Roll 5 Dice Simultaneously: Probabilities and Conditional Probability

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In summary: C2)*1. Then the three other dice can be any of the five other numbers. Here, the right method is 5*5*5, but why can't we use the choose function, 5C3, here? There are 5C3 ways of choosing the rest of the dice, and there are 5 choices for them, so (5C3)*5. Why is (5C3)*5=/=5*5*5?Using 5*5*5, P(two dice show 5)= [(5C2)*5^3]/[6^5]
  • #1
spacetimedude
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Homework Statement


Calculate
a) P(exactly three dice have the same number)
b) Calculate the conditional probability P(three of the dice shows six|two of the dice shows 5)

Homework Equations

The Attempt at a Solution


a) Say we have 1 as the number we get for three dice. There is 1/6 chance of getting a 1 and 5/6 of getting other numbers. The total sample space is 6^5 for five rolls.
Hence for P(getting three 1's)=[(1/6)^3*(5/6)^2]/[6^5]
But that was just for getting three 1's and there are total of 6 numbers that can be used.
So
P(exactly three dice have the same number)=6*[(1/6)^3*(5/6)^2]/[6^5].
This answer seems to be right, but I would like to know how you can solve this using the binomial coefficent.

My attempt:
3 in 5 rolls will result in a number, which can be expressed as 5C3. Each number has 1/6 chance of being that number. The others have the probability of 5/6. The sample space is still 6^5.
Putting this all together, [6*(5C3)*(1/6)^3*(5/6)^2]/[6^5], which yields a wrong answer.

b) The conditional probability P(three of the dice shows six|two of the dice shows 5) can be written as
P(three of the dice show six|two of the dice show 5)=P(three show 6, two show 5)/P(two show 5).
Why are the events not independent here? Each die is a different entity so can't we just say P(three shows 6, two show 5)=P(three shows 6)*P(two show 5)?
Anyway, P(two shows 5)=[(1/6)^2*(5/6)^3]/[6^5]
I am not sure how to find the probability of the intersection of three show 6 and two show 5. I tried using the propoerty that the numerator of conditional probability can be written as P(three show six)P(two show 5|three show 6) but that got me nowhere.

Any help will be appreciated!
 
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  • #2
spacetimedude said:
a) Say we have 1 as the number we get for three dice. There is 1/6 chance of getting a 1 and 5/6 of getting other numbers. The total sample space is 6^5 for five rolls.
Hence for P(getting three 1's)=[(1/6)^3*(5/6)^2]/[6^5]
But that was just for getting three 1's and there are total of 6 numbers that can be used.
So
P(exactly three dice have the same number)=6*[(1/6)^3*(5/6)^2]/[6^5].
This the probability for "the first 3 dice show the same number, the last two do not have this number" (or any other set of 3 specific dice). It does not matter which three dice show the same number, you have to take this into account.

Also, you include the factors of 1/6 twice. Use 1/6 and 5/6, or use 1 and 5 and divide by 65, but not both together. This is also a problem in all other formulas.
spacetimedude said:
My attempt:
3 in 5 rolls will result in a number, which can be expressed as 5C3. Each number has 1/6 chance of being that number. The others have the probability of 5/6. The sample space is still 6^5.
Putting this all together, [6*(5C3)*(1/6)^3*(5/6)^2]/[6^5], which yields a wrong answer.
Apart from the wrong factors of 6 (see above), this approach is right.

spacetimedude said:
Why are the events not independent here?
Same as above: the dice don't have a given order. A die that shows 5 cannot show 6 and vice versa.
spacetimedude said:
Anyway, P(two shows 5)=[(1/6)^2*(5/6)^3]/[6^5]
This is not right. Neither for "exactly two" nor for "at least two".
 
  • #3
mfb said:
This the probability for "the first 3 dice show the same number, the last two do not have this number" (or any other set of 3 specific dice). It does not matter which three dice show the same number, you have to take this into account.

Also, you include the factors of 1/6 twice. Use 1/6 and 5/6, or use 1 and 5 and divide by 65, but not both together. This is also a problem in all other formulas.
Apart from the wrong factors of 6 (see above), this approach is right.

Same as above: the dice don't have a given order. A die that shows 5 cannot show 6 and vice versa.This is not right. Neither for "exactly two" nor for "at least two".

