10
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ I think the error you are making is that you're not accounting for the first roll properly... $\endgroup$ – David Mitra Mar 8 '12 at 3:30
18
$\begingroup$

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 $6\cdot({ 1\over6})^3 =({1\over6})^2$.

$\endgroup$
  • $\begingroup$ Makes sense, thanks for clearing it up very succinctly! $\endgroup$ – switz Mar 8 '12 at 3:33
  • $\begingroup$ David Mitra Sir :Can we do it like this :$$\binom{6}{1}$$ to select which outcome will be same i.e $$\frac{\binom{6}{1}}{6^{3}}$$? $\endgroup$ – laura Sep 2 '17 at 14:41
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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...)!

$\endgroup$
0
$\begingroup$

$S$=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 =${6 \choose 1}=6$

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

$\endgroup$

protected by Zev Chonoles Mar 29 '16 at 6:55

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.