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Two Ones

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 

Puzzle ID:#44413
Fun:** (2.05)
Difficulty:*** (2.11)
Category:Probability
Submitted By:shenqiangAcn****!!

 

 

 



Which is more likely to happen?

1) Roll a fair die 4 times and get at least one 1;
2) Roll 2 fair dice 24 times and get at least one "double 1".





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Comments

Mill89Aus*
May 04, 2009

The latter equation should read:
(35/36)^6 (or 0.844) which is slightly larger than 5/6 (or 0.833).
xdbtcpAus**
May 04, 2009

So.....the answer is the first one or second one?
bradon182001*us*
May 07, 2009

Confusing.
Lil_Dork*ca*
May 15, 2009

weird...
javaguru*us*
Aug 12, 2009

The problem was easy enough, but the answer was not very clear. It's more likely to role the single one than the double one.

(35/36)^24 ≈ .5086
(5/6)^4 ≈ .4823

So the odds of rolling double ones is 1 - .5086 ≈ .4914 and the odds of rolling a single one are 1 - .4823 ≈ .5177.

dalfamnestAnz*
Aug 16, 2009

OK, I got it - but it's more of a textbook question than a teaser.
Juanito*
Jun 18, 2010

The probability to fail the first dice once is not 5/6 but 5/6+(1/6)*(5/6) for one time.
Juanito*
Jun 18, 2010

I was mentioning the two dice there....
mathemagicianus*
Jun 08, 2011

I teach university statistics and probablity.

javaguru is precisely correct.

X = successfully rolling a "1"
Y = successfully rolling a "double 1"

P(X>= 1) = 1 - P(X=0) = 1 - (5/6)^4

~ 0.5177 or 51.77%

P(Y >=1) = 1 - P(x=0) = 1 - (35/36)^24 ~ 0.4914 or 49.14%

So, rolling at least "one 1" is a more likely outcome than rolling at least "one double 1" by about 2.63%.
princess2007A*
Jul 27, 2011

ok. I gotta try this in about a year. XD I have no idea how you got your answers. I'm sure its good though. This goes into my favorites for now.
ron93
Jan 23, 2015

I got both 0.5177 and 0.4914038761 as the respective answers (not bad for a 21 year old guy like me)

Let X be the random variable which represents the number of 1 obtained in 4 rolls of a dice

then by simple combinatorics and noting that 6^4 = 1296, we have,

P(X=1) = {C(4,3)*5^3}/1296

P(X=2) = {C(4,2)*5^2}/1296

P(x = 3) = {C(4,1)*5}/1296

&

P(x=4) = 1/1296

then

P(at least one 1 in four rolls) = P(X>=1) = P(x=1)+P(x=2)+P(X=3)+P(x=4) = 0.51777

with the use of a scientific calculator

However, it takes a bit more patience to obtain the other answer which is 0.4914038761

Nevertheless we can represent how I got this answer with summations

((1/36)^24)*(Summation(1
ron93
Jan 23, 2015

my previous comment was incomplete for some reason, anyway,

the answer 0.4914038761.... is equivalent to

((1/36)^24)*Summation(from i=1 to 24)[C(24,i)*35^(24-i)]

if you are patient enough, you can check this with a calculator

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