Roll a fair die 5 times

  • #1
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Homework Statement


Roll a fair die 5 times, find the probability that the first two rolls have the same outcomes.

Homework Equations




The Attempt at a Solution


The total outcomes is 6^5, I think we have 6^2 * 6 choose 2 /6^5 since the first two numbers are fixed and we can choose 2 numbers from 1 to 6 be a pair. Where do I count it wrong?
 

Answers and Replies

  • #2
olivermsun
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1. What is the probability that two rolls of a die are the same?

2. What effect do the next 3 rolls have on the probability of certain things happening for the first two rolls...?
 
  • #3
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1. What is the probability that two rolls of a die are the same?

2. What effect do the next 3 rolls have on the probability of certain things happening for the first two rolls...?
I don't understand your second question. The probability of two rolls of a die are the same is 1/6
 
  • #4
olivermsun
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You roll the die 5 times. You know where you're at after 2 rolls. What effect do the next 3 rolls have on the outcome?
 
  • #5
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You roll the die 5 times. You know where you're at after 2 rolls. What effect do the next 3 rolls have on the outcome?
I can have a combination of any 3 number of 6 if I understand correctly
 
  • #6
PeroK
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I can have a combination of any 3 number of 6 if I understand correctly

Probability can be difficult if you just don't see these things. Look at it this way. If the first two are the same, let's call that "success" and if the first two are different, let's call that "failure". For example:

##44362## is a success

##21555## is a failure

Now, let's do the experiment again, slowly this time:

First, we get a ##6## then we get a ##5##. Now, if we stop there, do we already know whether we have success of failure? We still have to throw the die three more times, but it's difficult to see what could possibly happen now to make the first two the same. We already know this is a failure. Nothing can happen with the last three throws to change this. Let's say we throw them anyway and get:

##65461## - and this is a failure. The first two are different, but we knew that already.

Next time, suppose we get ##3## then ##3## for the first two throws. We know already that we have a success. It doesn't matter what the last three throws are, it cannot change the first two. Let's say we get:

##33226## - this is a success, which we already knew after the first two throws:

In other words, the probability that the first two are the same depends only on the first two throws. The remaining throws, whether you throw the die once more, three more times or a thousand more times can make no difference.
 
  • #7
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Probability can be difficult if you just don't see these things. Look at it this way. If the first two are the same, let's call that "success" and if the first two are different, let's call that "failure". For example:

##44362## is a success
Probability can be difficult if you just don't see these things. Look at it this way. If the first two are the same, let's call that "success" and if the first two are different, let's call that "failure". For example:

##44362## is a success

##21555## is a failure

Now, let's do the experiment again, slowly this time:

First, we get a ##6## then we get a ##5##. Now, if we stop there, do we already know whether we have success of failure? We still have to throw the die three more times, but it's difficult to see what could possibly happen now to make the first two the same. We already know this is a failure. Nothing can happen with the last three throws to change this. Let's say we throw them anyway and get:

##65461## - and this is a failure. The first two are different, but we knew that already.

Next time, suppose we get ##3## then ##3## for the first two throws. We know already that we have a success. It doesn't matter what the last three throws are, it cannot change the first two. Let's say we get:

##33226## - this is a success, which we already knew after the first two throws:

In other words, the probability that the first two are the same depends only on the first two throws. The remaining throws, whether you throw the die once more, three more times or a thousand more times can make no difference.
In the first two throw, I will have 6/36 to obtain two same outcomes. Then on third throw, forth throw anf fifth throw, I have 6 number to choose, the probability is 6^3/6^5.

So, in this problem, it doesn't matter wheat her first aND second roll habe same outcomes or second and third roll have same outcomes, the probability still the sam.

##21555## is a failure

Now, let's do the experiment again, slowly this time:

First, we get a ##6## then we get a ##5##. Now, if we stop there, do we already know whether we have success of failure? We still have to throw the die three more times, but it's difficult to see what could possibly happen now to make the first two the same. We already know this is a failure. Nothing can happen with the last three throws to change this. Let's say we throw them anyway and get:

##65461## - and this is a failure. The first two are different, but we knew that already.

Next time, suppose we get ##3## then ##3## for the first two throws. We know already that we have a success. It doesn't matter what the last three throws are, it cannot change the first two. Let's say we get:

##33226## - this is a success, which we already knew after the first two throws:

In other words, the probability that the first two are the same depends only on the first two throws. The remaining throws, whether you throw the die once more, three more times or a thousand more times can make no difference.
 
  • #8
Ray Vickson
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Homework Statement


Roll a fair die 5 times, find the probability that the first two rolls have the same outcomes.

Homework Equations




The Attempt at a Solution


The total outcomes is 6^5, I think we have 6^2 * 6 choose 2 /6^5 since the first two numbers are fixed and we can choose 2 numbers from 1 to 6 be a pair. Where do I count it wrong?

