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Bunuel
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What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

Similarly, What is the probability that you will get a sum of 22 when you throw four dice simultaneously?

What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

This sum of 8 can occur in the following cases:
1-1-6 --> 3!/2!=3 ways;
1-2-5 --> 3!=6 ways;
1-3-4 --> 3!=6 ways;
2-2-4 --> 3!/2!=3 ways;
2-3-3 --> 3!/2!=3 ways.

Total of 3+6+6+3+3=21 ways.

Now, since the total number of outcomes is 6^3, then the probability is 21/6^3.

What is the probability that you will get a sum of 22 when you throw four dice simultaneously?

This sum of 22 can occur in the following cases:
6-6-6-4 --> 4!/3!=4 ways;
6-6-5-5 --> 4!/(2!2!)=6 ways.

Total of 4+6=10 ways.

Now, since the total number of outcomes is 6^4, then the probability is 10/6^4.

Hope it's clear.

Hi Bunuel chetan2u

Small point that needs clarification. Shouldn't 'simultaneously' mean that the dices aren't thrown in any particular order and therefore the order doesn't matter ?
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What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

Similarly, What is the probability that you will get a sum of 22 when you throw four dice simultaneously?

What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

This sum of 8 can occur in the following cases:
1-1-6 --> 3!/2!=3 ways;
1-2-5 --> 3!=6 ways;
1-3-4 --> 3!=6 ways;
2-2-4 --> 3!/2!=3 ways;
2-3-3 --> 3!/2!=3 ways.

Total of 3+6+6+3+3=21 ways.

Now, since the total number of outcomes is 6^3, then the probability is 21/6^3.

What is the probability that you will get a sum of 22 when you throw four dice simultaneously?

This sum of 22 can occur in the following cases:
6-6-6-4 --> 4!/3!=4 ways;
6-6-5-5 --> 4!/(2!2!)=6 ways.

Total of 4+6=10 ways.

Now, since the total number of outcomes is 6^4, then the probability is 10/6^4.

Hope it's clear.

Hi Bunuel chetan2u

Small point that needs clarification. Shouldn't 'simultaneously' mean that the dices aren't thrown in any particular order and therefore the order doesn't matter ?

But we still have different dice, right? For example, for highlighted case above, #1 showed 1, #2 showed 1, and #3 showed 6 is different case from #1 showed 6, #2 showed 1, and #3 showed 1.
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Hi Bunuel ! I followed and got the correct answer by brute force, just as you have done. I had read some question which said find integers < 10000 who sum of digits is 5. In that question we had applied separators technique, can we do it over here ?
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Bunuel
But we still have different dice, right? For example, for highlighted case above, #1 showed 1, #2 showed 1, and #3 showed 6 is different case from #1 showed 6, #2 showed 1, and #3 showed 1.

Right. So what would the wording have to be in order to identify that the order doesn't matter ? Would it be correct to take order to always matter in case of probabilistic events involving dices, coins etc ?
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Bunuel
What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

This sum of 8 can occur in the following cases:
1-1-6 --> 3!/2!=3 ways;
1-2-5 --> 3!=6 ways;
1-3-4 --> 3!=6 ways;
2-2-4 --> 3!/2!=3 ways;
2-3-3 --> 3!/2!=3 ways.

Total of 3+6+6+3+3=21 ways.

I see you have started with
1-1-X goes all the way
Then
2-2-X goes all the way.

I went similar way. But, kept checking 33X,44X,55X and 66X. I had a gut feeling its all repetition. But, Still checked.

Is there any logical way to tell after 2-3-3, it will repeat?
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Bunuel
What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

This sum of 8 can occur in the following cases:
1-1-6 --> 3!/2!=3 ways;
1-2-5 --> 3!=6 ways;
1-3-4 --> 3!=6 ways;
2-2-4 --> 3!/2!=3 ways;
2-3-3 --> 3!/2!=3 ways.

Total of 3+6+6+3+3=21 ways.

I see you have started with
1-1-X goes all the way
Then
2-2-X goes all the way.

I went similar way. But, kept checking 33X,44X,55X and 66X. I had a gut feeling its all repetition. But, Still checked.

Is there any logical way to tell after 2-3-3, it will repeat?

In my solution, I first listed all combinations that include 1, followed by those that include 2 (but not 1). By the time I reached combinations starting with 3, they could only contain 3s or higher, resulting in a sum greater than 8. Therefore, there was no need to check beyond that point.
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Bunuel
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Bunuel
What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

This sum of 8 can occur in the following cases:
1-1-6 --> 3!/2!=3 ways;
1-2-5 --> 3!=6 ways;
1-3-4 --> 3!=6 ways;
2-2-4 --> 3!/2!=3 ways;
2-3-3 --> 3!/2!=3 ways.

Total of 3+6+6+3+3=21 ways.

I see you have started with
1-1-X goes all the way
Then
2-2-X goes all the way.

I went similar way. But, kept checking 33X,44X,55X and 66X. I had a gut feeling its all repetition. But, Still checked.

Is there any logical way to tell after 2-3-3, it will repeat?

In my solution, I first listed all combinations that include 1, followed by those that include 2. By the time I reached combinations starting with 3, they could only contain 3s or higher, resulting in a sum greater than 8. Therefore, there was no need to check beyond that point.
That makes sense, thank you.
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