If your friend happens to choose one of the colored dice, A, B, or C, then yes, you can choose a certain die to give yourself the upper hand. If your friend happens to choose the normal die, N, then you will be playing with equal chances of winning, no matter which die you choose. The reason why competing with just the colored dice does not provide equal chances of rolling the higher number is because they are nontransitive dice —a special set of dice for which the property of "rolling the higher number more than half the time" is not transitive:
When rolling this set of dice repeatedly, you will see that A beats B most of the time, B beats C, and surprisingly, C beats A.
So, the fact that A beats B more than 50% of the time and B beats C more than 50% of the time DOES NOT ensure that A beats C more than 50% of the time!

A Look at the Probabilities


By observing all possible combinations found by rolling two of the dice, we can determine how often a certain die will roll a higher number than the competing die. In the case that we roll two of the nontransitive dice against one another, we will see that the probability of one of them winning is greater than 50%, meaning it should win more often. But, in the case that we roll a nontransitive die against a normal die, we will see that the probability of each of them winning is equal, meaning that they are equally matched and so should win the same amount of times.
A vs. B
Faces

2

2

2

5

5

5

3

A

A

A

B

B

B

3

A

A

A

B

B

B

3

A

A

A

B

B

B

3

A

A

A

B

B

B

3

A

A

A

B

B

B

6

A

A

A

A

A

A




Probability of A beating B = $\frac{21}{36}\approx 58.3\%$
Probability of B beating A = $\frac{15}{36}\approx 41.7\%$
Therefore, A is more likely to roll the higher number.



B vs. C
Faces

2

2

2

5

5

5

1

B

B

B

B

B

B

4

C

C

C

B

B

B

4

C

C

C

B

B

B

4

C

C

C

B

B

B

4

C

C

C

B

B

B

4

C

C

C

B

B

B




Probability of B beating C = $\frac{21}{36}\approx 58.3\%$
Probability of C beating B = $\frac{15}{36}\approx 41.7\%$
Therefore, B is more likely to roll the higher number.
${}$



A vs. C
Faces

1

4

4

4

4

4

3

A

C

C

C

C

C

3

A

C

C

C

C

C

3

A

C

C

C

C

C

3

A

C

C

C

C

C

3

A

C

C

C

C

C

6

A

A

A

A

A

A




Probability of A beating C = $\frac{11}{36}\approx 30.6\%$
Probability of C beating A = $\frac{25}{36}\approx 69.4\%$
Therefore, C is more likely to roll the higher number.



A vs. N
Faces

1

2

3

4

5

6

3

A

A

/

N

N

N

3

A

A

/

N

N

N

3

A

A

/

N

N

N

3

A

A

/

N

N

N

3

A

A

/

N

N

N

6

A

A

A

A

A

/




Probability of A beating N = $\frac{15}{36}\approx 41.7\%$
Probability of N beating A = $\frac{15}{36}\approx 41.7\%$
Probability of a tie = $\frac{6}{36}\approx 16.6\%$
Therefore, A and N are equally likely to roll the higher number.



B vs. N
Faces

1

2

3

4

5

6

2

B

/

N

N

N

N

2

B

/

N

N

N

N

2

B

/

N

N

N

N

5

B

B

B

B

/

N

5

B

B

B

B

/

N

5

B

B

B

B

/

N




Probability of B beating N = $\frac{15}{36}\approx 41.7\%$
Probability of N beating B = $\frac{15}{36}\approx 41.7\%$
Probability of a tie = $\frac{6}{36}\approx 16.6\%$
Therefore, B and N are equally likely to roll the higher number.



C vs. N
Faces

1

2

3

4

5

6

1

/

N

N

N

N

N

4

C

C

C

/

N

N

4

C

C

C

/

N

N

4

C

C

C

/

N

N

4

C

C

C

/

N

N

4

C

C

C

/

N

N




Probability of C beating N = $\frac{15}{36}\approx 41.7\%$
Probability of N beating C = $\frac{15}{36}\approx 41.7\%$
Probability of a tie = $\frac{6}{36}\approx 16.6\%$
Therefore, C and N are equally likely to roll the higher number.



