Double up question.

I need a little help with this math/reasoning problem.

Let's say you have $1,100 and your goal is to win $1,000 sports betting but you have little to no knowledge of sports.

All bets are at a regular 10% vig and the only goal is to win $1,000 regardless of how much time it takes.

What would be the best way to approach this? One double up bet and a number of smaller bets if so how many?

Why?

[DougOzzzz]

This depends on whether or not you are a winning sports bettor. If you are a winner, then the way to win $1000 with the least risk is to do it with as small a unit as possible.

If you are a loser, then bet it all on 1 game.

[jedinite]

[ QUOTE ]
Let's say you have $1,100 and your goal is to win $1,000 sports betting but you have little to no knowledge of sports.

All bets are at a regular 10% vig and the only goal is to win $1,000 regardless of how much time it takes.

[/ QUOTE ]

Emphasis mine...

If you're risk averse, i.e. the goal is literally to win $1000 no matter how long it takes, and not lose your original bankroll under any circumstances, and assuming you have no ability to handicap or find your own lines, I'd take the $1100 and bet two units on the opposite side of all the Wagerline consensus picks (both the ATS and O/U picks). Purely fading the public across the board.

This would put roughly half your bankroll on the line in week one, but you're spread across 28 total plays.

Wagerline consensus picks are roughly 45% - 55% lifetime on the NFL and 49-63-1 (43.75%) overall for this year. Probably the easiest way to consistently pick at a small winning edge with zero knowledge.

If you really wanted to minimize risk you could go with $10 units but I think you'd take ten years to build your $1000.

Personally I'd go with the roulette method (i.e if it were imperitive that you got the $1000 profit, otherwise you're screwed regardless if you lose your bankroll or not): pick one game where you think you have the largest edge and put $500 on it. You'll need to win two of those in a row before you lose two in a row. This is supposedly the mathematically best approach to doubling your money in a near-coinflip scenario IIRC.

I guess this doesn't account for the vig, as you'd be +$950, after going 3-1, but personally if I were in that scenario I think I'd go that route for the quick hit and then make a $50 play thereafter to close that last gap. Making the $500/week play on my selected game of the week would have resulted in a +$1900 profit so far this season (6-2) and would have hit the +$950 in week four (3-1 to open the season)

[DougOzzzz]

[ QUOTE ]

Personally I'd go with the roulette method (i.e if it were imperitive that you got the $1000 profit, otherwise you're screwed regardless if you lose your bankroll or not): pick one game where you think you have the largest edge and put $500 on it. You'll need to win two of those in a row before you lose two in a row. This is supposedly the mathematically best approach to doubling your money in a near-coinflip scenario IIRC.

[/ QUOTE ]

I have read this before but I forget why this is. I almost thought that this was only for black jack, where if you bet your whole bankroll you lose options and thus lose EV. It doesn't make sense to me otherwise that betting anything less than 100% of your bankroll would be the best strategy if you are a loser.

Bet it all on red!