Probability
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I think they're saying that you're throwing a single die. So, probability of throwing a 6 the first time = 1/6.....probability of throwing a second 6 is again 1/6. There are six different sides (1, 2, 3, 4, 5, and 6) so the chances it lands on 6 are 1/6. The only place this is kind of tricky is the fact that they say you do it twice. So, in reality, the answer should be 1/3....(1/6 x 2). So basically you have double the chances of getting a six if you roll twice as opposed to if you only rolled the die once. Hmmm...I'm confused now lol. Is this a destroyer problem? Sounds like it haha
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The probability is depandant on the first outcome. So, you can roll either 1,2,3,4,5 (not 6) on the first die so the probability of rolling those is 5/6. For the second roll the probability of rolling the complimentary number lets say first one was 1 second one has to be 5 for the sum to equal 6. So that outcome is 1/6. So answer should be 5/36. Not sure how they go 1/6. Where did you get the question from?
I'm not that great at probability but here are my thoughts:
To roll a sum of 6 in two tosses of one die, there are a couple of ways you can do this:
Roll 1 and 5 (or 5 and 1)
Roll 2 and 4 (or 4 and 2)
Roll 3 and 3 (or 3 and 3)
That's 6 possible ways to roll a sum of 6. There are 6^2 = 36 possible combinations you could roll. So the probability is 6/36 = 1/6. Does this make sense? Can anyone confirm this?
To roll a sum of 6 in two tosses of one die, there are a couple of ways you can do this:
Roll 1 and 5 (or 5 and 1)
Roll 2 and 4 (or 4 and 2)
Roll 3 and 3 (or 3 and 3)
That's 6 possible ways to roll a sum of 6. There are 6^2 = 36 possible combinations you could roll. So the probability is 6/36 = 1/6. Does this make sense? Can anyone confirm this?
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I disagree and am inclined to go with the same answer as Jia7. Since the same die is thrown twice, from a sequential perspective, the only way to land a sum of 6 is by rolling 1-5 the first time, and then its complement the second time, so 5 pairs out of 36 total. With an answer of 6/36, you imply you count the 3-3 roll twice, which should not be the case, since 3-3' and 3'-3 is the exact same throw. Look at the possible ways the throws can land:
1-1...2-1...3-1...4-1...5-1...6-1
1-2...2-2...3-2...4-2...5-2...6-2
1-3...2-3...3-3...4-3...5-3...6-3
1-4...2-4...3-4...4-4...5-4...6-4
1-5...2-5...3-5...4-5...5-5...6-5
1-6...2-6...3-6...4-6...5-6...6-6
out of 36 possible throws only five of them add up to 6: 5/36 should be your answer.
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We're talking probability here. We cannot take out the 6-1, 6-2, 6-3, 6-4, 6-5, 6-6 because they are all part of the probability. If it was a 5 sided die then it would be ok. But you are skewing the numbers in your favor to get the answer. If you did take 6 out of the equation. The probability of rolling 1-5 on the first chance is 1/1 or 100% and the probability of rolling it's complimentary number to add up to 6 is 1/5. You multiply the probabilities of the first and second roll and you get the probability of rolling a number that will add up to 6. The logic in how you just take out 6 of the probability outcomes does not make sense.
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My answer would be 5/36 as well.
It's definitely 5/36
Like someone said earlier, to roll a sum of six you can roll anything you want on the first dice besides a 6 (odds are 5/6). Then on the second dice, assuming you didn't roll a 6, you only have one number available to get to 6 (odds 1/6). Multiply together to get 5/36.
The odds of rolling a 7 are the highest possible for two rolls and are 1/6. So the odds to roll a six have to be lower than that so the answer cannot be 1/6
Ah, best explanation, makes complete sense~ The odds are only 5/6 x 1/6 because you need an exact number to fit the first number pulled. Ex: Pulled a 4 first roll (5/6), you need exactly a 2 to match, so 1/6 chance for this to happen.. similarly with 3+3, 2+4, etc.
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*The only question still, is if 3-3' the same thing as rolling 3'-3.. I would think it's the same so you would only count it once, but I think Jia's question includes it as a second possibility (thus where the extra roll comes from).
Yea you got it. 3-3' is the same as 3'-3. If you think about the method you described above, each succesful sum of a 6 is due to the initial roll + it's "conjugate" roll to get to 6. And since there are only 5 initial rolls that can be paired with a "conjugate roll" to succesfully sum to 6, these are the 5 events that should be divided by the 36 possible events.
