Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Question:
6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?
I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?
Notice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum.
The first 5 dice don't matter.
Edit to formalize this and address concerns in the comments:
Statement:
Let $X_1,\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \mbox{ odd}) = 0.5$. Then $P(\sum X_i \mbox{ odd}) = 0.5$.
Proof:
Without loss of generality let $k=n$, and let $S = \sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$ \begin{eqnarray*} P\left(\sum X_i \mbox{ odd}\right) = P(S + X_n \mbox{ odd}) &=& P(S \mbox{ odd and } X_n \mbox{ even}) + P(S \mbox { even and } X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ even}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ odd}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& \left(P(S \mbox{ odd}) + P(S \mbox{ even})\right)P(X_n \mbox{ odd}) \\ &=& P(X_n \mbox{ odd}) \\ &=& 0.5 \end{eqnarray*} $$
The only thing about the numbers rolled that matters is their parity - whether they are even or odd. In order to get an even sum, an even number of the six dice must be even. In order to get an odd sum, an odd number of the six dice must be even.
Using O for odd and E for even, we can list out the possibilities.
Even sum:
This gives a total of $32$ arrangements with even sum.
Since there are $2^6 = 64$ total possibilities, we see that your intuition of $50\%$ is correct.
The generating function approach:
$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\sum a_i x^i$$
Then $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.
Now, to find the even terms, you can compute $$\frac{P(1)+P(-1)}{2}=\sum_i a_{2i}.$$
But $P(1)=6^6$ and $P(-1)=0$. So $$\frac{P(1)+P(-1)}{2}=\frac{6^6}{2},$$ or exactly half, as you conjectured.
For another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\equiv i\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:
$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$ then computing $$N_i=\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$
This gives the result:
$$N_i =\begin{cases}\frac{6^6+4}{5}&i\equiv 1\pmod 5\\ \frac{6^6-1}{5}&\text{otherwise} \end{cases}$$
More generally, if $N_{n,i}$ is the number of ways to get $\equiv i\pmod 5$ when $n$ dice are rolled, you get:
$$N_{n,i} =\begin{cases}\frac{6^n+4}{5}&i\equiv n\pmod 5\\ \frac{6^n-1}{5}&\text{otherwise} \end{cases}$$
It's this simple because of the fact that $6=5+1$.
If each die has $d$ sides, and you ask what are the number of ways to get a total $\equiv i\pmod {d-1}$, then you get:
$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\ \frac{d^n-1}{d-1}&\text{otherwise} \end{cases}$$
Using the same logic given by @Eric tessler but writing the maths behind it.
Let $P_x$ = We get an even sum on the roll of x dice and
$1 - P_x$ = We get an odd sum.
So we require $P_6$
Now $P_6 = P_5 \times $ P[even number on the last dice] + $(1 - P_5) \times$ P[odd number on the last dice]
$P_6 = P_5 \times 0.5 + (1-P_5) \times 0.5 $
$P_6 = 0.5 $
A lot of very informative answers but none of them do not explain your intuition, so I'll post mine.
There is a symmetry between all tosses with odd sum and those with even sum: just take the number $k$ on the first die and replace it with $7-k$.
So the numbers of odd tosses and even tosses are equal, therefore, the probability is $1/2$.
Your intuition is correct. There could be a proof based on this intuition (using the fact that each die has 50% chance of being even and we're tossing 6 of them) but I'm not so sure.
But there are ways to prove it. For example, the only ways we could get an even sum are if:
The probability of cases 1 and 4 is $(\frac{1}{2})^6$ and the probability of cases 2 and 3 is $\binom{6}{2}(\frac{1}{2})^6=15(\frac{1}{2})^6$ (because we need to order the dice which show odd numbers.) Add them up and we get the probability is $$32(\frac{1}{2})^6=\frac{1}{2}.$$
The problem is about fair dice.
Whether the number of dice is 6 (even) or 7 (odd) $Pr = \dfrac12$
The logic is based on parity (odd/even). Since the probability that any particular throw is even or odd is equal at $\dfrac12$
(a) For an even number of dice, if you interchange odd and even throws, the parity remains the same, thus there will always be an equal number of odd and even sums.
EEEEEE OOOOOO even
EEEEEO OOOOOE odd
EEEEOO OOOOEE even
EEEOOO OOOEEE odd
(b) For an odd number of dice, every such interchange changes the parity, but by symmetry, again there will be an equal number of odd and even sums.
EEEEEEE even OOOOOOO odd
EEEEEEO odd OOOOOOE even
EEEEEOO even OOOOOEE odd
EEEEOOO odd OOOOEEE even
All outcomes are equally likely. Half the outcomes will have an odd number of odd dice faces and half of the outcomes will have an even number of odd dice faces. The outcomes with an odd number will have an odd total and the outcomes with an even number of odd face will be even. As there are equal numbers of outcomes for each case the probability for each case is 50%.
Okay so how do we know that half the outcomes will have on odd number of odd faces and half the number will have an even number of even faces? Well,we can make a one to one correspondence between outcomes with an odd number of odd faces to outcomes to an even number of odd faces. If outcome has an even number of odd faces, map it to the exact same outcome but the 6th die is one number higher (let's assume 1 is one number higher than 6). This outcome has an odd number of odd faces. If an outcome has an odd number of odd faces map it the exact same outcome but the 6th die is one number lower. This is a one to one correspondence. So there are an equal number of outcomes with an odd number of odd faces and there are outcomes with an even number of odd faces.
And that's it. Equal number of odd comes as even outcomes, each outcome equally likely, odd outcomes and even outcomes equally likely.
