View Full Version : A DIE IS ROLLED 19 TIMES $10 TO CORRECT ANSWER
kurosh
10-11-2007, 04:35 PM
A DIE IS ROLLED 19 TIMES, WHAT IS THE STANDARD DEVIATION OF THE NUMBER OF 2'S ROLLED? YOU HAVE 5 MINUTES. WORK NEEDED
BigBiceps
10-11-2007, 04:41 PM
mean = 19/6
variance = 19*5/6/6
standard deviation = sqrt (variance)
sd = 1.62
BigBiceps
10-11-2007, 04:43 PM
ie.
sd = sqrt(npq)
where
n=19
p=1/6
q=5/6
blah_blah
10-11-2007, 04:44 PM
this is a binomial distribution so sd = sqrt(np(1-p))
p = 1/6
n = 19
sd = 1.624
e: the answer above is incorrect and tells you the variance and not the sd
e2: big biceps edited it to my answer
how many sides on the die?
budblown
10-11-2007, 05:26 PM
quit doing his homework for him
ImsaKidd
10-11-2007, 05:45 PM
The answer is 19. You can get 0 2's or 19 2's. 19-0=19.
stinkypete
10-11-2007, 06:55 PM
who cares, its not normally distributed
SmokeyRidesAgain
10-11-2007, 07:11 PM
The dice put you on AK.
ImsaKidd
10-11-2007, 08:04 PM
[ QUOTE ]
The dice put you on AK.
[/ QUOTE ]
The dice put you to a decision for all your chips.
prodonkey
10-11-2007, 08:09 PM
all 20 bb?
hra146
10-11-2007, 08:09 PM
50/50
either you have the answer or you dont.
sledghammer
10-11-2007, 08:19 PM
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
colonel81
10-11-2007, 08:26 PM
its not a fair die - ship
Alobar
10-11-2007, 08:50 PM
[ QUOTE ]
quit doing his homework for him
[/ QUOTE ]
yeah, hell never learn unless he does it himself!
ICMoney
10-11-2007, 08:57 PM
It depends if you're betting on the Pass Line or not.
Double Ice
10-11-2007, 09:09 PM
It's true that the answer is sqrt(npq) = sqrt(19*5)/6, where n is the number of trials, p the probability of success, and q = 1-p.
----
Often the derivation of the standard answer (stddev. = sqrt(npq) ) is omitted, and thus hard to find. In general, a proof of this fact can be found here http://planetmath.org/encyclopedia/BernoulliDistribution2.html , which requires knowledge of moment-generating functions.
Also, another way to do this problem is just to use the definition and grunge out terms. Since stddev is the square root of variance, and variance v = E( (X - m)^2 ) where m = EX, lets calculate m (it is 19/6).
Now, we can get X = k with probability p_k = (nCk) p^k q^(n-k), and so we want v = sum_{k=0 to n} p_k (k - np)^2. This adds to npq by calculation of the 20 terms, which would be time consuming for a human, but not for say, a computer.
stinkypete
10-11-2007, 09:09 PM
[ QUOTE ]
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
[/ QUOTE ]
please tell me what i "think i know"
Double Ice
10-11-2007, 09:16 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
[/ QUOTE ]
please tell me what i "think i know"
[/ QUOTE ]
Well, for a fixed probability of success P, and N trials, since N is discrete, clearly a binomial distribution is not a normal distribution technically. However, the normal distribution approximates it quite well, and also the binomial distribution converges to the normal one as N -> infinity. So while a binomial distribution is not actually a normal one, this has more to do with discreteness/continuity than not being in the "spirit" of a normal one.
stinkypete
10-11-2007, 09:19 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
[/ QUOTE ]
please tell me what i "think i know"
[/ QUOTE ]
Well, for a fixed probability of success P, and N trials, since N is discrete, clearly a binomial distribution is not a normal distribution technically. However, the normal distribution approximates it quite well, and also the binomial distribution converges to the normal one as N -> infinity. So while a binomial distribution is not actually a normal one, this has more to do with discreteness/continuity than not being in the "spirit" of a normal one.
[/ QUOTE ]
that's not what he was referring to at all, unless he's even dumber than i thought. please let sledghammer reply so i can tell him he's dumb.
sledghammer
10-11-2007, 10:01 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
[/ QUOTE ]
please tell me what i "think i know"
[/ QUOTE ]
Well, for a fixed probability of success P, and N trials, since N is discrete, clearly a binomial distribution is not a normal distribution technically. However, the normal distribution approximates it quite well, and also the binomial distribution converges to the normal one as N -> infinity. So while a binomial distribution is not actually a normal one, this has more to do with discreteness/continuity than not being in the "spirit" of a normal one.
