Roll 5 dice simultaneously

  • #1

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!
 

Answers and Replies

  • #2
35,656
12,219
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.
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.

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.
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
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
18,716
10,352
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.
 
  • Like
Likes spacetimedude
  • #5
35,656
12,219
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.
 
  • Like
Likes spacetimedude
  • #6
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!

"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
35,656
12,219
P(three show 6, two show 5)? This is easier than all the other probabilities you calculated so far.
 
  • #9
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
35,656
12,219
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
35,656
12,219
Right.
 
  • Like
Likes spacetimedude

Related Threads on Roll 5 dice simultaneously

  • Last Post
Replies
1
Views
27K
Replies
14
Views
2K
  • Last Post
Replies
6
Views
4K
Replies
4
Views
843
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
3K
Replies
8
Views
1K
Replies
4
Views
2K
  • Last Post
Replies
22
Views
1K
Top