# 3 Dices and the probabilty of me guessing 2-of them and 3-of them back to back.

A weird thing just happened and I tried to solve the problem but I couldn't. My friend had 3 dices, I guessed 2 of them as 2, 4 and he rolled '2,4,5'. So I guessed it correct. Than I guessed 3 of them as '1, 3, 5' and he rolled that. We discussed what are the chances of me guessing the first one correct, and right after that guessing the 2nd one correct back to back, and we are stuck. Any help would be appreciated.

For your first guess you would have been right if your friend had rolled a $1$ with the $2$ and $4$. Because you have three different dice, there are six ways this could have happened: $$1,2,4\quad\hbox{or}\quad 1,4,2\quad\hbox{or}\quad 2,1,4\quad\hbox{or}\quad 4,1,2\quad\hbox{or}\quad 2,4,1\quad\hbox{or}\quad 4,2,1\,.$$ The same thing happens if he had rolled a $3,5$ or $6$. If the rolls were $2,2,4$ it's a bit different and there are only three possibilities: $$2,2,4\quad\hbox{or}\quad 2,4,2\quad\hbox{or}\quad 4,2,2\,.$$ And the same for $2,4,4$. So altogether you would have been right in $30$ cases and the chance of your first guess being right is $30/6^3$. You can figure out the probability of the second guess being right in a similar way - it's actually easier than the first one - and then combine them to get your final answer.

• So, if I got it right, I'd be right on 6 cases, so the chance of me guessing 3 of them right is 6/216. How would I combine both answers? Chances of me getting that right after getting the first one right?
– cbt
Jan 28, 2014 at 23:53
• @cbt, $6/216$ is correct. Re: combining, you have the probabilities for two independent events, so how do you find the probability that both occur? Jan 29, 2014 at 0:13

2 out of 3 dices:

Probabilities: $$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} + \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = 0.06944444 ...$$

Binomial Distribution: $$X \sim B\left(n = 3, p = \frac{1}{6}\right), P(x = 2) = 0.069444444 ...$$

3 out of 3 dices

Probabilities: $$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = 0.0046296 ...$$

Binomial Distribution: $$X \sim B\left(n = 3, p = \frac{1}{6}\right), P(x = 3) = 0.0046296 ...$$

## 3 out of 3

you guessed x,y,z.

• probability of first roll match to x,y or z 1/2.x hit
• second roll y or z = 1/3.y hit
• third roll z = 1/6.

Probability of full match is (1/6)(1/3)(1/2) = 1/36

## 2 out of 3

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

1number (mod6) matching posibility with number(mod6)*number(mod6)SET. 11/36

so not matching is 25/36

third one fails 1/3 1/6
first one is 1/3 .there are two numbers to match. 3 roll is 1 because 2 of them already matched.

second one fails (1/3*5/6*1/6)

first one fails (25/36*1/3*1/6)

_____________________+

(1/3*1/6)+(1/3*5/6*1/6)+(25/36*1/3*1/6) = 0.0385802469...

Think about the number of combination that could occur and you still guess correctly.

So for your first guess the outcomes that result in a correct guess are below, where X is any number except 2 or 4:

Dice1   Dice2   Dice3
2       4       X
4       2       X
X       2       4
X       4       2
2       X       4
4       X       2
2       2       4
2       4       4
2       4       2
4       2       4
4       4       2
4       2       2

There are 6 combinations where you see a 2 and 4 and a different number, plus 6 combinations where we see only 4 and 2. The total number of combinations is therefore 6 * 4 = 24 (the six combinations above, times the 4 values that X can take) plus the 6 combinations of only 4 and 2. The total is 30 combinations where your guess is correct.

The total number of combinations for 3 dice is 1/6 * 1/6 * 1/6 = 1/216

So the probability of guessing 2 out of 3 dice correctly is 30/216

Hopefully you can see that for the case where you guess 3 out of 3 the probability is 6/216 (think about the above list but where X is a specific number and you cannot repeat any numbers).

So the total probability of your guesses both being correct is the probability of the first being correct multiplied by the probability of the second being correct

30/216 * 6/216 = 5/1296

• I got 3 answers and 3 of them are different, or am I wrong? :D
– cbt
Jan 29, 2014 at 0:10
• This answer is wrong because some outcomes are counted twice, e.g., $2,4,4$ is counted both in the first row and the fifth. Jan 29, 2014 at 0:11
• David's answer is correct, I have miscounted. Thanks David for pointing this out Jan 29, 2014 at 0:12
• Edited now to fix my mistake Jan 29, 2014 at 0:32