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"1 in 6, then one in 5. Unless you get 'it' on the first go then your odds are 1 in 1. Maybe." I wasn't very clear. I've got 5 dice, 4 of which are the ones I need and the other one I can roll twice in order to get the one outcome that I need. So it's 1 in 6 on the first roll and 1 in 6 on the second, but what would it be if I were to ask what are my chances of rolling a 1 (the only positive outcome) when I have two attempts. (If I rolled a 2 on the first roll, I could still roll a 2 on the second roll too) | |||
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"1 in 3? Or have I just made myself look stupid... Now i'm thinking of several ways that could be right or wrong and i've confused myself on a question about a simple dice roll. Clearly I'm not the target audience here "im saying nothing
apart from i have a a 1 in 4 lol. | |||
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"im saying nothing
apart from i have a a 1 in 4 lol."Hah, how'd you get to that? To be fair, staying up til 2am turns any question into a brain teaser. I've got 2-3 different answers to this one simple question and... yeah, it's probably bedtime  | |||
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"im saying nothing
apart from i have a a 1 in 4 lol.
Hah, how'd you get to that?
To be fair, staying up til 2am turns any question into a brain teaser. I've got 2-3 different answers to this one simple question and... yeah, it's probably bedtime "and drinking not good lol | |||
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"My brain hurts " Nothing new | |||
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"1 in 6, then one in 5. Unless you get 'it' on the first go then your odds are 1 in 1. Maybe. I wasn't very clear. I've got 5 dice, 4 of which are the ones I need and the other one I can roll twice in order to get the one outcome that I need. So it's 1 in 6 on the first roll and 1 in 6 on the second, but what would it be if I were to ask what are my chances of rolling a 1 (the only positive outcome) when I have two attempts. (If I rolled a 2 on the first roll, I could still roll a 2 on the second roll too)" OK... so you're rolling one dice, with 6 sides and have 2 chances to roll it, and you only want 1 particular outcome? The first roll, and presuming you don't get what you want first roll, it seems like you have a 2 in 12 chance? Ugh maths, i like puzzles though. | |||
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"My brain hurts Nothing new "Just because I pick up the thinking slack of the world from people like you who just can't manage on their own | |||
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"1 in 6, then one in 5. Unless you get 'it' on the first go then your odds are 1 in 1. Maybe. I wasn't very clear. I've got 5 dice, 4 of which are the ones I need and the other one I can roll twice in order to get the one outcome that I need. So it's 1 in 6 on the first roll and 1 in 6 on the second, but what would it be if I were to ask what are my chances of rolling a 1 (the only positive outcome) when I have two attempts. (If I rolled a 2 on the first roll, I could still roll a 2 on the second roll too) OK... so you're rolling one dice, with 6 sides and have 2 chances to roll it, and you only want 1 particular outcome? The first roll, and presuming you don't get what you want first roll, it seems like you have a 2 in 12 chance? Ugh maths, i like puzzles though. " Could it be 1 in 12 on the first roll and 1 in 6 on the second? 2 in 12 is the same as 1 in 6 isn't it? | |||
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"My brain hurts Nothing new
Just because I pick up the thinking slack of the world from people like you who just can't manage on their own "It's 6 to 1 no matter how many times you try. For the odds to change you'd have to take a number away each time you try. Derrrrr  | |||
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"1 in 6, then one in 5. Unless you get 'it' on the first go then your odds are 1 in 1. Maybe. I wasn't very clear. I've got 5 dice, 4 of which are the ones I need and the other one I can roll twice in order to get the one outcome that I need. So it's 1 in 6 on the first roll and 1 in 6 on the second, but what would it be if I were to ask what are my chances of rolling a 1 (the only positive outcome) when I have two attempts. (If I rolled a 2 on the first roll, I could still roll a 2 on the second roll too) OK... so you're rolling one dice, with 6 sides and have 2 chances to roll it, and you only want 1 particular outcome? The first roll, and presuming you don't get what you want first roll, it seems like you have a 2 in 12 chance? Ugh maths, i like puzzles though. " Yes correct - 2 in 12 chances - the "odds" are still 1 in 6 Divide 2 into 12 = 6:1 Like 4 in 16 chances is 1 in 4 or 4:1
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"My brain hurts Nothing new
Just because I pick up the thinking slack of the world from people like you who just can't manage on their own
It's 6 to 1 no matter how many times you try. For the odds to change you'd have to take a number away each time you try. Derrrrr "1 in 6 is the odds for each roll. I want the odds for the total amount of rolls. I don't even know if that calculation is possible.. | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again." Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something! | |||
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"1 in 6, then one in 5. Unless you get 'it' on the first go then your odds are 1 in 1. Maybe. I wasn't very clear. I've got 5 dice, 4 of which are the ones I need and the other one I can roll twice in order to get the one outcome that I need. So it's 1 in 6 on the first roll and 1 in 6 on the second, but what would it be if I were to ask what are my chances of rolling a 1 (the only positive outcome) when I have two attempts. (If I rolled a 2 on the first roll, I could still roll a 2 on the second roll too) OK... so you're rolling one dice, with 6 sides and have 2 chances to roll it, and you only want 1 particular outcome? The first roll, and presuming you don't get what you want first roll, it seems like you have a 2 in 12 chance? Ugh maths, i like puzzles though. Yes correct - 2 in 12 chances - the "odds" are still 1 in 6 Divide 2 into 12 = 6:1 Like 4 in 16 chances is 1 in 4 or 4:1
"Yeah thanks, i am so tired (up with a sick kid here and feeling a bit ill myself) but i managed to explain myself eventually. | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something! "Nope sweet you need to get a bloody life | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something!