Do you understand the concept of "independent events" in probability? Are the results of the first two tosses independent of the results of the last three tosses? If your answer is "yes", do you see how this simplifies the solution?
 
  • #9
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Do you understand the concept of "independent events" in probability? Are the results of the first two tosses independent of the results of the last three tosses? If your answer is "yes", do you see how this simplifies the solution?

I see that already, we have 6/36* 6^3/6^3=1/6
 
  • #10
PeroK
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I see that already, we have 6/36* 6^3/6^3=1/6

I would say we have ##6/36## and the rest is irrelevant. There is no need to involve the remaining three throws in your calculations.
 
  • #11
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I would say we have ##6/36## and the rest is irrelevant. There is no need to involve the remaining three throws in your calculations.

Now , suppose we roll a die n times, would the probability that any ith, jth, kth, lth roll have the same outcomes still 1/6?
 
  • #12
PeroK
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Now , suppose we roll a die n times, would the probability that any ith, jth, kth, lth roll have the same outcomes still 1/6?

Any two specific rolls, ##i## and ##j##, yes. You can ignore the rest. The probability that the ##i##th throw equals the ##j##th is ##1/6##.
 
  • #13
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Any two specific rolls, ##i## and ##j##, yes. You can ignore the rest. The probability that the ##i##th throw equals the ##j##th is ##1/6##.
I mean _ _ _... I _ _ _..._ j _ _ _...._ _k... something like this.
 
  • #14
PeroK
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I mean _ _ _... I _ _ _..._ j _ _ _...._ _k... something like this.

The probablity that any three rolls are the same is ##1/36##. It doesn't matter whether it's the first three, the last three or any three.
 
  • #15
Ray Vickson
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I mean _ _ _... I _ _ _..._ j _ _ _...._ _k... something like this.

You tell us. What do YOU think the answer would be?
 
  • #16
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The probablity that any three rolls are the same is ##1/36##. It doesn't matter whether it's the first three, the last three or any three.

If we let R_j,k be event that jth and kth rolls have the same outcome, then events R_j,k are't pairwise independent.
 
  • #17
PeroK
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If we let R_j,k be event that jth and kth rolls have the same outcome, then events R_j,k are't pairwise independent.

Suppose we roll a die 100 times. Let's assume that the probability that the first three are the same is ##1/36##. Do you agree with that?

Now, give me three numbers (##i, j, k##) where the probability that the ##i##th, ##j##th and ##k##th rolls are the same is not equal to ##1/36##.

Maybe ##21, 32## and ##99##? Maybe there's something special about those rolls? And, is it more or less likely than the first three?
 
  • #18
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Suppose we roll a die 100 times. Let's assume that the probability that the first three are the same is ##1/36##. Do you agree with that?

Now, give me three numbers (##i, j, k##)
where the probability that the ##i##th, ##j##th and ##k##th rolls are the same is not equal to ##1/36##.




Maybe ##21, 32## and ##99##? Maybe there's something special about those rolls?


And, is it more or less likely than the first three?

Want to make sure i understand. The order of rolls doesn't matter. They are pairwise independent. However, not mutually independent be because P (R_j,k , R_a,b) is 1/6^3 and P (R_j,k) is 1/6 with district indices j,k,a,b
 
  • #19
PeroK
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Want to make sure i understand. The order of rolls doesn't matter. They are pairwise independent. However, not mutually independent be because P (R_j,k , R_a,b) is 1/6^3 and P (R_j,k) is 1/6 with district indices j,k,a,b

What does P (R_j,k , R_a,b) mean?
 
  • #20
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What does P (R_j,k , R_a,b) mean?
j,k,a,b are distinct indices, P (R_j,k and R_a,b) means j,k,a,bth rolls have the same outcomes
 
  • #21
PeroK
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j,k,a,b are distinct indices, P (R_j,k and R_a,b) means j,k,a,bth rolls have the same outcomes

That's four different rolls?

Or, is it that ##j, k## are the same and ##a, b## are the same, but ##j## may not be equal to ##a##?
 
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  • #22
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That's four different rolls?

Or, is it that ##j, k## are the same and ##a, b## are the same, but ##j## may not be equal to ##a##?

Yes, j,k,a,b gives the same outputs, we have 4 rolls have the same outputs.
 
  • #23
PeroK
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Yes, j,k,a,b gives the same outputs, we have 4 rolls have the same outputs.

A better notation would be ##R_{jkab}##.

Want to make sure i understand. The order of rolls doesn't matter. They are pairwise independent. However, not mutually independent be because P (R_j,k , R_a,b) is 1/6^3 and P (R_j,k) is 1/6 with district indices j,k,a,b

Yes, with your notation.
 

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