probability between two 6-sided dice and a 12-sided die

Question

So if we have a game with player A who roll 2 dice with 6-sides and player B with one 12-sided die what is the probability to win for the player A and the probability to have a draw game.

I have make a thing but I don't know if it is true I have found : probability for A to win : 6/144.

Answer 1

For the probability that A wins, add up the following probabilities:

• $P(A= 2,B< 2)=\frac{1}{36}\cdot\frac{ 1}{12}=\frac{ 1}{432}$
• $P(A= 3,B< 3)=\frac{2}{36}\cdot\frac{ 2}{12}=\frac{ 4}{432}$
• $P(A= 4,B< 4)=\frac{3}{36}\cdot\frac{ 3}{12}=\frac{ 9}{432}$
• $P(A= 5,B< 5)=\frac{4}{36}\cdot\frac{ 4}{12}=\frac{16}{432}$
• $P(A= 6,B< 6)=\frac{5}{36}\cdot\frac{ 5}{12}=\frac{25}{432}$
• $P(A= 7,B< 7)=\frac{6}{36}\cdot\frac{ 6}{12}=\frac{36}{432}$
• $P(A= 8,B< 8)=\frac{5}{36}\cdot\frac{ 7}{12}=\frac{35}{432}$
• $P(A= 9,B< 9)=\frac{4}{36}\cdot\frac{ 8}{12}=\frac{32}{432}$
• $P(A=10,B<10)=\frac{3}{36}\cdot\frac{ 9}{12}=\frac{27}{432}$
• $P(A=11,B<11)=\frac{2}{36}\cdot\frac{10}{12}=\frac{20}{432}$
• $P(A=12,B<12)=\frac{1}{36}\cdot\frac{11}{12}=\frac{11}{432}$

So the answer is $\frac{1+4+9+16+25+36+35+32+27+20+11}{432}=\frac{1}{2}$.

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For the probability of a draw, add up the following probabilities:

• $P(A= 2,B= 2)=\frac{1}{36}\cdot\frac{1}{12}=\frac{1}{432}$
• $P(A= 3,B= 3)=\frac{2}{36}\cdot\frac{1}{12}=\frac{2}{432}$
• $P(A= 4,B= 4)=\frac{3}{36}\cdot\frac{1}{12}=\frac{3}{432}$
• $P(A= 5,B= 5)=\frac{4}{36}\cdot\frac{1}{12}=\frac{4}{432}$
• $P(A= 6,B= 6)=\frac{5}{36}\cdot\frac{1}{12}=\frac{5}{432}$
• $P(A= 7,B= 7)=\frac{6}{36}\cdot\frac{1}{12}=\frac{6}{432}$
• $P(A= 8,B= 8)=\frac{5}{36}\cdot\frac{1}{12}=\frac{5}{432}$
• $P(A= 9,B= 9)=\frac{4}{36}\cdot\frac{1}{12}=\frac{4}{432}$
• $P(A=10,B=10)=\frac{3}{36}\cdot\frac{1}{12}=\frac{3}{432}$
• $P(A=11,B=11)=\frac{2}{36}\cdot\frac{1}{12}=\frac{2}{432}$
• $P(A=12,B=12)=\frac{1}{36}\cdot\frac{1}{12}=\frac{1}{432}$

So the answer is $\frac{1+2+3+4+5+6+5+4+3+2+1}{432}=\frac{1}{12}$.