Last updated on November 27th, 2019 at 10:02 am
At each and every Joe Bass Team Trail event if you are the highest finishing participating Skeeter boat owner fishing from your Skeeter boat you could win up to $1000.00 with a minimum of $250.00 GUARANTEED!
1) So what’s the up to part?
- $1000 if you qualify in a current model year.
- $500 for a 2017 Skeeter boat or newer.
- $250 for a 2016 Skeeter boat or older.
2) So who gets the money?
Prize money goes to the Skeeter owner. (have proof of ownership)
3) So what’s the fine print?
- Begins 1/10/2013
- Must be the highest participating Skeeter boat.
- Must be present to win.
- Must fish from your own Skeeter.
- Must weigh in a minimum of 1 bass.
- Must complete an entry form with type of boat filled in. (Important: Joe Bass Team Trail Director will not be responsible for errors if entry form is not completed at each event.)
- Must have a paid Joe Bass membership.
4) What makes the Joe Bass Team Trail Skeeter Incentive so special?
- No limit on the number of times an owner can win.
- No minimum number of boats to qualify.
5) So where does the money come from?
- SportBoats USA provides the Skeeter Owner incentive money as part of their sponsorship. No money is taken from the entry fees.
Visit SportBoats USA for your new or used Skeeter boat.
Big Fish! Big Checks! Skeeter Boats!
Who doesn’t want more money? Beginning in February of 2017 Sport Boats USA will add an additional $500 to the highest finishing Skeeter owner if their boat was purchased from Sport Boats USA and is of the current Joe Bass Team Trail season year or newer. This is in addition to the current Highest Fishing Skeeter Owner Incentive. Is there any fine print you ask? Generosity only goes so far this can only be won 3 times per season per Sport Boats USA Skeeter owner, in addition tournaments must have 30 boats.
Example; 2019 Joe Bass Season. $500 Bonus paid on 2019 & 2020 model Skeeters. Once the new year model comes out the years of additional $500 moves to current Joe Bass Team Trail plus 1 year model.