Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

RektTheNoob
Feb 2, 2018

#1**+3 **

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

P(2 twos and 2 ones and the other two different ) + P(2 sixes and 2 ones and 2 of some other number)

\( =\frac{6!}{2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{3}{6}) + \frac{6!}{2!2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{1}{6})\\ =180*\frac{12}{6^6} + 90*\frac{4}{6^6}\\ =\frac{2160}{6^6} + \frac{360}{6^6}\\ =\frac{2520}{46656}\\ =\frac{35}{648} \\\approx 0.054\)

**I do not think this is correct, I think 2160 should be half the size at 1080.**

**BUT I do not know what I did wrong. Probablility is like that ... it's sneaky!!**

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Here is another way to look at it

You have six squares an each square represents a number rolled.

There will be 6^6 = 46656 possible outcomes but I have to work out how many of those are favourable.

First there are 6C2 = 15 places where the 1s can go.

Now there are four spots left so

There are 4C2= 6 places where the 2s can go.

So far that is 15*6=90

Now there are 2 spots left.

If the the last two numbers are different then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 3 fro the last place so that is 90*4*3 = 1080

If the the last two numbers are the same then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 1 for the last place so that is 90*4*1 =360

So that is 1080+360 = 1440

**So the probablility is **\(\frac{1440}{6^6}=\frac{1440}{46656}=\frac{5}{162}\)

This is why I think that the second answer is correct. I counted the possibilities.

Melody
Feb 2, 2018

#3**+3 **

**Solution: **

\(\text {With six dice }\\ \small \text {There are } \dbinom{6}{2} \text { ways to chooses two dice for the “ones” }\\ \small \text{There are then} \dbinom{4}{2} \text { ways to choose two dice for the “twos” }\\ \small \text{There are } 6^2 = 36 \text{ ways to roll two dice. There is one way to roll two dice for two 1s (snake eyes) (1/36). }\\ \small \text{There is one way to roll two dice for two 2s (1/36).} \\ \small \text{Of the remaining two dice, there are 4 of 6 ways to roll each die without a 1 or 2 (2/3). }\\ \dbinom{6}{2} * (1/36) * \dbinom{4}{2} * (1/36) * (2/3) * (2/3) = 5/162 \\ \)

We really do get it right if we monkey around with it enough.

GA

GingerAle
Feb 2, 2018