If this position had not been presented as a problem, I would undoubtedly make the twelve point, 15/12, 13/12, saving both outside blots. As things stand, the only roll White has that fails to hit one or both of them is 21. Any four, five, six or seven will hit. The real danger for Black is that he could end up with two blots up, while White cruises home unopposed. The pip count is 149 to 108, with gammon chances that are low, but still significant.
In the context of a problem, that answer seems a bit too obvious. Is this some kind of a trap? Unstacking the six point, 6/5, 6/3, would certainly be nice, but can Black afford the risk? I think not. So, what about hitting White, 13/9*, offering a return hit with 11, 12, 13, 15, 36, 45 and 55 (12 shots)? This seems premature, leaving the six point badly stacked, and offering little chance to contain White.
All things considered, I'll stick with the small play. Let's make the twelve point, save the gammon and resign ourselves to limited, late-game chances for a hit.
My solution: 15/12, 13/12
For the Record
I am so often wrong that I like to post my record in these messages. It's kind of a truth-in-advertising thing. I have been answering these problems without the use of a bot, and before checking the excellent solutions of others, since Problem 28. Including the 39(a) tossup, my record at this writing is 55% correct.
18 Correct: 28a, 29, 30, 32, 35, 36, 38, 39a, 39b, 42b, 43, 44, 45, 47, 48a, 48b, 50, 52b. 15 Incorrect: 28b, 31, 33a, 33b, 34, 37, 40a, 40b, 41, 42a, 46, 48c, 49, 51, 52a.[/I][/B]