement when asking this question, but I would be very curious what can be concluded if it is allowed to be used. – ThomasL Dec 30 '19 at 15:49

I think:

These dice can't exist.

Here's my proof:

Let $$a$$, $$b$$, $$c$$, $$d$$ be the probabilities of the first die being 1, the second die being 1, the second die being 6, the first die being 6, respectively.

Since each sum from 2-12 has equal probability, each sum has 1/11 probability.

There is only one way to roll a 2 or 12: 1-1 and 6-6, so we know:

$$ab = 1/11$$
$$cd = 1/11$$

Now consider the ways to make 7. Among the ways to make 7 are 1-6 and 6-1, along with 4 other ways. So we know:

$$ac + bd \leq 1/11$$

Substituting $$1/(11a)$$ for $$b$$ and $$1/(11c)$$ for $$d$$, we get:

$$ac + 1/(121ac) \leq 1/11$$

Multiplying by $$ac$$ on both sides (note $$ac$$ can't be negative so the $$\leq$$ can't flip):

$$a^2c^2 + 1/121 \leq ac/11$$

Subtracting $$ac/11$$ from both sides:

$$a^2c^2 - ac/11 + 1/121 \leq 0$$

Now at this point you can plug this into some equation solver and find out that there are no real solutions. You can also see this from the quadratic equation. Solving for $$a$$, the square root part of the quadratic equation is this:

$$\sqrt{c^2/121 - 4c^2 * X}$$, where $$X \geq 1/121$$ based on the inequality. Since the value inside the square root is always $$\leq 0$$, the equation doesn't have a real solution unless $$c = 0$$, which can't be true because $$cd = 1/11$$.

• "So we know..." Shouldn't the inequality that follows be ad + bc <= 1/11? – Warren Weckesser Dec 30 '19 at 13:15
• @WarrenWeckesser You're right. I'll edit the answer so to switch what d and c mean. – JS1 Dec 31 '19 at 1:40

It can be:

True!

If:

The dice explode when you roll them so the probability of rolling any number is 0.

• @Downvoter: This answer was posted when there was a lateral-thinking tag on the OP. – Paul Evans Dec 30 '19 at 18:02

We can test this statement by the following:

We know for this statement to be true the dice must be weighted as the described distribution doesn't match the distribution of the sum of two six sided dice. Therefore we can roll the dice repeatedly, record the results and run Kolmogorov-Smirnov or Chi-square test.

However I doubt this is the "lateral thinking" the OP has in mind.

• You can prove/disprove it analytically without actually rolling a single die – smci Dec 30 '19 at 9:09

You can't make it in a fair way, as JS1 answered.

But given lateral thinking tag - you could have dice where one influences the other. Perhaps by magic or magnets. For example, first die is fair (1/6 chance of any number) and when it rolls 1, there is 6/11 chance of rolling 1 on the second one and 1/11 to roll any other number. Symmetry tells us 6 should behave the same as 1. Now keep working through the list (sums 3 and 11 next; you get 5/11 to roll 1 to get sum 3).

In the example above where remaining probability is distributed equally, 1 and 6 on the second die would be more likely than the middle numbers. I believe it should be possible to tweak "roll table" probabilities in a way the second die behaves as a fair one (1/6 probability of any number).

• Given the lateral thinking tag we could also change and reduce the numbers on the dice to {0, 1, 2, 3} and {1, 5, 9} to get equal probabilities for the numbers 1-12 which may be close enough to the desired range of 2-12. (3-sided dice are already proven to exist). – nitzel Dec 30 '19 at 16:03

From JS1's answer we already see that this is impossible for normal dice. But assuming the dice's result are dependent on each other as if they could communicate,

there is indeed a way to realize this.

This is how it works:

Since the probabilities are dependent on the outcome of the other dice $$P(\color{red}{i}, \color{green}{j}) \neq P(\color{red}{i}) \cdot P(\color{green}{j})$$. This means we have all 36 $$P(\color{red}{i}, \color{green}{j})$$ as free variables and only 11 conditions. Ok, admittedly, we must make sure that all probabilities are within $$\left[ 0, 1 \right]$$, but this is just restricting the hypervolume of the 25-dimensional solution space and not reducing its dimensionality.
So there are infinitely many possibilities for solutions.

Let's pick a "nice" solution:

As additional criterion I want that

1. the dice are equal: $$P(\color{red}{i}, \color{green}{j}) = P(\color{red}{j}, \color{green}{i})$$
2. and they should behave like normal dice, i.e. it shows each side with equal probability if you ignore the outcome of the other dice: $$\color{green}{\sum_{j = 1}^6} P(\color{red}{i}, \color{green}{j}) = \frac{1}{6}$$
Like this they appear as if they are normal dice. Only if you look at the correlations you find that something is weird.