[/ QUOTE ]
that's not what he was referring to at all, unless he's even dumber than i thought. please let sledghammer reply so i can tell him he's dumb.
[/ QUOTE ]
I assumed you were implying that variance isn't important for non-normal distributions, because all you know to do with variance is run a t-test/ci.
iloveny161
10-11-2007, 10:04 PM
answer is 7
stinkypete
10-11-2007, 10:41 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
who cares, its not normally distributed
[/ QUOTE ]
Spoken like someone who thinks they know what they are talking about.
[/ QUOTE ]
please tell me what i "think i know"
[/ QUOTE ]
Well, for a fixed probability of success P, and N trials, since N is discrete, clearly a binomial distribution is not a normal distribution technically. However, the normal distribution approximates it quite well, and also the binomial distribution converges to the normal one as N -> infinity. So while a binomial distribution is not actually a normal one, this has more to do with discreteness/continuity than not being in the "spirit" of a normal one.
[/ QUOTE ]
that's not what he was referring to at all, unless he's even dumber than i thought. please let sledghammer reply so i can tell him he's dumb.
[/ QUOTE ]
I assumed you were implying that variance isn't important for non-normal distributions, because all you know to do with variance is run a t-test/ci.
[/ QUOTE ]
why would you be so retarded? i have no idea what a t-test/ci is.
jessica1994
10-11-2007, 10:46 PM
pi
sledghammer
10-12-2007, 07:46 AM
I refuse to be leveled, but you still don't know what you're talking about.
StrongArm
10-12-2007, 07:53 AM
wheres will hunting when u need him
Pugwash
10-12-2007, 08:14 AM
LOL DICEAMENTS
stinkypete
10-12-2007, 02:05 PM
[ QUOTE ]
I refuse to be leveled, but you still don't know what you're talking about.
[/ QUOTE ]
i challenge thee to a statistics competition! you're a moron, and i know exactly what i'm talking about. hint: don't take my original reply so literally.
shootaa
10-12-2007, 08:22 PM
poker is more +ev
Cardgrrl
10-13-2007, 12:49 AM
42.
Isn't the answer always 42?
whorasaurus
10-13-2007, 12:50 AM
[ QUOTE ]
42.
Isn't the answer always 42?
[/ QUOTE ]
its been awhile, is this hitchhikers guide?
Seth Money
10-13-2007, 12:52 AM
too nerdy ; didn't read
[ QUOTE ]
3.16666666666667
[/ QUOTE ]
moving shapes
10-13-2007, 01:10 AM
oh babay
pnazari
10-13-2007, 01:35 AM
[ QUOTE ]
LOL DICEAMENTS
[/ QUOTE ]
TurdFerguson
10-13-2007, 02:41 AM
10+
Bulletproof Monk
10-13-2007, 02:52 AM
depends on which line you bet it on
TheFeeCees
10-13-2007, 03:22 AM
Eat the die preflop
RikaKazak
10-13-2007, 03:43 AM
7.....always pick 7
Banzai
10-13-2007, 04:11 AM
But how much did you lose?
JokersAttack
10-13-2007, 04:22 AM
a die is rolled 19 times and all of them are twos.
the moron gambler would say that because there have been so many twos, the other numbers are due to come up and would bet on a number other than two to come up next!
the dumb maths gambler would say that because every event is independent, the chance of a two is still 1/6 and would bet only if he is getting better than 5:1 odds.
the ship it crucial gambler would conclude that there must be something wrong with the dice and would bet with someone else (not the dice roller) that a two would come up next!
10 dollars please.
[ QUOTE ]
a die is rolled 19 times and all of them are twos.
the moron gambler would say that because there have been so many twos, the other numbers are due to come up and would bet on a number other than two to come up next!
the dumb maths gambler would say that because every event is independent, the chance of a two is still 1/6 and would bet only if he is getting better than 5:1 odds.
the ship it crucial gambler would conclude that there must be something wrong with the dice and would bet with someone else (not the dice roller) that a two would come up next!
10 dollars please.
[/ QUOTE ]
This answer seems to be worth more than $10
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