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 Message Boards » » ridiculously stupid probability question. Page [1]  
joe_schmoe
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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
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oh wow

2/15/2010 3:48:52 AM

DPK
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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
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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
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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
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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 = 671/1,296

2/15/2010 8:54:00 AM

neolithic
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^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
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but what if you roll 8 dice, 1 at a time?

2/16/2010 8:55:30 PM

EuroTitToss
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what I want to know is how you get a 7 on normal dice

2/16/2010 10:35:03 PM

neolithic
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Dice = 2 die. 3+4 = 7 or 5+2=7 or 6+1=7

2/16/2010 11:07:58 PM

EuroTitToss
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jesus christ, I'm illiterate

2/17/2010 6:44:00 AM

Colemania
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.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
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^ 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
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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 7s

2/22/2010 6:38:15 PM

0EPII1
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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
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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
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It turns out this thread was very appropriately titled.

2/23/2010 1:49:02 PM

Firehoze
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3/3/2010 10:30:05 AM

EuroTitToss
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3/3/2010 8:59:19 PM

joe_schmoe
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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

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