probability dice question

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nicholaskong100

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Since having a number be a certain number show up a dice is 1/6. And I am rolling 8 of these dices.

Would the probability be (1/6) to the power of 8?
 
Suppose I asked you what was the probability of rolling eight 2’s.

How might you compute that?

Do the two answers seem logically consistent!
 
Did you take into account that the first die can be anything?
In how many ways can you roll eight dice and get all the dice to show the same number?
 
JeffM said:
Suppose I asked you what was the probability of rolling eight 2’s.

How might you compute that?

Do the two answers seem logically consistent!
Click to expand...
{2,2,2,2,2,2,2,2} only shows up one time out of the 6^8 outcomes. Thus, it would be 1/(6^8)= 1/1679616
 
nicholaskong100 said:
{2,2,2,2,2,2,2,2} only shows up one time out of the 6^8 outcomes. Thus, it would be 1/(6^8)= 1/1679616
Click to expand...
That is correct. But it is not the question you were asked.
You were asked What is the probability that all eight dice show the same number?
 
Jomo said:
Did you take into account that the first die can be anything?
In how many ways can you roll eight dice and get all the dice to show the same number?
Click to expand...
Oh so that what the question meant...Well in that case, same number would be the subset {1,1,1,1,1,1,1,1}, {2,2,2,2,2,2,2,2},{3,3,3,3,3,3,3,3},{4,4,4,4,4,4,4,4},{5,5,5,5,5,5,5,5},{6,6,6,6,6,6,6,6}. Which is a totally of 6 of these things out of the 6^8 outcomes. So probability would be 6/(6^8)
 
nicholaskong100 said:
{2,2,2,2,2,2,2,2} only shows up one time out of the 6^8 outcomes. Thus, it would be 1/(6^8)= 1/1679616
Click to expand...
True, but the original question did not ask about eight 2’s, but 8 of any of the six possible numbers, which are mutually exclusive so the correct answer is

[math]\dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} = 6 * \dfrac{1}{6^8} = \dfrac{1}{6^7}.[/math]
The arithmetic in probability theory is usually simple. Fully grasping the question is what may be tricky.
 
JeffM said:
True, but the original question did not ask about eight 2’s, but 8 of any of the six possible numbers, which are mutually exclusive so the correct answer is

[math]\dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} + \dfrac{1}{6^8} = 6 * \dfrac{1}{6^8} = \dfrac{1}{6^7}.[/math]
The arithmetic in probability theory is usually simple. Fully grasping the question is what may be tricky.
Click to expand...
I noticed you added 1/(6^8) six times. But aren't we rolling eight dices, shouldn't we be adding 1/(6^8) eight times?
 
nicholaskong100 said:
I noticed you added 1/(6^8) six times. But aren't we rolling eight dices, shouldn't we be adding 1/(6^8) eight times?
Click to expand...
Whoa there. You need to think about what the different numbers mean. Often a good way to think about complex problems is to start by thinking about a simpler problem. Suppose the problem is what is the probability of rolling two fours from one throw of two fair dice.

Each of the 36 possible outcomes is equally likely.

Exactly one of those outcomes is two fours, the specified outcome.

So the answer to that very simple problem is [imath]\dfrac{1}{6^2} = \dfrac{1}{36}[/imath].

Any question about the answer to that problem?

If you are good so far, let's try a different problem. What is the probability of rolling two ones or two twos or two threes?

Each of the 36 possible outcomes is equally likely.

Three of those outcomes give two ones, two twos, or two threes, the specified outcomes. But they are mutually exclusive so we can just add the individual probabilities together.

[math]\dfrac{1}{6^2} + \dfrac{1}{6^2} + \dfrac{1}{6^2} = 3 * \dfrac{1}{6^2} = \dfrac{1}{12}.[/math].

Make sense? If not, write out all 36 possible outcomes using a red die and a green die. How many of them give the specified outcome? Three.

The number of probabilities we add up tells us how many individual (mutually exclusive) cases are specified. (In this simple problem, three). The exponent on the six tells us how many dice we are using. The six tells us that each die is of the normal kind with six distinct faces.

Now in your problem, there are indeed eight dice. That is where the exponent of 8 comes in. How many distinct outcomes are specified if we say all the dice show the same number of spots. That is six. The fact that there are eight dice is fully accounted for in that exponent of 8. We add up six fractions because six is the number of specified cases.

Does that clarify things?
 
JeffM said:
Whoa there. You need to think about what the different numbers mean. Often a good way to think about complex problems is to start by thinking about a simpler problem. Suppose the problem is what is the probability of rolling two fours from one throw of two fair dice.

Each of the 36 possible outcomes is equally likely.

Exactly one of those outcomes is two fours, the specified outcome.

So the answer to that very simple problem is [imath]\dfrac{1}{6^2} = \dfrac{1}{36}[/imath].

Any question about the answer to that problem?

If you are good so far, let's try a different problem. What is the probability of rolling two ones or two twos or two threes?

Each of the 36 possible outcomes is equally likely.

Three of those outcomes give two ones, two twos, or two threes, the specified outcomes. But they are mutually exclusive so we can just add the individual probabilities together.

[math]\dfrac{1}{6^2} + \dfrac{1}{6^2} + \dfrac{1}{6^2} = 3 * \dfrac{1}{6^2} = \dfrac{1}{12}.[/math].

Make sense? If not, write out all 36 possible outcomes using a red die and a green die. How many of them give the specified outcome? Three.

The number of probabilities we add up tells us how many individual (mutually exclusive) cases are specified. (In this simple problem, three). The exponent on the six tells us how many dice we are using. The six tells us that each die is of the normal kind with six distinct faces.

Now in your problem, there are indeed eight dice. That is where the exponent of 8 comes in. How many distinct outcomes are specified if we say all the dice show the same number of spots. That is six. The fact that there are eight dice is fully accounted for in that exponent of 8. We add up six fractions because six is the number of specified cases.

Does that clarify things?
Click to expand...
I was redoing the problem using the simple problem thinking. It is such a great technique to use. And I understand how the solution was arrived now. Thank you, Jeff.
 
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