Dice game keep rolling [closed]
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There is a dice game where both players roll a fair dice once. If player A rolls a 1, then he keeps rolling. And in the end, if the score is tied, player B wins. What is the probability of player A winning?
probability puzzle
asked Aug 27 at 16:43
user587649
61
closed as off-topic by amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants Aug 28 at 0:35
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants
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There is a dice game where both players roll a fair dice once. If player A rolls a 1, then he keeps rolling. And in the end, if the score is tied, player B wins. What is the probability of player A winning?
probability puzzle
asked Aug 27 at 16:43
user587649
61
closed as off-topic by amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants Aug 28 at 0:35
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
1
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06
 |Â
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up vote
1
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up vote
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favorite
There is a dice game where both players roll a fair dice once. If player A rolls a 1, then he keeps rolling. And in the end, if the score is tied, player B wins. What is the probability of player A winning?
probability puzzle
asked Aug 27 at 16:43
user587649
61
There is a dice game where both players roll a fair dice once. If player A rolls a 1, then he keeps rolling. And in the end, if the score is tied, player B wins. What is the probability of player A winning?
probability puzzle
asked Aug 27 at 16:43
user587649
61
asked Aug 27 at 16:43
user587649
61
asked Aug 27 at 16:43
user587649
61
asked Aug 27 at 16:43
user587649
61
61
closed as off-topic by amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants Aug 28 at 0:35
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants
closed as off-topic by amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants Aug 28 at 0:35
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Adrian Keister, Xander Henderson, Strants
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
1
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06
 |Â
show 1 more comment
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
1
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
1
1
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06
 |Â
show 1 more comment
1 Answer
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Make yourself a grid of all possibilities:
$beginarraycccccchline
colorblue2>1&2leq 2&2leq 3&2leq 4&2leq 5&2leq 6\
colorblue3>1&colorblue3>2&3leq 3&3leq 4&3leq 5&3leq 6\
vdots\
colorblue6>1&dots&&dots&colorblue6>5&6leq 6endarray$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=frac5cdot 62=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
Now, taking the ratio of the number of "good" scenarios to the total number of scenarios gives the probability:
$$Pr(A~textwins)=frac1530=frac12$$
answered Aug 27 at 17:26
JMoravitz
44.7k33582
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Make yourself a grid of all possibilities:
$beginarraycccccchline
colorblue2>1&2leq 2&2leq 3&2leq 4&2leq 5&2leq 6\
colorblue3>1&colorblue3>2&3leq 3&3leq 4&3leq 5&3leq 6\
vdots\
colorblue6>1&dots&&dots&colorblue6>5&6leq 6endarray$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=frac5cdot 62=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
Now, taking the ratio of the number of "good" scenarios to the total number of scenarios gives the probability:
$$Pr(A~textwins)=frac1530=frac12$$
answered Aug 27 at 17:26
JMoravitz
44.7k33582
add a comment |Â
up vote
1
down vote
Make yourself a grid of all possibilities:
$beginarraycccccchline
colorblue2>1&2leq 2&2leq 3&2leq 4&2leq 5&2leq 6\
colorblue3>1&colorblue3>2&3leq 3&3leq 4&3leq 5&3leq 6\
vdots\
colorblue6>1&dots&&dots&colorblue6>5&6leq 6endarray$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=frac5cdot 62=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
Now, taking the ratio of the number of "good" scenarios to the total number of scenarios gives the probability:
$$Pr(A~textwins)=frac1530=frac12$$
answered Aug 27 at 17:26
JMoravitz
44.7k33582
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Make yourself a grid of all possibilities:
$beginarraycccccchline
colorblue2>1&2leq 2&2leq 3&2leq 4&2leq 5&2leq 6\
colorblue3>1&colorblue3>2&3leq 3&3leq 4&3leq 5&3leq 6\
vdots\
colorblue6>1&dots&&dots&colorblue6>5&6leq 6endarray$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=frac5cdot 62=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
Now, taking the ratio of the number of "good" scenarios to the total number of scenarios gives the probability:
$$Pr(A~textwins)=frac1530=frac12$$
answered Aug 27 at 17:26
JMoravitz
44.7k33582
Make yourself a grid of all possibilities:
$beginarraycccccchline
colorblue2>1&2leq 2&2leq 3&2leq 4&2leq 5&2leq 6\
colorblue3>1&colorblue3>2&3leq 3&3leq 4&3leq 5&3leq 6\
vdots\
colorblue6>1&dots&&dots&colorblue6>5&6leq 6endarray$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=frac5cdot 62=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
Now, taking the ratio of the number of "good" scenarios to the total number of scenarios gives the probability:
$$Pr(A~textwins)=frac1530=frac12$$
answered Aug 27 at 17:26
JMoravitz
44.7k33582
answered Aug 27 at 17:26
JMoravitz
44.7k33582
answered Aug 27 at 17:26
JMoravitz
44.7k33582
answered Aug 27 at 17:26
JMoravitz
44.7k33582
44.7k33582
add a comment |Â
add a comment |Â
This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied?
– paw88789
Aug 27 at 16:50
B wins if the score is tied. Other outcomes, A wins.
– user587649
Aug 27 at 16:54
1
What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$.
– JMoravitz
Aug 27 at 17:02
A's final score is the first roll that is not 1.
– user587649
Aug 27 at 17:05
Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster)
– JMoravitz
Aug 27 at 17:06