Okay, I am having a bit of trouble understanding the choose function. It would be great if you can help me clear a problem:

P(two dice show 5)
So there are 5C2 possible ways to choose the two dice. And those dice have one choice (that they are showing 5), so for that part, it will be (5C2)*1. Then the three other dice can be any of the five other numbers. Here, the right method is 5*5*5, but why can't we use the choose function, 5C3, here? There are 5C3 ways of choosing the rest of the dice, and there are 5 choices for them, so (5C3)*5. Why is (5C3)*5=/=5*5*5?
Using 5*5*5, P(two dice show 5)= [(5C2)*5^3]/[6^5]
 
  • #4
spacetimedude said:
Okay, I am having a bit of trouble understanding the choose function. It would be great if you can help me clear a problem:

P(two dice show 5)
So there are 5C2 possible ways to choose the two dice. And those dice have one choice (that they are showing 5), so for that part, it will be (5C2)*1. Then the three other dice can be any of the five other numbers. Here, the right method is 5*5*5, but why can't we use the choose function, 5C3, here? There are 5C3 ways of choosing the rest of the dice, and there are 5 choices for them, so (5C3)*5. Why is (5C3)*5=/=5*5*5?
Using 5*5*5, P(two dice show 5)= [(5C2)*5^3]/[6^5]

To try to clear up this confusion, think of 3 dice, rather than 5. Suppose you want precisely two dice to show 5. There are ##3C2 = 3## ways to choose the two dice that are the same. These are:

55X
5X5
X55

Now, as we have chosen the two places with the 5's, there is no independent choice for the remaining place. It's already decided.

It's the same when you have 5 dice. There are ##5C2## choices for where the two 5's are. But, once you have chosen these, the remaining three places are chosen also:

55XXX
5X5XX
etc.
 
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  • #5
spacetimedude said:
P(two dice show 5)= [(5C2)*5^3]/[6^5]
"exactly two", yes.

Note that you can also choose the 3 non-5 dice instead of the 2 dice that are 5, as (5 choose 3) = (5 choose 2). It is still just one choice that fixes both groups.
 
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  • #6
PeroK said:
To try to clear up this confusion, think of 3 dice, rather than 5. Suppose you want precisely two dice to show 5. There are ##3C2 = 3## ways to choose the two dice that are the same. These are:

55X
5X5
X55

Now, as we have chosen the two places with the 5's, there is no independent choice for the remaining place. It's already decided.

It's the same when you have 5 dice. There are ##5C2## choices for where the two 5's are. But, once you have chosen these, the remaining three places are chosen also:

55XXX
5X5XX
etc.
That makes it much more clear! Thanks so much!

mfb said:
"exactly two", yes.

Note that you can also choose the 3 non-5 dice instead of the 2 dice that are 5, as (5 choose 3) = (5 choose 2). It is still just one choice that fixes both groups.
Okay :) thank you!
 
  • #7
I'm still a bit puzzled about part b of the original question. I do not know how to find the probability of the intersection (the numerator) in the conditional probability. Any hints to how I should approach this?
 
  • #8
P(three show 6, two show 5)? This is easier than all the other probabilities you calculated so far.
 
  • #9
mfb said:
P(three show 6, two show 5)? This is easier than all the other probabilities you calculated so far.
Yes, that one. I can do them individually but not sure how you can combine them.
 
  • #10
What do you mean by individually?

How many options are there to have "6 6 6 5 5" as result?
 
  • #11
Doh! So it's just 5C2 or 5C3?? So the probability is (5C3)/(6^5)?
 
  • #12
Right.
 
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Related to Roll 5 Dice Simultaneously: Probabilities and Conditional Probability

1. How does rolling 5 dice simultaneously affect the probability of getting a specific outcome?

Rolling 5 dice simultaneously does not affect the probability of getting a specific outcome. The probability of getting a specific outcome is determined by the number of possible outcomes divided by the total number of outcomes.

2. Is rolling 5 dice simultaneously more efficient than rolling them one at a time?

Rolling 5 dice simultaneously is not necessarily more efficient than rolling them one at a time. The efficiency depends on the context and the purpose of the dice roll. For example, in a game, rolling 5 dice simultaneously may save time, but in a scientific experiment, rolling them one at a time may be more accurate.

3. How many possible combinations can be obtained when rolling 5 dice simultaneously?

When rolling 5 dice simultaneously, there are 6 possible outcomes for each dice, resulting in a total of 6^5 = 7776 possible combinations.

4. Does the order of the dice matter when rolling them simultaneously?

Yes, the order of the dice does matter when rolling them simultaneously. For example, rolling a 2 on the first dice and a 3 on the second dice is considered a different outcome than rolling a 3 on the first dice and a 2 on the second dice.

5. How does the number of dice rolled simultaneously affect the overall outcome?

The number of dice rolled simultaneously can greatly affect the overall outcome. As the number of dice increases, the number of possible outcomes also increases, which can greatly impact the probability of getting a specific outcome. Additionally, rolling more dice simultaneously can also affect the distribution of the outcomes, making some outcomes more likely than others.

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