joe_schmoe All American 18758 Posts user info edit post |
so i've got two normal dice. to roll a 7 is a 16.67% chance.
what if i roll them 4 times? is the overall chance of getting a 7 just 16.67% * 4 = 66.67% chance? doesn't seem right, because using that logic, then rolling them 6 times is a 100% chance.
excuse my ignorance. i got a C in ST370, and that was like 8 years ago. 2/15/2010 2:50:29 AM |
johnny57 All American 624 Posts user info edit post |
oh wow 2/15/2010 3:48:52 AM |
DPK All American 2390 Posts user info edit post |
Ignoring the law of averages or other crazy number theories, your odds are still the same for every roll of a set of dice. 2/15/2010 5:43:46 AM |
EuroTitToss All American 4790 Posts user info edit post |
I want to say 1-(1-p)^n
but it's been like 5 years since I took statistics
in other words, getting a 7 within two rolls is the chance of getting it once plus the chance of NOT getting it once and getting it the second time. maybe
[Edited on February 15, 2010 at 7:32 AM. Reason : asdf] 2/15/2010 7:31:07 AM |
jethromoore All American 2529 Posts user info edit post |
16.67%^4= .077% (1 in 1300 chance to roll 4 7's in a row)
^I think that's right if you are looking for the chance of getting atleast one 7 in 4 rolls
[Edited on February 15, 2010 at 8:35 AM. Reason : ] 2/15/2010 8:34:13 AM |
0EPII1 All American 42541 Posts user info edit post |
Quote : | "what if i roll them 4 times? is the overall chance of getting a 7" |
question is ambiguous.
you will roll 2 dice, and then repeat this 3 more times, right?
what do you want the probability of? a sum of 7 each of the 4 times? a sum of 7 at least once?
4 times: (1/6)^4 = 1/1,296
at least once: 1 - (prob of not a sum of 7 in 4 repetitions) = 1 - (5/6)^4 = 671/1,2962/15/2010 8:54:00 AM |
neolithic All American 706 Posts user info edit post |
^Is correct.
Phrasing on this is pretty ambiguous.
[Edited on February 15, 2010 at 2:10 PM. Reason : ] 2/15/2010 2:09:01 PM |
qntmfred retired 40726 Posts user info edit post |
but what if you roll 8 dice, 1 at a time? 2/16/2010 8:55:30 PM |
EuroTitToss All American 4790 Posts user info edit post |
what I want to know is how you get a 7 on normal dice 2/16/2010 10:35:03 PM |
neolithic All American 706 Posts user info edit post |
Dice = 2 die. 3+4 = 7 or 5+2=7 or 6+1=7 2/16/2010 11:07:58 PM |
EuroTitToss All American 4790 Posts user info edit post |
jesus christ, I'm illiterate 2/17/2010 6:44:00 AM |
Colemania All American 1081 Posts user info edit post |
.1667^4 = 0.0772222% chance of it happening
Assuming that the previous rolls probability doesnt alter your current probability on your next roll.
A more simple example: Youre an 80% free throw shooter, chances of making two in a row are: .8^2 = 64%, assuming that making the first doesnt increase or decrease the odds of making the next one. 2/22/2010 3:15:06 PM |
neolithic All American 706 Posts user info edit post |
^ The OP didn't say whether it was for seeing one 7 or four sevens in a row. 2/22/2010 5:48:07 PM |
EuroTitToss All American 4790 Posts user info edit post |
kind of ridiculous that we're arguing over the OP's intentions as if he were one of the founding fathers or some shit, but...
Quote : | "the overall chance of getting a 7" |
seems like he meant at least one 7 to me, not all 7s2/22/2010 6:38:15 PM |
0EPII1 All American 42541 Posts user info edit post |
Quote : | "kind of ridiculous that we're arguing over the OP's intentions as if he were one of the founding fathers or some shit, but..." |
^ Huh? You are being weird... we are arguing over it because it really is not clear what he meant (as he used non-standard language), and he is the one who needs the answer. What's ridiculous about that?
And if you really want to guess from his non-standard language ("the overall chance of getting a 7"), I don't see how you can construe "a 7" to mean "at least one 7" if you are taking it literally.
To recap:
One 7: 4*(1/6)*(5/6)^3 = 500/1,296 ~ 39%
>= One 7: 1 - (5/6)^4 = 671/1,296 ~ 52%
All 7s: (1/6)^4 = 1/1,296 ~ 0.077%
But yeah, we really are the idiots for arguing over this... joe_schmoe posted his question and disappeared. He needs to show up and explain what he meant .
[Edited on February 23, 2010 at 6:31 AM. Reason : ]2/23/2010 6:21:54 AM |
EuroTitToss All American 4790 Posts user info edit post |
the OP probably stopped giving a fuck a week ago
but we're still debating his confusion like it's holy scripture or the constitution 2/23/2010 9:27:39 AM |
neolithic All American 706 Posts user info edit post |
It turns out this thread was very appropriately titled. 2/23/2010 1:49:02 PM |
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EuroTitToss All American 4790 Posts user info edit post |
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joe_schmoe All American 18758 Posts user info edit post |
lol.... sorry. i forgot all about this thread. i think i was half-drunk playing Settlers of Catan online. and as for my "original intent", i was trying to ask
-- what is the probability that I will *eventually* get a 7, when rolling two dice, 'n' number of times ?
IOW, a score of 7 has a 16.66% chance of occurring on any given roll of these two dice. if i roll these two dice 4 times, what's the probability that at least one of those throws will result in a score of 7. 3/16/2010 12:35:52 PM |