If three dice are rolled, what is the probability that all three are the same number?

The dice are fair.

You have a $1\over6$ chance of getting the first number. A $1\over6$ chance of the second and so on. Is it just $({1\over6})^3$ (1/216) or is that not accounting for the second and third roll properly?

Solution 1:

It's just $({1\over6})^2$ It's the probability that the second roll is the same as the first (1/6) multiplied by the probability that the third roll is the same as the second (1/6).

Or, think of it this way. The desired outcomes are $(1,1,1)$, $(2,2,2)$, ... ,$(6,6,6)$. Each of these outcomes has probability $({1\over6})^3$. Sum these the probabilities of these mutually exclusive outcomes to get $6\cdot({ 1\over6})^3 =({1\over6})^2$.

Solution 2:

total no of outcomes will be 6x6x6=216.....favourable otcomes are (1,1,1), (2,2,2), ... ,(6,6,6)...i.e. 6 favouracle outcomes.....so probability will be 6/216=1/36

Solution 3:

Well, the probability would be $1\cdot$$1\over 6$$\cdot{1\over 6}$ because for three dice, there are six outcomes for the same number because there are six numbers, so put in a 1 for the first factor and every other outcome will be different numbers and you're talking about six-sided dice, so use $1\over 6$s for the next two factors. Also, the odds is 1 in 36 because of this. Hope this helps! Good luck also trying to beat the odds (if you have any dice, that is...)!

Solution 4:

$S$=sample space

$n(S)$=number of total outcome of sample space =$6^3=216$

$E$=event all the faces are same

$n(E)$=number of ways have same faces =${6 \choose 1}=6$

$P(E)=n(E)/n(S)=6/216=1/36$