### Brain Teasers

# Two Ones

Probability
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".

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".

### Hint

They seem to be equal, but they aren't.### 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

The latter equation should read:

(35/36)^6 (or 0.844) which is slightly larger than 5/6 (or 0.833).

(35/36)^6 (or 0.844) which is slightly larger than 5/6 (or 0.833).

So.....the answer is the first one or second one?

Confusing.

weird...

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 â‰ˆ .5086

(5/6)^4 â‰ˆ .4823

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

(35/36)^24 â‰ˆ .5086

(5/6)^4 â‰ˆ .4823

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

OK, I got it - but it's more of a textbook question than a teaser.

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

I was mentioning the two dice there....

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%.

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%.

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.

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

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

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

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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