Understanding the Odds of Rolling 4 Dice Together

[deep519]

Hi guys,

I have a very simple but confusing problem that I have with die.

It's not a homework problem, more a general understanding question.

Now, question (im my own words)

Four six-sided die are rolled together and independently to each other.
Is the chance to roll a sum of (even) or (odd) the same or is rolling even higher.


Now, i think that rolling four die, the chances that their sum are even is higher than rolling odd for the reason that the possible combination's - reminded that it doesn't matter which order the dice are rolled are:

Where 1 means odd and 0 is even
0 0 0 0 = Even
0 0 0 1 = Odd
0 0 1 1 = Even
0 1 1 1 = Odd
1 1 1 1 = Even

Unless I am missing something, there is a higher chance to roll even since 3E > 2O.

Maybe this is a very fundamental stats problem, but I'm a little confused.

Me and friend are arguing about it, so if somebody could clarify the correct answer.

 

[Outlined]

you missed something, look at all the possible outcomes

0 0 0 0 : even
0 0 0 1 : odd
0 0 1 0 : odd
0 0 1 1 : even
0 1 0 0 : odd
0 1 0 1 : even
0 1 1 0 : even
0 1 1 1 : odd
1 0 0 0 : odd
1 0 0 1 : even
1 0 1 0 : even
1 0 1 1 : odd
1 1 0 0 : even
1 1 0 1 : odd
1 1 1 0 : odd
1 1 1 1 : even

So 8 times even so chance you have even is 50% so equal chances.

 

[deep519]

Well, the thing is order should not matter since if four dice are rolled, you will get only those 5 combinations.

0110 or 1100 would still mean two dice were even and two were odd, thus why would the order matter.

1110 or 1011 same thing... ect..

We are just concerned with the chance that the sum is even or odd.

 

[chiro]

Well, the thing is order should not matter since if four dice are rolled, you will get only those 5 combinations.

0110 or 1100 would still mean two dice were even and two were odd, thus why would the order matter.

1110 or 1011 same thing... ect..

We are just concerned with the chance that the sum is even or odd.

It is true that you are interested in the frequency and not the order but you have to take into account the frequency which is what the above poster has pointed out.

When interested in unordered sets we use nCr so for the frequency we get

0 0 0 0 - x 1
0 0 0 1 - x 4
0 0 1 1 - x 6
0 1 1 1 - x 4
1 1 1 1 - x 1

Total frequency - 16
Total even - 1 + 6 + 1 = 8 = 50%
Total odd - 4 + 4 = 8 = 50%

 

[blue_raver22]

Hello,

This is a great question and one that can be easily answered with some basic statistical understanding. To start, let's define our terms. In this case, we are looking at the probability of rolling an even or odd sum when four six-sided dice are rolled independently.

To answer your question, the chances of rolling an even or odd sum are actually the same. This is because the probability of rolling an even or odd number on any one die is 1/2 or 50%. Therefore, when rolling four dice independently, the probability of getting an even or odd sum is still 50%.

To understand this better, let's look at the possible outcomes when rolling four dice. As you mentioned, there are 16 possible combinations (2^4). Out of these 16 combinations, 8 of them will result in an even sum (0 0 0 0, 0 0 1 1, 0 1 0 1, 1 0 0 1, 0 1 1 0, 1 1 0 0, 1 0 1 0, 1 1 1 1) and 8 of them will result in an odd sum (0 0 0 1, 0 0 1 0, 0 1 0 0, 1 0 0 0, 0 1 1 1, 1 1 0 1, 1 0 1 1, 1 1 1 0). Therefore, the probability of rolling an even or odd sum is 8/16 or 1/2.

It is important to note that the order in which the dice are rolled does not affect the outcome, as you mentioned. This is because each die has an equal chance of rolling an even or odd number, regardless of the numbers rolled on the other dice.

In summary, the chances of rolling an even or odd sum when four dice are rolled independently are the same. This is because each die has an equal probability of rolling an even or odd number, resulting in an overall probability of 1/2 for both even and odd sums.

I hope this helps to clarify the correct answer for you and your friend. If you have any further questions, please don't hesitate to ask.

Best,