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?


8 Answers 8


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:


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 $P(\sum X_i \mbox{ odd}) = 0.5$.


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, $$ \begin{eqnarray*} P\left(\sum X_i \mbox{ odd}\right) = P(S + X_n \mbox{ odd}) &=& P(S \mbox{ odd and } X_n \mbox{ even}) + P(S \mbox { even and } X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ even}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ odd}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& \left(P(S \mbox{ odd}) + P(S \mbox{ even})\right)P(X_n \mbox{ odd}) \\ &=& P(X_n \mbox{ odd}) \\ &=& 0.5 \end{eqnarray*} $$

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    $\begingroup$ o.O That's also a nice approach. Will the same apply for a dice with 7 sides? $\endgroup$ Sep 26, 2015 at 5:18
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    $\begingroup$ @Gummybears no, because then you have inequal probabilities coming from the last die, so the parity of the sum you had before you roll it matters. However, this sort of thinking does apply in a pretty wide variety of cases, and it saves a lot of work when you notice it happening. $\endgroup$ Sep 26, 2015 at 5:22
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    $\begingroup$ @Gummybears: sorry, I missed that. However, what is not mentioned is that with an odd number of seven sided dice, there is a higher chance of an odd sum. Whereas with an even number of seven sided dice, there is a higher chance of an even sum. $\endgroup$
    – robjohn
    Sep 26, 2015 at 5:50
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    $\begingroup$ @JoseAntonioDuraOlmos, the first 5 dice matter as much as the last one. They all matter the same here. It is not our neglect of the first 5 dice which leads us to the conclusion that the probability is $1/2$ but it is the symmetry of odd/even tosses. There is a great danger in basing the argument on the neglect, as one might be tempting to write it for the case where all dice are biased. $\endgroup$
    – zhoraster
    Sep 26, 2015 at 8:53
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    $\begingroup$ There is no neglect. If I am presented a box with an unknown number of dices of unknown shapes all of them labeled with integers. I take only one out of the box. The dice has 6 sides, from 1 to 6, and is unbiased. Then I can claim that if I take all the dices, roll them and add their results the sum has a 50% probability of being even. I can claim it without examining the other dices in the box. And the proof of such claim is like the one in this answer. As for the case where all dice are biased, yeah this proof scheme does not work for that, don't fall into those temptations. $\endgroup$ Sep 26, 2015 at 9:08

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:

  • OOOOOO $$\binom{6}{6}=1 \text{ arrangement}$$
  • OOOOEE $$\binom{6}{4}=15 \text{ arrangements}$$
  • OOEEEE $$\binom{6}{2}=15 \text{ arrangements}$$
  • EEEEEE $$\binom{6}{0}=1 \text{ arrangement}$$

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.

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    $\begingroup$ Lovely answer. Although slightly lengthy, it is simple to understand. Thank you. $\endgroup$ Sep 26, 2015 at 5:24
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    $\begingroup$ +1 Good answer. I just have a wording suggestion: instead of saying "to get an even/odd sum, an even/odd number of the six dice must be even", I think wording it in terms of the number of odd dice gives more intuition and can generalize to any number of dice. The number of even dice doesn't really matter, the number of odd dice does. $\endgroup$
    – jadhachem
    Sep 28, 2015 at 5:39

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:

$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\ \frac{d^n-1}{d-1}&\text{otherwise} \end{cases}$$

  • $\begingroup$ I'm afraid I lost you at the last sentence. $\endgroup$ Sep 26, 2015 at 5:03
  • $\begingroup$ That's okay, it is an advanced approach, but a useful one for problems like this. If you rolled $n$ dice, each with $7$ sides, you'd get the number of even results is: $$\frac{7^n + (-1)^n}{2}$$ would be even. $\endgroup$ Sep 26, 2015 at 5:09
  • $\begingroup$ Aha..... That's a good approach. However, I'm not sure if I completely get it. Not sure how you used the probability of 1 and $-1$ to calculate the even terms. $\endgroup$ Sep 26, 2015 at 5:16
  • $\begingroup$ If $i$ is odd, then $a_i\cdot 1^i + a_i(-1)^{i} = 0$. If $i$ is even, then $a_u\cdot 1^i + a_i(-1)^i = 2a_i$. So $P(1)+P(-1)=\sum_{i} 2a_{2i}$, which is twice the value you are looking for. @Gummybears $\endgroup$ Sep 26, 2015 at 5:19
  • $\begingroup$ Aha.... I'm stupid, aren't I? This is perfect! It holds true for any sided dice and for any number of dice. $\endgroup$ Sep 26, 2015 at 5:23

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 $

  • $\begingroup$ Exactly what @erictessler is missing - a rigurous proof. One by induction could work for all sorts of dice. $\endgroup$
    – Tibos
    Sep 26, 2015 at 23:22
  • $\begingroup$ More upvotes needed! $\endgroup$ Sep 27, 2015 at 4:14
  • $\begingroup$ Nicely done! +1 $\endgroup$ Sep 27, 2015 at 17:27

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 $1/2$.

  • $\begingroup$ I'm not getting you much.... elaborate please? $\endgroup$ Sep 26, 2015 at 5:22
  • $\begingroup$ Same here. Please elaborate. $\endgroup$ Sep 26, 2015 at 5:23
  • $\begingroup$ See the answer by @EricTressler, posted the same time. $\endgroup$
    – zhoraster
    Sep 26, 2015 at 5:24

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:

  1. All 6 dice show even numbers.
  2. 4 dice show even and 2 dice show odd.
  3. 2 dice show even and 4 dice show odd.
  4. All 6 dice show odd.

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.





(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.






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.

  • $\begingroup$ Oops, Zhoraster beat me to it and did a much clearer and shorter job explaining it. Oh, well. $\endgroup$
    – fleablood
    Sep 26, 2015 at 23:27
  • $\begingroup$ It was asked above about 7 sided die. I don't have enough reputation to comment there, so I'm going to comment here: If a 7-sided die has 4 odd faces and 3 even faces the probability of an even result with n die is 1/2 + (- 1/7)^n. This can be shown by induction by noting that each extra die has a 4/7 chance of changing parity. So if Pn is probability of an even total with n die. Then P(n+1) = Pn*3/7 + (1 - Pn)*4/7. as P1 = 3/7 = 1/2 + (- 1/7), the result follows by induction $\endgroup$
    – fleablood
    Sep 26, 2015 at 23:51

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