Okay, so this is more complicated than you first thought! In the table below, the top row shows the possible outcomes from 2 to 12 when you throw two dice. For each outcome, the second row shows the probability of that outcome occurring, and the bottom row shows the probability of the outcome being at least that much.

$\displaystyle \begin{array}{c|ccccccccccc}k&2&3&4&5&6&7&8&9&10&1 1&12\\ \\P(k)& \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36}& \frac4{36}& \frac3{36} & \frac2{36}& \frac1{36}\\ \\ P(\geqslant k)& \frac{36}{36} & \frac{35}{36} & \frac{33}{36} & \frac{30}{36} & \frac{26}{36} & \frac{21}{36} & \frac{15}{36} & \frac{10}{36} & \frac6{36} & \frac3{36} & \frac1{36} \end{array}$

Now, if the first player (the one with a current score of 5) throws the dice and scores k, then the second player (currently with 7) needs to score at least k–1 in order for their total to remain greater than that of the other player. If the first player rolls the dice and gets 2 or 3, there is no way that they can go ahead. But if they get 4, then the second player must get at least 3. The probability of that happening is 3/36 (= probability of the first player scoring 4) times 35/36 (= probability of the second player scoring at least 3).

Adding up the probability for each of these combinations, the total probability of the second player remaining ahead after both players have rolled the dice is

$\displaystyle \frac{\parbox{5in}{$1\times36 + 2\times 36 + 3\times35 + 4\times 33 + 5\times 30 + 6\times26$\\ {\color{white}.}\hfill ${}+ 5\times21 + 4\times 15 + 3\times10+2\times6+ 1\times3$}}{36\times36}\approx 0.66435...$

If you want it as a percentage then the answer is approx. 66.4%.