Actual numbers:

All possibilities which lead to a sum of $$\color{#26F}{2}$$ are $$\left\{ (\color{red}{1}, \color{green}{1}) \right\}$$, so we know immediately that the corresponding probability must be $$\frac{1}{11}$$. The same is true for a sum of $$\color{#26F}{12}$$, because here we also have only $$1$$ possible way $$\left\{ (\color{red}{6}, \color{green}{6}) \right\}$$. $$P(\color{red}{1}, \color{green}{1}) = \frac{1}{11} = P(\color{red}{6}, \color{green}{6})$$ For a sum of $$\color{#26F}{3}$$ there are $$2$$ possibilities: $$\left\{ (\color{red}{1}, \color{green}{2}), (\color{red}{2}, \color{green}{1}) \right\}$$, but since the dice are equal they appear with equal probability. Furthermore they must add up to $$\frac{1}{11}$$. The same holds for a sum of $$\color{#26F}{11}$$. $$P(\color{red}{1}, \color{green}{2}) = P(\color{red}{2}, \color{green}{1}) = \frac{1}{11} \cdot \frac{1}{2} = P(\color{red}{5}, \color{green}{6}) = P(\color{red}{6}, \color{green}{5})$$ $$3$$ different outcomes lead to sums of $$\color{#26F}{4}$$ and $$\color{#26F}{10}$$, respectively. The probabilities for $$(\color{red}{1}, \color{green}{3})$$ and $$(\color{red}{3}, \color{green}{1})$$ are equal, but the probability for $$(\color{red}{2}, \color{green}{2})$$ can be different from the other two. Let's parametrize this degree of freedom by $$a$$. $$P(\color{red}{1}, \color{green}{3}) = P(\color{red}{3}, \color{green}{1}) = \frac{1}{11} \cdot a = P(\color{red}{4}, \color{green}{6}) = P(\color{red}{6}, \color{green}{4}) \\ P(\color{red}{2}, \color{green}{2}) = \frac{1}{11} \cdot (1-2a) = P(\color{red}{5}, \color{green}{5})$$ It continues similarly for sums of $$\color{#26F}{5}$$ and $$\color{#26F}{9}$$, except that here we have $$4$$ possibilities each, with pairs of them being equal. $$P(\color{red}{1}, \color{green}{4}) = P(\color{red}{4}, \color{green}{1}) = \frac{1}{11} \cdot b = P(\color{red}{3}, \color{green}{6}) = P(\color{red}{6}, \color{green}{3}) \\ P(\color{red}{2}, \color{green}{3}) = P(\color{red}{3}, \color{green}{2}) = \frac{1}{11} \cdot \frac{1}{2} (1-2b) = P(\color{red}{4}, \color{green}{5}) = P(\color{red}{5}, \color{green}{4})$$ For sums of $$\color{#26F}{6}$$ and $$\color{#26F}{8}$$ there are $$5$$ possibilities each, hence $$3$$ independent numbers, which means $$2$$ more parameters. $$P(\color{red}{1}, \color{green}{5}) = P(\color{red}{5}, \color{green}{1}) = \frac{1}{11} \cdot c = P(\color{red}{2}, \color{green}{6}) = P(\color{red}{6}, \color{green}{2}) \\ P(\color{red}{2}, \color{green}{4}) = P(\color{red}{4}, \color{green}{2}) = \frac{1}{11} \cdot d = P(\color{red}{3}, \color{green}{5}) = P(\color{red}{5}, \color{green}{3}) \\ P(\color{red}{3}, \color{green}{3}) = \frac{1}{11} \cdot (1 - 2c - 2d) = P(\color{red}{4}, \color{green}{4})$$ Finally, for a sum of $$\color{#26F}{7}$$ there are $$6$$ possibilities, hence another $$2$$ parameters. $$P(\color{red}{1}, \color{green}{6}) = P(\color{red}{6}, \color{green}{1}) = \frac{1}{11} \cdot e \hphantom{= P(\color{red}{1}, \color{green}{6}) = P(\color{red}{6}, \color{green}{1})} \\ P(\color{red}{2}, \color{green}{5}) = P(\color{red}{5}, \color{green}{2}) = \frac{1}{11} \cdot f \hphantom{= P(\color{red}{2}, \color{green}{5}) = P(\color{red}{5}, \color{green}{2})} \\ P(\color{red}{3}, \color{green}{4}) = P(\color{red}{4}, \color{green}{3}) = \frac{1}{11} \cdot \frac{1}{2} (1 - 2c - 2d) \hphantom{= P(\color{red}{4}, \color{green}{4})}$$

Now let's implement the additional condition 2. \begin{align} \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{1}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{1}) = \color{green}{\sum_{j=1}^6} P(\color{red}{6}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{6}) \\ &= \frac{1}{11} \left( 1 + \frac{1}{2} + a + b + c + e \right) \\ \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{2}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{2}) = \color{green}{\sum_{j=1}^6} P(\color{red}{5}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{5}) \\ &= \frac{1}{11} \left( \frac{1}{2} + (1 - 2a) + \frac{1}{2} (1 - 2b) + c + d + f \right) \\ \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{3}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{3}) = \color{green}{\sum_{j=1}^6} P(\color{red}{4}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{4}) \\ &= \frac{1}{11} \left( a + \frac{1}{2} (1 - 2b) + (1 - 2c - 2d) + \frac{1}{2} (1 - 2e - 2f) + c + d \right) \end{align} This can be further reduced to \begin{align} e &= \frac{1}{3} - a - b - c \\ f &= 2a + b - c - d - \frac{1}{6} \text{.} \end{align} By fixing 4 of these parameters we can find a concrete solution. For example $$a = \frac{1}{3}$$, $$b = c = d = e = 0$$, $$f = \frac{1}{2}$$. This gives the following probabilities:

I would say that he’s correct, because:

The chance of getting 3-4 vs 4-3 is the same since there are always 36 possible outcomes.

1/6 * 1/6 = 1/36. So the chance is 1/36 in both cases, therefore the chance is always the same.

Hope it makes sense.

• The question says "each sum", though. 3+4 and 4+3 and 2+5 and 5+2 etc. all have the same sum (whereas there is no other outcome having the same sum as 1+1 or 6+6). – Ilmari Karonen Dec 29 '19 at 17:15