### Brain Teasers

# Dice

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

If you roll two dice, what is the probability the sum of the numbers you get is odd?

### Hint

It's not 1/3.### Answer

1/2The first die you roll - whether it's even or odd doesn't matter. The second die you roll has to be the opposite.

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## Comments

How many times did you toss the dice? I think you would have to specify how many times one could roll the dice to come up with your answer. It is still just the luck of the roll. Cute idea tho.

Smarty this didn't work when I started rolling die. I might not understand your explanation. I like anything that is die related.

(I rolled double sixes, followed by two ones; followed by a four and five and a pair of twos. I rolled for a while and was pretty shocked at the pairs I kept rolling. Never got a pair of fours - oh well.)

(I rolled double sixes, followed by two ones; followed by a four and five and a pair of twos. I rolled for a while and was pretty shocked at the pairs I kept rolling. Never got a pair of fours - oh well.)

i think the answer is 1/4.. i get it wrong.. wew.. i really like dice things .. haha..

Nice solution, its perfectly correct.

Basically the parity of the second dice must be different to the parity of the first dice, so it has a 50,50 chance.

Hence probabililty is half.

For those who don't believe it, just draw a possibility space.

...1.2.3.4.5.6

--------------

1|2.3.4.5.6.7

2|3.4.5.6.7.8

3|4.5.6.7.8.9

4|5.6.7.8.9.10

5|6.7.8.9.10.11

6|7.8.9.10.11.12

Half the numbers are odd, and all 36 cases are equally likely, so probability is half

Basically the parity of the second dice must be different to the parity of the first dice, so it has a 50,50 chance.

Hence probabililty is half.

For those who don't believe it, just draw a possibility space.

...1.2.3.4.5.6

--------------

1|2.3.4.5.6.7

2|3.4.5.6.7.8

3|4.5.6.7.8.9

4|5.6.7.8.9.10

5|6.7.8.9.10.11

6|7.8.9.10.11.12

Half the numbers are odd, and all 36 cases are equally likely, so probability is half

Way to go CMS. Perfect explanation of probability and proof of the answer. Remember, probability is constant, short term runs in either direction are possible.

No, the probability is 44.44%. If you actually count out all the combinations of two dice, and count the number of ways to get odds, it comes out differently.

All the combinations are listed in the possibility space above, so its 50%.

Or use the other proof that the parity of the second dice must be the opposite parity of the first.

Either way, its 50%.

Or use the other proof that the parity of the second dice must be the opposite parity of the first.

Either way, its 50%.

Herman, it appears you miscounted, and counted 16 out of 36 had an odd sum.

It should be 18 out of 36.

It should be 18 out of 36.

i actually made CMS' chart to solve this as well.

Can't be right!How the heck is that possible....!!!!!!!!!!!!!!!!!!!!!!!!

lui, it is correct. look at cms' chart.. the answer is 18/36 = 1/2

This one has a mathematical spring to it, so i like it!

I made a chart like cvs's anyways to figure it out.

I made a chart like cvs's anyways to figure it out.

Another commonsense way to think about it is that both dice are equally likely to come up either odd or even so its just like flipping 2 coins marked odd and even. Now ODD + ODD = EVEN and EVEN + EVEN = EVEN and ODD + EVEN = ODD and EVEN + ODD = ODD so 2 out of the four ways the coins could fall make EVEN and 2 of the four ways make ODD so 50% is the answer. In other words since the numbers on each die are expected to turn up odd or even on a 50/50 basis and the sums of LIKE/UNLIKE turn out odd or even on a 50/50 basis then we don't expect any bias in the resulting sum!

I had to work this one out the slow way with a pencil listing all possible combinations to get the answer. It surprised me a bit as I never have done much with dice probabilities before. The statement "The second die you roll has to be the opposite." is hardly correct if you think about it, but the final answer of 1/2 agreed with mine.

Nice one!

Nice one!

I believe what was meant was that if you want the sum to be odd then the second die you roll has to be the opposite parity of the first.

It is Not a simple 1/2. Rolling 2 dice will give you a total of between 2 and 12, with possibility for 6 even totals and 5 odd totals so the simple odds are 6 to 5 in favor of even

Part 2. You asked about the SUM in your question, Not the odds of Matching, which are 1/2.

Probability on the SUM being ODD are 6/5 against- there are 6 chances for an EVEN sum and 5 chances of an ODD sum on any roll of 2 Dice.

