- My Forums
- Tiger Rant
- LSU Recruiting
- SEC Rant
- Saints Talk
- Pelicans Talk
- More Sports Board
- Fantasy Sports
- Golf Board
- Soccer Board
- O-T Lounge
- Tech Board
- Home/Garden Board
- Outdoor Board
- Health/Fitness Board
- Movie/TV Board
- Book Board
- Music Board
- Political Talk
- Money Talk
- Fark Board
- Gaming Board
- Travel Board
- Food/Drink Board
- Ticket Exchange
- TD Help Board
Customize My Forums- View All Forums
- Show Left Links
- Topic Sort Options
- Trending Topics
- Recent Topics
- Active Topics
Started By
Message
re: Help me figure out a dice/math question
Posted on 10/31/16 at 11:53 am to Nuts4LSU
Posted on 10/31/16 at 11:53 am to Nuts4LSU
quote:
Of course rolling more dice adds to the probability that ONE of them will turn up a 1 or a 5. Do you seriously think you could roll 1000 dice and there'd be a 66% chance that NONE of them would come up either 1 or 5?
The probability is the same for each die, but the more dice you roll, the better your chances that ONE of them will be what you need (in this case, 1 or 5).
Yea I should have never ventured into explaining this but I get what you are saying the probability that the probability of rolling a 1 or a 5 will increase when you add more die. I was wrong, but I knew that the odds did not double because he doubled the sample.
1
Posted on 10/31/16 at 11:54 am to FutureMikeVIII
36 possible combinations of two dice.
16 of them don't contain a 1 or a 5.
20 of them do.
Pretty easy to figure out from there.
16 of them don't contain a 1 or a 5.
20 of them do.
Pretty easy to figure out from there.
Posted on 10/31/16 at 11:57 am to Winston Cup
quote:
This sounds like the kind of problem that would never get figured out after a few beers.
We gave up on it. But I kept rolling when I was down to two dice and it sure as hell didn't feel like I was getting a 1 or 5 55% of the time.
Posted on 10/31/16 at 11:57 am to LSU_Saints_Hornets
quote:
I knew that the odds did not double because he doubled the sample.
As did we.
Posted on 10/31/16 at 11:58 am to Chucktown_Badger
33% first toss
33% second toss
33% third toss
Roughly 58% that you will roll a 1 or a 5 over three consecutive tosses
33% second toss
33% third toss
Roughly 58% that you will roll a 1 or a 5 over three consecutive tosses
This post was edited on 10/31/16 at 12:00 pm
Posted on 10/31/16 at 12:06 pm to Chucktown_Badger
quote:
We gave up on it. But I kept rolling when I was down to two dice and it sure as hell didn't feel like I was getting a 1 or 5 55% of the time.
Well risk/reward you should probably quit when down to two dice, because there's only a 1/9 chance you're going to roll a 1 or 5 on both dice. So you're risking whatever you have built up for an extra 50 or 100 points... not usually worth it.
Posted on 10/31/16 at 12:06 pm to Chucktown_Badger
2 dice. 12 total faces. 4 "good faces".
Probability of NO 1s or 5s is 2/3rds on each die.
2/3 X 2/3 - .445 = The probability that no ones or fives come up on both die. So subtract that from 1 and you get a prob of .555 that when you roll two dice a one or five will show up on one of the two.
Probability of NO 1s or 5s is 2/3rds on each die.
2/3 X 2/3 - .445 = The probability that no ones or fives come up on both die. So subtract that from 1 and you get a prob of .555 that when you roll two dice a one or five will show up on one of the two.
Posted on 10/31/16 at 12:08 pm to Fred Farkle
quote:
2 dice. 12 total faces. 4 "good faces".
Probability of NO 1s or 5s is 2/3rds on each die.
2/3 X 2/3 - .445 = The probability that no ones or fives come up on both die. So subtract that from 1 and you get a prob of .555 that when you roll two dice a one or five will show up on one of the two.
Farkle is in your name, of course you'd have the correct answer as well.
Posted on 10/31/16 at 12:14 pm to Chucktown_Badger
For those who need to visualize it, these are the possibilities:
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
20 of the 36 possible outcomes contain a 1 or a 5.
20/36 = 55.5%
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
20 of the 36 possible outcomes contain a 1 or a 5.
20/36 = 55.5%
Posted on 10/31/16 at 12:24 pm to Chucktown_Badger
The easiest way to do a this is to determine the odds of rolling neither on each dice then subtracting that number from 1. If you always remember that, then you don't need to memorize a formula
So 1-4/6*4/6
So 1-4/6*4/6
Posted on 10/31/16 at 12:25 pm to pointdog33
quote:six upvotes
I'm pretty sure it's still 1/3.
4 chances of 12 to get one of those numbers.
Posted on 10/31/16 at 1:30 pm to Chucktown_Badger
288
OT, I am disappoint.
OT, I am disappoint.
Posted on 10/31/16 at 5:29 pm to Chucktown_Badger
Prob of (1 or 5 on 1st die ) plus prob ( 1 or 5 on 2nd ) - prob of (1 or 5 on both dir ) equals
2/6 plus 2/6 minus 4/36 equals 20/36.
Castorinfo also correct. 1 minus prob of no 1s or 5s on two tosses equals 1 - 16/36
Equals. 20/36.
Fair betting odds on at least one 1 or 5 on two rolls is 1 to 1.2. i.e. Bet 12 to win 10.
2/6 plus 2/6 minus 4/36 equals 20/36.
Castorinfo also correct. 1 minus prob of no 1s or 5s on two tosses equals 1 - 16/36
Equals. 20/36.
Fair betting odds on at least one 1 or 5 on two rolls is 1 to 1.2. i.e. Bet 12 to win 10.
This post was edited on 10/31/16 at 5:38 pm
Posted on 10/31/16 at 6:33 pm to 81Tiger
That is about as clear as it gets.
Posted on 10/31/16 at 6:42 pm to Chucktown_Badger
I love farkle.
It is 33% chance to roll a 1 or 5 on one die.
In fact is it 200% chance you will hit 1 or 5 on roll 1!!
It is 33% chance to roll a 1 or 5 on one die.
In fact is it 200% chance you will hit 1 or 5 on roll 1!!
This post was edited on 10/31/16 at 6:44 pm
Posted on 10/31/16 at 6:43 pm to Pectus
As a rule, I don't roll on 2 left.
I do roll on triple 2s.
I do roll on triple 2s.
Posted on 10/31/16 at 7:01 pm to Nuts4LSU
quote:
To get the odds that it will happen, you have to calculate the odds that it won't happen, and subtract that from 100%.
2/3 x 2/3 = 4/9 chance that you won't roll a 1 or 5 on either, or a 5/9 chance that you will.
This right here.
quote:
Unless, of course, you are me playing backgammon and need a 1 or a 5 to get out of jail, in which case the probability of rolling a 1 or a 5 is precisely zero.
And also this.
Posted on 10/31/16 at 7:19 pm to foshizzle
Yes, 5/9 = 20/36
Page 2 of 2
Popular
Back to top
Follow TigerDroppings for LSU Football News