Probability of rolling 6 dice with 1,2 & 3

Question

lets say all our dice is fair and the rules is we win if we got 1,2 or 3 and lose if we got 4,5 or 6. so the probability for 1 dice will be - 3/6 or 50% of winning/losing

how much the probability will be if we hv 6 dices and get a minimum of 3 winning dices(with faces 1, 2 or 3)?

look up on the net found something like this - 3/6 x 3/6 x 3/6 x 3/6 x 3/6 x 3/6 = 729/46656 or 1.5625% but thats not correct as we just need min.3 winning dices to consider a win

p/s - we think 9/216 also not correct

thanks in advance,

Answer 1

So we have a success rate of $p=\frac{1}{2}$ per experiment, we do $n=6$ experiments and if $X$ counts the number of succeses ; we need $P(X \ge 3)$. $X$ is distributed according to a binomial distibution and so

$$P(X \ge 3) = \sum_{i=3}^6 \binom{i}{6}\left(\frac{1}{2}\right)^6 = \frac{1}{64} (20 + 15+ 6+1) = \frac{21}{32}$$

Answer 2

Since the probability of winning when throwing dice with these rules is $\frac{1}{2}$, this can be equated to flipping a coin. Therefore, the question goes: When flipping a coin 6 times, what is the probability that there will be at least 3 Heads? (The probability of Heads is $\frac{1}{2}$)

The probability can be calculated like this: $$ \begin{split} P(\text{win}) &= P(3~\text{Heads}) + P(4~\text{Heads}) + P(5~\text{Heads}) +P(6~\text{Heads}) \\ &= {6 \choose 3} \left( \frac{1}{2} \right)^3\left( \frac{1}{2} \right)^3 + {6 \choose 4} \left( \frac{1}{2} \right)^4\left( \frac{1}{2} \right)^2 + {6 \choose 5} \left( \frac{1}{2} \right)^5\left( \frac{1}{2} \right)^1 + {6 \choose 6} \left( \frac{1}{2} \right)^6\left( \frac{1}{2} \right)^0 \\ &= \left( \frac{1}{2} \right)^6 \left[{6 \choose 3} + {6 \choose 4} + {6 \choose 5} +{6 \choose 6}\right] \\ &= \frac{21}{32} \end{split} $$