Expected number of “ones” in the rolls
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A game is played by repeatedly rolling a fair die until the die turns up "six" four times, then the game stops.
Given that the last game lasted greater than or equal to 20 rolls, find the expected number of "ones" in the rolls.
probability
asked Dec 4 at 1:01
Ryan Riley
11
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A game is played by repeatedly rolling a fair die until the die turns up "six" four times, then the game stops.
Given that the last game lasted greater than or equal to 20 rolls, find the expected number of "ones" in the rolls.
probability
asked Dec 4 at 1:01
Ryan Riley
11
What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
1
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26
add a comment |
up vote
0
down vote
favorite
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up vote
0
down vote
favorite
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1
A game is played by repeatedly rolling a fair die until the die turns up "six" four times, then the game stops.
Given that the last game lasted greater than or equal to 20 rolls, find the expected number of "ones" in the rolls.
probability
asked Dec 4 at 1:01
Ryan Riley
11
A game is played by repeatedly rolling a fair die until the die turns up "six" four times, then the game stops.
Given that the last game lasted greater than or equal to 20 rolls, find the expected number of "ones" in the rolls.
probability
probability
asked Dec 4 at 1:01
Ryan Riley
11
asked Dec 4 at 1:01
Ryan Riley
11
asked Dec 4 at 1:01
Ryan Riley
11
asked Dec 4 at 1:01
Ryan Riley
11
asked Dec 4 at 1:01
Ryan Riley
11
11
What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
1
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26
add a comment |
What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
1
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26
What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
1
1
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26
add a comment |
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What have you tried? Can you find the chance that you get four sixes before twenty rolls? That will give you a rescaling factor for removing those cases.
– Ross Millikan
Dec 4 at 1:04
Can you also find the expected number of ones if the game lasts, say, 30 rolls? How about $n$ rolls?
– Ross Millikan
Dec 4 at 1:21
I have found that basically what the problem is asking is "how many ones will you get if you roll a dice until you get four sixes". I believe the E(X) of the number of rolls would be k/p until you get four sixes since the distribution follows a negative binomial. After that, I am not sure how to derive how many ones we will see.
– Ryan Riley
Dec 4 at 1:34
1
You know there are four sixes. All the rest of the rolls are equally distributed between the other numbers.
– Ross Millikan
Dec 4 at 2:26