I roll two dice, where the first die gets a +1 bonus to it's roll
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Essentially, I'm playing a game, where two players each roll a 6 sided die, where the higher die roll wins. For some reason my die roll gets a +1 to whatever it lands on. What is the probability that my die roll is higher than my opponents die roll (assuming they get no bonuses)? We each get one roll per game.
probability
asked Dec 4 at 19:25
Christopher Hart
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Essentially, I'm playing a game, where two players each roll a 6 sided die, where the higher die roll wins. For some reason my die roll gets a +1 to whatever it lands on. What is the probability that my die roll is higher than my opponents die roll (assuming they get no bonuses)? We each get one roll per game.
probability
asked Dec 4 at 19:25
Christopher Hart
1
2
Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36
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favorite
up vote
0
down vote
favorite
Essentially, I'm playing a game, where two players each roll a 6 sided die, where the higher die roll wins. For some reason my die roll gets a +1 to whatever it lands on. What is the probability that my die roll is higher than my opponents die roll (assuming they get no bonuses)? We each get one roll per game.
probability
asked Dec 4 at 19:25
Christopher Hart
1
Essentially, I'm playing a game, where two players each roll a 6 sided die, where the higher die roll wins. For some reason my die roll gets a +1 to whatever it lands on. What is the probability that my die roll is higher than my opponents die roll (assuming they get no bonuses)? We each get one roll per game.
probability
probability
asked Dec 4 at 19:25
Christopher Hart
1
asked Dec 4 at 19:25
Christopher Hart
1
asked Dec 4 at 19:25
Christopher Hart
1
asked Dec 4 at 19:25
Christopher Hart
1
asked Dec 4 at 19:25
Christopher Hart
1
1
2
Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36
add a comment |
2
Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36
2
2
Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36
Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
HINT:
If you get the $+1$ bonus, but the other person doesn't, you can simplify that to being wherever your score is at least the value of theirs.
In other words, ignoring the bonus, if $A$ is your score and $B$ is theirs, you want:
$$P(Age B)$$
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
add a comment |
up vote
1
down vote
If $2$ die are rolled there are 36 different outcomes. $(1,1),(1,2) dots (1,6),(2,1), dots$ and so on. It is an empirical fact that for rolls with $2$ "normal" die all of these outcomes will occur with equal probability. So now if you count all the outcomes where you win and divide it by $36$ you will have the probability of winning.
edited Dec 4 at 20:22
Mason
1,8881527
answered Dec 4 at 19:40
Jagol95
555
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
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up vote
1
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HINT:
If you get the $+1$ bonus, but the other person doesn't, you can simplify that to being wherever your score is at least the value of theirs.
In other words, ignoring the bonus, if $A$ is your score and $B$ is theirs, you want:
$$P(Age B)$$
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
add a comment |
up vote
1
down vote
HINT:
If you get the $+1$ bonus, but the other person doesn't, you can simplify that to being wherever your score is at least the value of theirs.
In other words, ignoring the bonus, if $A$ is your score and $B$ is theirs, you want:
$$P(Age B)$$
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT:
If you get the $+1$ bonus, but the other person doesn't, you can simplify that to being wherever your score is at least the value of theirs.
In other words, ignoring the bonus, if $A$ is your score and $B$ is theirs, you want:
$$P(Age B)$$
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
HINT:
If you get the $+1$ bonus, but the other person doesn't, you can simplify that to being wherever your score is at least the value of theirs.
In other words, ignoring the bonus, if $A$ is your score and $B$ is theirs, you want:
$$P(Age B)$$
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
answered Dec 4 at 19:46
Rhys Hughes
4,6651327
4,6651327
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
add a comment |
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
To give an example of what I'm saying, suppose you both roll a $6$ on the die. You will win because of the $+1$ bonus
– Rhys Hughes
Dec 4 at 19:48
add a comment |
up vote
1
down vote
If $2$ die are rolled there are 36 different outcomes. $(1,1),(1,2) dots (1,6),(2,1), dots$ and so on. It is an empirical fact that for rolls with $2$ "normal" die all of these outcomes will occur with equal probability. So now if you count all the outcomes where you win and divide it by $36$ you will have the probability of winning.
edited Dec 4 at 20:22
Mason
1,8881527
answered Dec 4 at 19:40
Jagol95
555
add a comment |
up vote
1
down vote
If $2$ die are rolled there are 36 different outcomes. $(1,1),(1,2) dots (1,6),(2,1), dots$ and so on. It is an empirical fact that for rolls with $2$ "normal" die all of these outcomes will occur with equal probability. So now if you count all the outcomes where you win and divide it by $36$ you will have the probability of winning.
edited Dec 4 at 20:22
Mason
1,8881527
answered Dec 4 at 19:40
Jagol95
555
add a comment |
up vote
1
down vote
up vote
1
down vote
If $2$ die are rolled there are 36 different outcomes. $(1,1),(1,2) dots (1,6),(2,1), dots$ and so on. It is an empirical fact that for rolls with $2$ "normal" die all of these outcomes will occur with equal probability. So now if you count all the outcomes where you win and divide it by $36$ you will have the probability of winning.
edited Dec 4 at 20:22
Mason
1,8881527
answered Dec 4 at 19:40
Jagol95
555
If $2$ die are rolled there are 36 different outcomes. $(1,1),(1,2) dots (1,6),(2,1), dots$ and so on. It is an empirical fact that for rolls with $2$ "normal" die all of these outcomes will occur with equal probability. So now if you count all the outcomes where you win and divide it by $36$ you will have the probability of winning.
edited Dec 4 at 20:22
Mason
1,8881527
answered Dec 4 at 19:40
Jagol95
555
edited Dec 4 at 20:22
Mason
1,8881527
edited Dec 4 at 20:22
Mason
1,8881527
edited Dec 4 at 20:22
Mason
1,8881527
1,8881527
answered Dec 4 at 19:40
Jagol95
555
answered Dec 4 at 19:40
Jagol95
555
answered Dec 4 at 19:40
Jagol95
555
555
add a comment |
add a comment |
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Welcome to MSE! Did you make a table like this one? That should make it easier to visualize what's happening.
– Mason
Dec 4 at 19:36