Mixing
What is the probability of at least one out of three players rolling “1” two times in a row on 6 sided...
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More importantly, what is the generalized equation for calculating the probability?
probability statistics discrete-mathematics dice
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
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closed as off-topic by Shailesh, José Carlos Santos, Adrian Keister, Leucippus, Claude Leibovici Jan 18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Leucippus, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
More importantly, what is the generalized equation for calculating the probability?
probability statistics discrete-mathematics dice
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
$endgroup$
closed as off-topic by Shailesh, José Carlos Santos, Adrian Keister, Leucippus, Claude Leibovici Jan 18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Leucippus, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
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– JMoravitz
Jan 17 at 20:22
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@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
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– JMoravitz
Jan 17 at 20:22
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@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
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– Chris Custer
Jan 17 at 20:28
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@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
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– JMoravitz
Jan 17 at 20:32
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@ChrisCuster each player rolls only 2 times.
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– KidBlondie
Jan 17 at 20:34
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show 6 more comments
$begingroup$
More importantly, what is the generalized equation for calculating the probability?
probability statistics discrete-mathematics dice
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
$endgroup$
More importantly, what is the generalized equation for calculating the probability?
probability statistics discrete-mathematics dice
probability statistics discrete-mathematics dice
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
asked Jan 17 at 20:10
KidBlondieKidBlondie
6
6
closed as off-topic by Shailesh, José Carlos Santos, Adrian Keister, Leucippus, Claude Leibovici Jan 18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Leucippus, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, José Carlos Santos, Adrian Keister, Leucippus, Claude Leibovici Jan 18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, Leucippus, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
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– JMoravitz
Jan 17 at 20:22
$begingroup$
@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
$endgroup$
– JMoravitz
Jan 17 at 20:22
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@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
$endgroup$
– Chris Custer
Jan 17 at 20:28
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@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
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– JMoravitz
Jan 17 at 20:32
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@ChrisCuster each player rolls only 2 times.
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– KidBlondie
Jan 17 at 20:34
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show 6 more comments
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
$endgroup$
– Chris Custer
Jan 17 at 20:28
$begingroup$
@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
$endgroup$
– JMoravitz
Jan 17 at 20:32
$begingroup$
@ChrisCuster each player rolls only 2 times.
$endgroup$
– KidBlondie
Jan 17 at 20:34
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
$endgroup$
– Chris Custer
Jan 17 at 20:28
$begingroup$
@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
$endgroup$
– Chris Custer
Jan 17 at 20:28
$begingroup$
@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
$endgroup$
– JMoravitz
Jan 17 at 20:32
$begingroup$
@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
$endgroup$
– JMoravitz
Jan 17 at 20:32
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@ChrisCuster each player rolls only 2 times.
$endgroup$
– KidBlondie
Jan 17 at 20:34
$begingroup$
@ChrisCuster each player rolls only 2 times.
$endgroup$
– KidBlondie
Jan 17 at 20:34
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show 6 more comments
1 Answer
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Recognize that for a specific individual to roll two "1"s in a row, this occurs with probability $frac{1}{6}timesfrac{1}{6}=frac{1}{36}$
Now, this implies that the probability for a specific individual to have the opposite occur, for him to roll something other than specifically two "1"s in a row (whether that means the person rolled only one "1" and something else, or both results to be something other than "1") this occurs with probability:
$ 1- frac{1}{36} = frac{35}{36}$
Consider then what the probability is that none of the players rolled a result of two "1"s.
$frac{35}{36}timesfrac{35}{36}timesfrac{35}{36}$
Finally, recognize that this is the opposite outcome from what we were originally looking for. That is, the opposite of "at least one person did..." is "no people did..."
Generalizing this should be clear
$1 - frac{35^n}{36^n}$
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recognize that for a specific individual to roll two "1"s in a row, this occurs with probability $frac{1}{6}timesfrac{1}{6}=frac{1}{36}$
Now, this implies that the probability for a specific individual to have the opposite occur, for him to roll something other than specifically two "1"s in a row (whether that means the person rolled only one "1" and something else, or both results to be something other than "1") this occurs with probability:
$ 1- frac{1}{36} = frac{35}{36}$
Consider then what the probability is that none of the players rolled a result of two "1"s.
