You roll two fair dice. What is the probability to win the game based on restrictions over the sum of the results?

Question

You roll two fair dice. If the sum of the numbers shown is $7$ or $11$, you win; if it is $2$, $3$, or $12$, you lose. If it is any other number $j$, you continue to roll two dice until the sum is $j$ or $7$, whichever is sooner. If it is $7$, you lose; if it is $j$, you win.

(a) What is the probability $p$ that you win?

(b) What is the probability that you win on or before the second roll?

(c) What is the probability that you win on or before the third roll?

(d) What is the probability that you win if, on the first roll

(d.1) The first die shows 2?

(d.2) The first die shows 6?

(e) If you could fix the number to be shown by one die of the two on the first roll, what number would you choose?

MY ATTEMPT

Unfortunately, I do not know how tackle the exercise. I am aware that it may be needed to use conditional probability, but I am unable to describe properly the involved events. However, it is not a homework problem. I am really interested in the response of these questions. Thank you in advance.

Answer 1

(a) $$\frac{6}{36}+\frac{2}{36}+2*(\frac{3}{36}*\frac{3}{3+6}+\frac{4}{36}*\frac{4}{4+6}+\frac{5}{36}*\frac{5}{5+6})=244/495$$ (b) $$\frac{6}{36}+\frac{2}{36}+2*(\frac{3}{36}*\frac{3}{36}+\frac{4}{36}*\frac{4}{36}+\frac{5}{36}*\frac{5}{36})=\frac{97}{324}$$ (c) $$\frac{6}{36}+\frac{2}{36}+2*(\frac{3}{36}*\frac{36-6-3}{36}*\frac{3}{36}+\frac{4}{36}*\frac{36-6-4}{36}*\frac{4}{36}+\frac{5}{36}*\frac{36-6-5}{36}*\frac{5}{36})=\frac{539}{1944}$$ (d) (d.1) $$p_2=\frac{1}{6}+\frac{1}{6}*(\frac{3}{3+6}+\frac{4}{4+6}+2*\frac{5}{5+6})=\frac{218}{495}$$ (d.2) $$p_6=\frac{2}{6}+\frac{1}{6}*(\frac{3}{3+6}+\frac{4}{4+6}+\frac{5}{5+6})=\frac{263}{495}$$ (e) Intuitively $5$ is the optimal choice. Just calculate other 4 probability like what part(d) does to verify.

1. $p_1<p_2$ obviously;
2. $p_3=p_4>p_2$ obviously;
3. $$p_4=\frac{1}{6}+\frac{1}{6}*(\frac{3}{3+6}+2*\frac{4}{4+6}+2*\frac{5}{5+6})=\frac{251}{495}$$
4. $$p_5=\frac{2}{6}+\frac{1}{6}*(\frac{3}{3+6}+\frac{4}{4+6}+2*\frac{5}{5+6})=\frac{601}{990}$$
5. $p_5>p_6$ obviously.

So I would choose $5$.