Probability on the SUM being ODD are 6/5 against- there are 6 chances for an EVEN sum and 5 chances of an ODD sum on any roll of 2 Dice.

BillB- One would THINK that... But it's not true. There are Six even sums, and five odd sums, but there's more ways to GET the odd sums than there are the evens. For instance, the only way you can get 2 is 1+1, and the only way you can get 12 is 6+6, but you can get seven three different ways. (6+1, 5+2, 3+4)

So it does balance out to be a fifty percent chance. Here's my chart-- all the odd possibilities have a star.

1+1=2

1+2=3*

1+3=4

1+4=5*

1+5=6

1+6=7*

2+1=3*

2+2=4

2+3=5*

2+4=6

2+5=7*

2+6=8

3+1=4

3+2=5*

3+3=6

3+4=7*

3+5=8

3+6=9*

4+1=5*

4+2=6

4+3=7*

4+4=8

4+5=9*

4+6=10

5+1=6

5+2=7*

5+3=8

5+4=9*

5+5=10

5+6=11*

6+1=7*

6+2=8

6+3=9*

6+4=10

6+5=11*

6+6=12

Probability: 18/36

So it does balance out to be a fifty percent chance. Here's my chart-- all the odd possibilities have a star.

1+1=2

1+2=3*

1+3=4

1+4=5*

1+5=6

1+6=7*

2+1=3*

2+2=4

2+3=5*

2+4=6

2+5=7*

2+6=8

3+1=4

3+2=5*

3+3=6

3+4=7*

3+5=8

3+6=9*

4+1=5*

4+2=6

4+3=7*

4+4=8

4+5=9*

4+6=10

5+1=6

5+2=7*

5+3=8

5+4=9*

5+5=10

5+6=11*

6+1=7*

6+2=8

6+3=9*

6+4=10

6+5=11*

6+6=12

Probability: 18/36

Ok I get the idea but somebody tell me why this isnt correct.

There are ACTUALLY only 21 combinations of the two dice.

If you roll two dice it would be assumed you rolled them together, otherwise you could roll one twice.

So the combination 2 1 is the same as the combination 1 2, both have an odd sum but equal one combinbation not two.

Following then the combinations factor to 6 + 5 + 4 + 3 + 2 + 1 = 21.

Of the combinations 9 have odd sums and 12 have even sums.

Therefore you are 9 of 21 to be even or 9 : 12.

How many two card combinations are there in a standard 52 card deck for Texas Hold'em?

There are ACTUALLY only 21 combinations of the two dice.

If you roll two dice it would be assumed you rolled them together, otherwise you could roll one twice.

So the combination 2 1 is the same as the combination 1 2, both have an odd sum but equal one combinbation not two.

Following then the combinations factor to 6 + 5 + 4 + 3 + 2 + 1 = 21.

Of the combinations 9 have odd sums and 12 have even sums.

Therefore you are 9 of 21 to be even or 9 : 12.

How many two card combinations are there in a standard 52 card deck for Texas Hold'em?

I assume you are kidding when you say there are 21 possible combinations and not 36. Image one die is red and the other is white. A red 2 and a white 1 is a distinctly different roll from a red 1 and a white 2. To use your betting mechanism from another comment you posted, I would be more than happy to play you 100 times, where you pay me $1.10 if it is even, and I will pay you $1.00 if it is odd. (And to answer your HE question, there are 1326 distinct starting hands.)

Okay, after reading the answer and comments, I understand the nature of the question.

However, upon first reading it... I was under the impression that '"if you roll 2 dice", it meant that you rolled both at the same time.

If that was the case... then the possible sum of both dice would be -

2,3,4,5,6,7,8,9,10,11,and 12

5 possible odds, and 6 possible evens.

However, upon first reading it... I was under the impression that '"if you roll 2 dice", it meant that you rolled both at the same time.

If that was the case... then the possible sum of both dice would be -

2,3,4,5,6,7,8,9,10,11,and 12

5 possible odds, and 6 possible evens.

late664 -- those are indeed the possible sums, but they are not all equally possible. A 7 comes up 6 times, while a 12 only comes up once. You have to take this into account, as well. Otherwise, you could look at your results from buying a lottery ticket as either being down $1 or up $1,000,000, and conclude that you are therefore 50% to win a million!

This is way too easy. I read the hint and I don't understand how one can even come up with 1/3

Wow I must have gone dumber. I did 2 * 1/4 this time via casework odd + even and even + odd instead of fixing the first roll....

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