well according to the manga the two previously had a match two times already (the 1st one officially in the kantou finals, atobe was not able to keep up with rai and lost. the 2nd one was not finished because it was interrupted by yukimura). so if these 2 were to meet for the 3rd time (of course in their current state) who would win? of course most will say it will be sanada because of his win against tezuka (well atobe defeated tezuka too but tezuka was still injured at that moment) but tecnhique-wise atobe has a chance with his tannhauser serve and the world of ice. here's my opinion on each other's chances:
he can easily pull a win off using his famed fuurinkainzanrai, especially the killer "rai", which even tezuka was not able to return even once. in their previous meeting he was able to dominate atobe with his "zan" (mountain) until the world of ice which he was unable to return. IMO the world of ice would be his greatest nemesis in the match because he doesn't have the tezuka zone to counter it (he admitted it, being the only technique he was unable to copy in his muga form, so as atobe), so he needs to become more offensive in their match and hit the rai early on for an easy win. i don't know about "in" (shadow), if it can hide the blind spots world of ice can see but if it can, then we already know who the mangaka sides with.
his plays must only concentrate on his 2 greatest techniques, the tannhauser serve and the world of ice. he can easily win his service games with his tannhauser serves (with the nonbouncing ball it's an automatic ace). and come sanada's service play he must pull off the world of ice before sanada hits the "rai". IMO that match will be "rai or world of ice, whichever comes first".
The bottom line: Atobe has a lesser chance to win, but with enough determination and focus (continuously hit tannhauser serves to secure his service games and break sanada's service game even once with his world of ice) he still stands a chance. your opinions please.