Is the probability of rolling a single die...

[Hades]

which outcome has a greater occurence: having 4 rolls to land a six with one die, or having 36 rolls to land (2) 6's with 2 die?

[EverCurious]

Correct my math if I'm wrong, I always screw up simple equations...especially ones with exponential probabilities:

 

Having 36 rolls to land a double six with a pair of dice yields a greater chance of occuring as opposed to four rolls with one die to roll a six.

 

The probability of a die rolling a six is 1/6. With four rolls, the probability becomes 4/6.

 

The probability of a pair of dice rolling a double 6 is 1/21, since there are 21 possible combinations. With 36 chances the probability of rolling a double 6 becomes 36/21 or 12/7.

[JustStuit]

I think you have to use factorials or somethin because a fraction higher than one would imply an inevitable outcome.

[EverCurious]

Very true...I cannot believe I overlooked that...

[EverCurious]

i'm assuming you've taken prob/stat

[starbug1]

one die with four rolls yields a 2/3 probability for any one number.

 

With two dice you always take the number of sides (6) and raise an exponent with the number of dice (2). Therefore, you get a 6^2, which is 36. So, then, the probability of getting two sixes is 1/36.

 

2/3 > 1/36

[EverCurious]

you missed the part where you have 36 chances at rolling a 36 and with a pair of dice there aren't 36 possible outcomes, only 21...check your math, or maybe you just read the question wrong, or maybe i'm wrong (i've double checked now..don't see where i'm going wrong)

[J.branco]

as the probability of getting a 6 in a roll is 1/6, with 6rolls it's:

4

∑ (5^p-1)/(6^p) = 1/6 + 5/36 + 25/216 + 125/1296

p=1

 

following the same idea, ifthe probability of getting a double 6 in a roll with 2dies is 1/36, in 36rolls it's:

36

∑ (35^p-1)/(36^p)

p=1

 

right?

as they are geometric progressions,

for the first one the probabilty is (1/6)*(1-(5/6)^4)/(1-5/6) =.5177

for the secnd the probability is (1/36)*(1-(35/36)^36)/(1-35/36)=.6373

 

so you get better chances with the double 6:D

[J.branco]

sum for geometric progressions being u1*(1-r^n/1-r)

and r= 5/6 for the first and r=35/36 for the second

[starbug1]

sum for geometric progressions being u1*(1-r^n/1-r)

and r= 5/6 for the first and r=35/36 for the second

 

 

does it really have to be that complicated?

[BigMoosie]

It's easier to find out with 4 dice what are the odds of not getting any six: (5/6)^4, then subtract it from 1 to get the complimentary = 0.5177

 

With one roll of two die the odds of not getting double six are 35/36, using the same idea as before: 1-(35/36)^36 = 0.6373

[J.branco]

yeah, it's reallymuch easier that way;)