Probability question: What's the chance of exactly 1 red die matching the blue die?

Question

I'm currently working on a probability problem and could use some assistance. The problem involves three six-sided dice: one blue die and two red dice. I'm trying to determine the probability of exactly one of the red dice showing the same number of dots as the blue die.

Here's my attempt at solving it:

To find the probability, I started by considering the possible outcomes and favorable outcomes:

Possible outcomes: Each die has six possible outcomes since they are six-sided dice.

Favorable outcomes: To have exactly one of the red dice match the blue die, I need to choose one of the two red dice and match its dots with the blue die. Each red die has six possible outcomes, considering the six sides. So, there are a total of $2 \cdot 6 = 12$ favorable outcomes.

Using the formula for probability (favorable outcomes divided by possible outcomes), I attempted to calculate the probability as follows:

Probability = favorable outcomes / possible outcomes = 12 / (2 * 6) = 12 / 12 = 1

Therefore, my initial calculation suggests that the probability is 1 or 100%. This would mean it's certain that exactly one of the red dice will match the number of dots on the blue die.

Could someone please verify my approach and calculations? If I made any mistakes or if there's a more accurate method to solve this problem, I would greatly appreciate your guidance.

Thank you in advance for your help!

Answer 1

Let blue die be $B$, and red dice are $R_1, R_2$, for each dice, it can take $1\sim6$, six outcomes, and they are independent, so the total possible outcomes are $6\cdot6\cdot6=6^3$

Next, let's count the favored cases. If you want the blue die equals one of the red die, say $B=R_1$, which implies the remaining red die $R_2\neq R_1$, so you have six outcomes for $B=R_1=1, 2,\dots,6$. For each "tie", $R_2$ can take remaining five outcomes, hence $6\cdot 5$. Finally, since $R_1$ and $R_2$ are symmetric, we multiply a factor $2$ in front. So we get:

$$P=\frac{6\cdot 5\cdot 2}{6^3}=\frac{5}{18}$$

Answer 2

Let the blue die toss anything. One of the red dice must match and the other not match

Total outcomes of red dice = $6*6$,

Favorable outcomes include (Red die $1$ matches and red die $2$ doesn't)+(Red die $2$ matches and red die $1$ doesn't)

So Pr $= \frac16\frac56 +\frac56\frac16 =\frac5{18}$

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I don't know why you think there are $2*6$, favorable outcomes neither how you get an identical total outcome !

Answer 3

You could match with the first throw, and miss with the second ($\frac16\frac56$), or miss with the first throw, and match with the second ($\frac56\frac16$),for a total of $\frac{10}{36}=\frac{5}{18}$.