Probability: Rolling six, six-sided dice a single time
$begingroup$
My specific question (to settle a score-keeping argument in my family's never-ending Farkle tournament) is this:
- when rolling six, six-sided dice a single time, is it more likely to roll three pairs of three different numbers, or three dice of a single number?
For instance, in my family's game of Farkle, if you roll three pairs you get 200 points. However, if you get three of any number in particular you get that specific digit times x 100, except if you get three one's, then you get 1,000, or three fives, then you get 5,000.
- As the stickler that I am, I just want to make sure we're scoring relative to correctly weighted probability of rolling outcomes.
Feel free to also include other six, six-sided dice probability answers to help out other folks who stumble on this thread in search of help.
Thank you!
probability
asked Jan 2 at 14:52
user631109user631109
61
$endgroup$
|
show 2 more comments
$begingroup$
My specific question (to settle a score-keeping argument in my family's never-ending Farkle tournament) is this:
- when rolling six, six-sided dice a single time, is it more likely to roll three pairs of three different numbers, or three dice of a single number?
For instance, in my family's game of Farkle, if you roll three pairs you get 200 points. However, if you get three of any number in particular you get that specific digit times x 100, except if you get three one's, then you get 1,000, or three fives, then you get 5,000.
- As the stickler that I am, I just want to make sure we're scoring relative to correctly weighted probability of rolling outcomes.
Feel free to also include other six, six-sided dice probability answers to help out other folks who stumble on this thread in search of help.
Thank you!
probability
asked Jan 2 at 14:52
user631109user631109
61
$endgroup$
$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18
|
show 2 more comments
$begingroup$
My specific question (to settle a score-keeping argument in my family's never-ending Farkle tournament) is this:
- when rolling six, six-sided dice a single time, is it more likely to roll three pairs of three different numbers, or three dice of a single number?
For instance, in my family's game of Farkle, if you roll three pairs you get 200 points. However, if you get three of any number in particular you get that specific digit times x 100, except if you get three one's, then you get 1,000, or three fives, then you get 5,000.
- As the stickler that I am, I just want to make sure we're scoring relative to correctly weighted probability of rolling outcomes.
Feel free to also include other six, six-sided dice probability answers to help out other folks who stumble on this thread in search of help.
Thank you!
probability
asked Jan 2 at 14:52
user631109user631109
61
$endgroup$
My specific question (to settle a score-keeping argument in my family's never-ending Farkle tournament) is this:
- when rolling six, six-sided dice a single time, is it more likely to roll three pairs of three different numbers, or three dice of a single number?
For instance, in my family's game of Farkle, if you roll three pairs you get 200 points. However, if you get three of any number in particular you get that specific digit times x 100, except if you get three one's, then you get 1,000, or three fives, then you get 5,000.
- As the stickler that I am, I just want to make sure we're scoring relative to correctly weighted probability of rolling outcomes.
Feel free to also include other six, six-sided dice probability answers to help out other folks who stumble on this thread in search of help.
Thank you!
probability
probability
asked Jan 2 at 14:52
user631109user631109
61
asked Jan 2 at 14:52
user631109user631109
61
asked Jan 2 at 14:52
user631109user631109
61
asked Jan 2 at 14:52
user631109user631109
61
asked Jan 2 at 14:52
user631109user631109
61
61
$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18
|
show 2 more comments
$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18
$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18
|
show 2 more comments
1 Answer
1
active
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$begingroup$
Let's just compute. Of course there are $6^6$ unrestricted ways to throw the six dice.
Case: Three Pair.
We pick the three paired values,$binom 63=20$. We then place the two appearances of the least value, $binom 62= 15$ and then place the two appearances of the middle value, $binom 42=6$. Thus there are $$20times 15times 6=1800$$ ways to throw three pair. So the probability is $$frac {20times 15times 6}{6^6}=.03858$$
Case: three of a kind.
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $AAABCD$ or permutations of that then there are far more than $1800$ cases. For that pattern there are $$6times binom 63times 5times 4times 3=7200$$ ways to do it. So already you are $4$ times more likely to get this than to get a permutation of $AABBCC$
answered Jan 2 at 15:08
lulululu
43.3k25080
$endgroup$
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
add a comment |
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$begingroup$
Let's just compute. Of course there are $6^6$ unrestricted ways to throw the six dice.
Case: Three Pair.
We pick the three paired values,$binom 63=20$. We then place the two appearances of the least value, $binom 62= 15$ and then place the two appearances of the middle value, $binom 42=6$. Thus there are $$20times 15times 6=1800$$ ways to throw three pair. So the probability is $$frac {20times 15times 6}{6^6}=.03858$$
Case: three of a kind.
