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i think that orton might face taker again some wrestlemania i mean orton didnt beat taker or kane had like 2 times and lost so what if orton chllange taker at wrestleamania thats if batista lose but orton had taker tho at mania 21 so now that would be a wrestlemania match orton vs taker for world title and nobody cant beat taker orton might challange him again u know he will

2007-02-13 12:52:15 · 5 answers · asked by ortonchampion 1 in Sports Wrestling

5 answers

orton suxs ..that world title he won prolly his last

2007-02-13 13:20:02 · answer #1 · answered by ToddMan 5 · 0 0

I'm going to go on record in saying that no one will break Taker's streak at Mania, especially not Orton. Orton's put himself in a lot of hot water in the past, and any big push for him would have to be in the far off future, and the feeling in the air this year is that it's Taker's last mania, and he's going to end the legacy with a title victory. I assume he'll hold onto it for a few months, be dominant for at least 3 months, defending only at PPV's and fighting only at PPV's, and then hopefully he'll lose it mid year at a big event like Summer Slam to Ken Kennedy. They've primed Kennedy lately, he just needs that final big step. Of all of the youngsters to climb the ranks over the years, Double K's earned his spot at the top.

2007-02-13 14:19:21 · answer #2 · answered by Candle 7 · 0 0

No he wont. There on different shows, Taker is likely to retire soon, and the wwe wont have ANOTHER orton vs taker match. They already had a big feud. The wwe wont bring back that same feud

2007-02-13 12:56:28 · answer #3 · answered by Matt: aka; Edge is world champ!! 6 · 0 0

it may well be plausible. Randy has that flow he did to Shane and Vince and he would do it to the Undertaker in the adventure that they do attempt hostile to at least one yet another. He may even take him out like he did to Vince.

2016-12-04 03:46:15 · answer #4 · answered by duperne 4 · 0 0

its a frickin show its all planned out

2007-02-13 15:13:59 · answer #5 · answered by brandon m 2 · 0 0

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