[Math] Chevalier de mere paradox with game with three dice

probability

Chevalier de Mere asked Blaise Pascal why in a game with three dice the sum $11$ is more favorable than $12$, when both sums have exactly the same possible combinations:

For $11$ we have $(5,5,1), (5,4,2), (5,3,3), (4, 4, 3), (6,4,1), (6,3,2)$ and for $12$ we have $(6,5,1), (6,4,2), (6,3,3), (5,4,3), (4,4,4), (5,5,2)$, so both sums should be equiprobable.

My attempt: I think Chevalier de Mere made the mistake of thinking all the dice are indistinguishable. I tried to compute the exact probabilities.

Let $$\Omega = \left\{(x,y,z) \mid 1 \leq x,y,z \leq 6, \quad 3 \leq x + y + z \leq 18\right\} $$ be the sample space. We are interested in the events $$A = \left\{(x,y,z) \mid x + y + z = 11 \right\} $$ and $$ B = \left\{(x,y,z) \mid x + y + z = 12 \right\}. $$ For the sum $11$, we have $27$ possible permutations of all the triples. For the sum $12$, there are two less, that is $25$. So $\#A = 27$ and $\#B = 25$. Since $\#\Omega = 6 \cdot 6 \cdot 6 = 216$, we have $$ P(A) = \frac{27}{216} = 0.125, \qquad P(B) = \frac{25}{216} = 0.1157. $$
Is this reasoning correct?

Best Answer

You're making the very common mistake of confusing issues of distinguishability with issues of equiprobability. If you roll three indistinguishable dice, the probabilities of the various sums are exactly the same as if you roll three distinguishable dice. The dynamics of the dice are not influenced by your ability to distinguish among them.

Chevalier de Méré's error was not to think of the dice as indistinguishable, but to regard the wrong sorts of events as equiprobable. Absent prior information, events should be regarded as equiprobable when there is symmetry among them. In the present case the six possible results of each individual die should be regarded as equiprobable, since there is symmetry among the six sides. It then follows that each combination of these independent equiprobable events, that is, each ordered triple of dice results should also be considered equiprobable, as you did in your calculation. Instead, Chevalier de Méré regarded each unordered triple as equiprobable. This is wrong since unordered triples with different numbers of coinciding dice correspond to different numbers of ordered triples, and thus to different numbers of equiprobable elementary events.

Related Solutions

[Math] Dice Rolling Probability Space

The dice are distinct objects. By coloring the dice, it might be easier see how the numbers are counted. Here are the different ways to roll $11$ on $3$ dice:

RGB    RGB    RGB
641    461    335
632    452    326
623    443    263
614    434    254
551    425    245
542    416    236
533    362    164
524    353    155
515    344    146

Each of these has a $\frac1{216}$ chance of being rolled. Thus, an $11$ has a $\frac{27}{216}$ chance of being rolled.

Probability of Sum of Three Dice Being Eleven

The generating function approach is to ask for the coefficients of $x^{n}$ in the expansion of:

$$(x+x^2+x^3+x^4+x^5+x^6)^{3}=x^3\left(\frac{1-x^6}{1-x}\right)^3$$

Now, $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k$$

And $$(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$

So the product of these and $x^3$ has, for the coefficient $x^n$:

$$\binom{n-1}{2}-3\binom{n-7}{2}+3\binom{n-13}{2}-\binom{n-19}{2}$$

Trick is to treat $\binom{j}{2}=0$ when $j<2.$

So, for example, when $n=11$, you get $\binom{10}{2}-3\binom{4}{2}=45-18=27$.

This also gives you a hint of the value when dealing with $k$ dice. Then:

$$c_n = \sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{n-(1+6i)}{k-1}$$

If you don't like generating functions, this can be proven via inclusion-exclusion.

There's another approach that is a little faster for computing all values.

If $(1+x+x^2+\cdots+x^5)^3=a_0+a_1x+\cdots+a_nx^n+\cdots$ then you get that:

$$(a_0+a_1x+\cdots+a_nx^n+\cdots)(1-3x+3x^2-x^3)=(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$

What this means is that (setting $a_n=0$ when $n<0$: $$a_{n}-3a_{n-1}+3a_{n-2}-a_{n-3}=\begin{cases} 1&n=0\\ -3&n=6\\ 3&n=12\\ -1&n=18\\ 0&\text{otherwise} \end{cases}$$

or:

$$a_{n}=3\left(a_{n-1}-a_{n-2}\right)+a_{n-3}+\begin{cases} (-1)^k\binom{3}{k}&n=6k\\ 0&\text{otherwise} \end{cases}$$

Then the final coefficient of $x^n$, after we multiply by $x^3$ again, is $a_{n-3}$.

So you get:

$$\begin{align}a_0&=0+(-1)^{0}\binom{3}{0}=1\\ a_1&=3\left( a_0-a_{-1}\right)=3\\ a_2&=3\left( a_1 - a_0\right)=6\\ a_3&=3\left(a_2 - a_1\right)+a_0=10\\ a_4&=3\left(a_3-a_2\right)+a_1=15\\ a_5&=3\left(a_4-a_3\right)+a_2=21\\ a_6&=3\left(a_5-a_4\right)+a_3+(-1)^{1}\binom{3}{1}=25\\ &\cdots \end{align}$$

There's a special trick that can be applied here: $a_{n}=3a_{n-1}-3a_{n-2}+a_{n-3}$ is known to be a quadratic polynomial in $n$. So, when $n$ is not a multiple of $6$, you get that:

$$a_{n-2}-a_{n-3},a_{n-1}-a_{n-2},a_{n}-a_{n-1}$$ must be an arithmetic progression for any $n$ not a multiple of $6$.

So we have that $a_5-a_4 = 6, a_6-a_5=4$ and thus $a_7-a_6=2$, or $a_7=a_6+2=27.$ $a_8=27+0=27, a_9=27-2=25,a_{10}=25-4=21,a_{11}=21-6=15,a_{12}=15-8+(-1)^2\binom{3}{2}=10$.

Best Answer

Related Solutions

[Math] Dice Rolling Probability Space

Probability of Sum of Three Dice Being Eleven

Related Question