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $AAABCD$ or permutations of that then there are far more than $1800$ cases. For that pattern there are $$6times binom 63times 5times 4times 3=7200$$ ways to do it. So already you are $4$ times more likely to get this than to get a permutation of $AABBCC$
answered Jan 2 at 15:08
lulululu
43.3k25080
$endgroup$
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
add a comment |
$begingroup$
Let's just compute. Of course there are $6^6$ unrestricted ways to throw the six dice.
Case: Three Pair.
We pick the three paired values,$binom 63=20$. We then place the two appearances of the least value, $binom 62= 15$ and then place the two appearances of the middle value, $binom 42=6$. Thus there are $$20times 15times 6=1800$$ ways to throw three pair. So the probability is $$frac {20times 15times 6}{6^6}=.03858$$
Case: three of a kind.
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $AAABCD$ or permutations of that then there are far more than $1800$ cases. For that pattern there are $$6times binom 63times 5times 4times 3=7200$$ ways to do it. So already you are $4$ times more likely to get this than to get a permutation of $AABBCC$
answered Jan 2 at 15:08
lulululu
43.3k25080
$endgroup$
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
add a comment |
$begingroup$
Let's just compute. Of course there are $6^6$ unrestricted ways to throw the six dice.
Case: Three Pair.
We pick the three paired values,$binom 63=20$. We then place the two appearances of the least value, $binom 62= 15$ and then place the two appearances of the middle value, $binom 42=6$. Thus there are $$20times 15times 6=1800$$ ways to throw three pair. So the probability is $$frac {20times 15times 6}{6^6}=.03858$$
Case: three of a kind.
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $AAABCD$ or permutations of that then there are far more than $1800$ cases. For that pattern there are $$6times binom 63times 5times 4times 3=7200$$ ways to do it. So already you are $4$ times more likely to get this than to get a permutation of $AABBCC$
answered Jan 2 at 15:08
lulululu
43.3k25080
$endgroup$
Let's just compute. Of course there are $6^6$ unrestricted ways to throw the six dice.
Case: Three Pair.
We pick the three paired values,$binom 63=20$. We then place the two appearances of the least value, $binom 62= 15$ and then place the two appearances of the middle value, $binom 42=6$. Thus there are $$20times 15times 6=1800$$ ways to throw three pair. So the probability is $$frac {20times 15times 6}{6^6}=.03858$$
Case: three of a kind.
Now there are a number of patterns to consider. But even if we consider the strongest constraint, namely patterns of the form $AAABCD$ or permutations of that then there are far more than $1800$ cases. For that pattern there are $$6times binom 63times 5times 4times 3=7200$$ ways to do it. So already you are $4$ times more likely to get this than to get a permutation of $AABBCC$
answered Jan 2 at 15:08
lulululu
43.3k25080
answered Jan 2 at 15:08
lulululu
43.3k25080
answered Jan 2 at 15:08
lulululu
43.3k25080
answered Jan 2 at 15:08
lulululu
43.3k25080
43.3k25080
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
add a comment |
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
$begingroup$
numerically, the probability of any three die being equal (including quads, trip+pair etc) is around $0.3675$, compared to that $0.03858$ for three pairs. It's almost 10 times more likely
$endgroup$
– John Doe
Jan 2 at 15:16
add a comment |
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$begingroup$
Not sure this is clear. When you say three dice of a single color would you also include all six of that number? Would you include something like $(1,1,1,2,2,5)$?
$endgroup$
– lulu
Jan 2 at 15:04
$begingroup$
You ask about "scoring relative to correctly weighted probability" but you already have six events with the same probability (rolling three of any one of the six numbers) with scores ranging from $200$ to $5000,$ so what exactly is the point you want to be a stickler about? (Idle curiosity is still a very good reason to ask which is more probable, so the question is fine, I just wonder if we're missing something.)
$endgroup$
– David K
Jan 2 at 15:08
$begingroup$
- I believe i said three dice of a single number, just to be clear. So yes, if rolling six dice altogether a single time, what are the odds that three of those dice land on the same number. So yes, a 1-1-1-2-2-5 roll would qualify.
$endgroup$
– user631109
Jan 2 at 15:15
$begingroup$
Usually in betting or scoring situations, the payout scenarios are carefully isolated. In Poker, say, "three of a kind" usually precludes "four of a kind" or "full house". Anyway, as my post below indicates, even with the restrictive definition, it is far more likely to get three matching values than it is to get three pairs.
$endgroup$
– lulu
Jan 2 at 15:17
$begingroup$
The argument began on New Year's Eve so my memory of it is a little hazy, but I believe I argued that it was more likely to roll three ones than three pairs, even though three pairs is two-hundred points and three ones is a thousand points. I just thought that the scoring was a little bogus.
$endgroup$
– user631109
Jan 2 at 15:18