$frac{35}{36}timesfrac{35}{36}timesfrac{35}{36}$
Finally, recognize that this is the opposite outcome from what we were originally looking for. That is, the opposite of "at least one person did..." is "no people did..."
Generalizing this should be clear
$1 - frac{35^n}{36^n}$
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
$begingroup$
Recognize that for a specific individual to roll two "1"s in a row, this occurs with probability $frac{1}{6}timesfrac{1}{6}=frac{1}{36}$
Now, this implies that the probability for a specific individual to have the opposite occur, for him to roll something other than specifically two "1"s in a row (whether that means the person rolled only one "1" and something else, or both results to be something other than "1") this occurs with probability:
$ 1- frac{1}{36} = frac{35}{36}$
Consider then what the probability is that none of the players rolled a result of two "1"s.
$frac{35}{36}timesfrac{35}{36}timesfrac{35}{36}$
Finally, recognize that this is the opposite outcome from what we were originally looking for. That is, the opposite of "at least one person did..." is "no people did..."
Generalizing this should be clear
$1 - frac{35^n}{36^n}$
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
$begingroup$
Recognize that for a specific individual to roll two "1"s in a row, this occurs with probability $frac{1}{6}timesfrac{1}{6}=frac{1}{36}$
Now, this implies that the probability for a specific individual to have the opposite occur, for him to roll something other than specifically two "1"s in a row (whether that means the person rolled only one "1" and something else, or both results to be something other than "1") this occurs with probability:
$ 1- frac{1}{36} = frac{35}{36}$
Consider then what the probability is that none of the players rolled a result of two "1"s.
$frac{35}{36}timesfrac{35}{36}timesfrac{35}{36}$
Finally, recognize that this is the opposite outcome from what we were originally looking for. That is, the opposite of "at least one person did..." is "no people did..."
Generalizing this should be clear
$1 - frac{35^n}{36^n}$
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
$endgroup$
Recognize that for a specific individual to roll two "1"s in a row, this occurs with probability $frac{1}{6}timesfrac{1}{6}=frac{1}{36}$
Now, this implies that the probability for a specific individual to have the opposite occur, for him to roll something other than specifically two "1"s in a row (whether that means the person rolled only one "1" and something else, or both results to be something other than "1") this occurs with probability:
$ 1- frac{1}{36} = frac{35}{36}$
Consider then what the probability is that none of the players rolled a result of two "1"s.
$frac{35}{36}timesfrac{35}{36}timesfrac{35}{36}$
Finally, recognize that this is the opposite outcome from what we were originally looking for. That is, the opposite of "at least one person did..." is "no people did..."
Generalizing this should be clear
$1 - frac{35^n}{36^n}$
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
answered Jan 17 at 20:43
JMoravitzJMoravitz
47.9k33886
47.9k33886
add a comment |
add a comment |
$begingroup$
How many times does each player roll a die? Twice each? Is the die a fair die? It is easiest to calculate the opposite event, no players rolling "1" two times in a row and subtracting this away from $1$. Do you know how to find the probability of rolling two "1"s in a row if there was only one player?
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@ChrisCuster ouch, please delete that comment. It is incorrect. Apply inclusion-exclusion or use the complementary event. If you need to be convinced of this, then consider what happens if you have a large number of players, such as $1000$. Your proposed formula would result in a probability much greater than $1$, a clear impossibility.
$endgroup$
– JMoravitz
Jan 17 at 20:22
$begingroup$
@JMoravitz. It may be incorrect, but I don't see why yet. If each player rolls twice...
$endgroup$
– Chris Custer
Jan 17 at 20:28
$begingroup$
@ChrisCuster what you calculated is the expected number of players who roll "1" two times in a row (when there are three players and each player rolls a fair die two times). This is different than the probability of at least one player doing so. $Pr(Acup Bcup C) = Pr(A)+Pr(B)+Pr(C)color{red}{-Pr(Acap B)-Pr(Acap C)-Pr(Bcap C)+Pr(Acap Bcap C)}$. You neglected to take any of the red terms into account.
$endgroup$
– JMoravitz
Jan 17 at 20:32
$begingroup$
@ChrisCuster each player rolls only 2 times.
$endgroup$
– KidBlondie
Jan 17 at 20:34