-
We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.
Tricky math problem
- Thread starter Mark R
- Start date
Frosty3799
Diamond Member
at least one of each value, so that would have to be 6 times
mugs
Lifer
Flipping the coin - twice.
If you flip it two times you have the following possibilities:
HH
HT
TH
TT
50% of those have one head and one tail.
The die I believe is higher than 6. Just a quick sanity check tells me that if you roll the die 6 times, odds are you won't roll one of each number.
If you flip it two times you have the following possibilities:
HH
HT
TH
TT
50% of those have one head and one tail.
The die I believe is higher than 6. Just a quick sanity check tells me that if you roll the die 6 times, odds are you won't roll one of each number.
DrPizza
Administrator Elite Member Goat Whisperer
Wow, I'm tempted to, but I *really* hate these types of problems. (and probability in general)
I hate probability problems.
(in case this is a homework assignment, I'm not giving away the answer, just a suggested path.)
Obviously, it has to be at least 6 rolls of the die.
After 6 rolls, what's the probability that at least 2 of the rolls came up the same?
After 7 rolls, what's the probability that at least 2 of the rolls came up the same?
-once you get 50% (or the closest number less than 50%) you have your answer.
A very similar problem, with the solution published in many places is "how many people in a room do you need to have before there is a 50% probability that 2 of them have the same birthday."
I hate probability problems.
(in case this is a homework assignment, I'm not giving away the answer, just a suggested path.)
Obviously, it has to be at least 6 rolls of the die.
After 6 rolls, what's the probability that at least 2 of the rolls came up the same?
After 7 rolls, what's the probability that at least 2 of the rolls came up the same?
-once you get 50% (or the closest number less than 50%) you have your answer.
A very similar problem, with the solution published in many places is "how many people in a room do you need to have before there is a 50% probability that 2 of them have the same birthday."
I was not kidding. He said that the coin probability is 2^2. 4 is not a possible value for a probability. Though while typing up this response, I think I understand what logic1485 meant with that post (which wasn't clear the first time I read it). He meant that it would take 2^2 times to get the probability to be at least 50%. I read it as the probability is 2^2 which clearly is not allowed. My bad.
-Tom
There is no (simple) scientific way to solve the second problem. You must use a guess and check method.
If you roll the die six times, the event that you get one of each side would be you rolling each side on each of the six turns, i.e. starting with a 5, then a 4, then a 6, then 2, 3, and 1.
The probability of that is 1 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6..since it doesn't matter what the first die you roll is, and in the second roll you have 5 choices out of 6 to go, then you have 4 values that haven't been rolled yet, etc.. Equivalently that is 5!/6^5 = 6!/6^6 (a pair of 6's cancel out)
The prob of that is 0.015. What we want though is a probability that is >= 50%. It is clearly a lot more than 6.
Let's roll it 7 times then. Then to get one of each value you have to get 6 specific rolls, while the 7th roll is irrelevant. So, you can roll a 6, 5, 2, 1, 6, 3, 4. Try to figure out probability of that.
If you roll the die six times, the event that you get one of each side would be you rolling each side on each of the six turns, i.e. starting with a 5, then a 4, then a 6, then 2, 3, and 1.
The probability of that is 1 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6..since it doesn't matter what the first die you roll is, and in the second roll you have 5 choices out of 6 to go, then you have 4 values that haven't been rolled yet, etc.. Equivalently that is 5!/6^5 = 6!/6^6 (a pair of 6's cancel out)
The prob of that is 0.015. What we want though is a probability that is >= 50%. It is clearly a lot more than 6.
Let's roll it 7 times then. Then to get one of each value you have to get 6 specific rolls, while the 7th roll is irrelevant. So, you can roll a 6, 5, 2, 1, 6, 3, 4. Try to figure out probability of that.
TRENDING THREADS
-
Discussion Zen 5 Speculation (EPYC Turin and Strix Point/Granite Ridge - Ryzen 9000)- Started by DisEnchantment
- Replies: 25K
-
Discussion Intel Meteor, Arrow, Lunar & Panther Lakes + WCL Discussion Threads
- Started by Tigerick
- Replies: 25K
-
Discussion Intel current and future Lakes & Rapids thread- Started by TheF34RChannel
- Replies: 24K
-
-
