Two Ones
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
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".
HintThey seem to be equal, but they aren't.
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Answer
Rolling a fair die once and failing to get a 1 has probability 5/6, and rolling 2 fair dice 6 times and failing to get a double 1 has probability (35^36)^6, which is slightly larger than 5/6.
Therefore, if you do each of the above 4 times and still fail to get a 1, the latter has a larger probability, therefore, the latter has a smaller succeeding probability than the former.
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Comments
Mill89
May 04, 2009
 The latter equation should read:
(35/36)^6 (or 0.844) which is slightly larger than 5/6 (or 0.833). 
xdbtcp
May 04, 2009
 So.....the answer is the first one or second one? 
bradon182001
May 07, 2009
 Confusing. 
Lil_Dork
May 15, 2009
 weird... 
javaguru
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.

dalfamnest
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.... 
mathemagician
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%. 
princess2007
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^(24i)]
if you are patient enough, you can check this with a calculator 
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