Nope sweet you need to get a bloody life "Fuck off  I'm just playing Yahtzee and wanted to know the answer to a bloody question. Stop picking on me | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something!
Nope sweet you need to get a bloody life
Fuck off I'm just playing Yahtzee and wanted to know the answer to a bloody question. Stop picking on me "You no I love ya really. Come here and give us a kiss xxx | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something!
Nope sweet you need to get a bloody life
Fuck off I'm just playing Yahtzee and wanted to know the answer to a bloody question. Stop picking on me
You no I love ya really. Come here and give us a kiss xxx"No fuck off. Don't like you no more *strop* | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something!
Nope sweet you need to get a bloody life
Fuck off I'm just playing Yahtzee and wanted to know the answer to a bloody question. Stop picking on me "he's nowt else to do up there in that home, all the other residents asleep after bingo hours ago. it's just fab he has for company and his little mobility scooter (he thinks it's a motorbike bless him) | |||
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"Same as a roulette table .. if the marble lands on black 12 times in a row, what are the odds of it landing on red next time? The odds are exactly the same as it will land on black because each spin is an individual occurrence and the dice is the same. Each roll is a new even so you will always have a 1 in 6 chance of getting the desired outcome and the amount of rolls you make is irrelevant. I think!" So the second roll is irrelevant unless the first one fails? | |||
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"Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something! "You can only ever have a 1 in 6 chance of rolling what you want, seeing as there's 6 numbers. Every time you roll one dice you always have a 1 in 6 chance. Took me about a year to understand the Monty hall puzzle tbh as it didn't make sense to me for a long time. The probability of something happening is always 1 or 0. In your example of rolling 1 dice twice this means you only use addition to work out the probability. Maybe think of it this way..you are rolling the dice once, then roll it an additional time, so you can add the probability (like i did with 2 in 12). I'm a shit teacher but it's my favourite role play. | |||
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"Same as a roulette table .. if the marble lands on black 12 times in a row, what are the odds of it landing on red next time? The odds are exactly the same as it will land on black because each spin is an individual occurrence and the dice is the same. Each roll is a new even so you will always have a 1 in 6 chance of getting the desired outcome and the amount of rolls you make is irrelevant. I think! So the second roll is irrelevant unless the first one fails? " If you don't need to roll the dice twice then you rolled the first dice once, which is a 1 in 6 chance. There's no more than one dice and no more than 6 sides. Probability laws don't change unless some variable makes it more complicated (like 2 dices being rolled at once and you want 2 different numbers). They aren't really 'maths', they're more predictions and they don't even have to correlate with actual data/results. | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something!
Nope sweet you need to get a bloody life
Fuck off I'm just playing Yahtzee and wanted to know the answer to a bloody question. Stop picking on me
he's nowt else to do up there in that home, all the other residents asleep after bingo hours ago.
it's just fab he has for company and his little mobility scooter (he thinks it's a motorbike bless him)"That so made me laff  | |||
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"No. Like the guy above said the odds of rolling something one one particular dice is always the same. There are always 6 sides on that dice and always a 1 in 6 chance to get 1 outcome. 2 in 12 is just the raw answer. You're rolling 1 dice 2 times. So you have a 2 in 12 chance but usually you make fractions the smallest version of themselves and the smallest version of 2 in 12 is 1 in 6 again. Apparently it's this 1:6 [first roll] + (5:6)[probability of first roll not being a 1] x(1:6) [probability of second roll being a 1] = 11/36 It sounds clever. And it work out at just under 1 in 3. I don't know though. I need a maths professor or something! "Have you thought about playing Cluedo instead? | |||
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