Why Are You Laying 11-To-10, And Where Is The Break-Even Point?

When wagering on a football game or basketball game, the bettor MUST lay out $11 to win $10. This 10% difference (which in this case is $1) is the vigorish, or vig, or juice, which, in effect is the fee the book receives in exchange for handling the wager. Think of it in terms of the book being a broker. If you wanted to bet $100 on a game, for example, you would actually have to put up $110 to win the $100. If you WIN, you collect your $110 original wager, plus your $100 win. If you LOSE, you’ll lose the $110 you put up in the beginning. So as you can see, the vig is only retained by the book when you lose. It’s the difference between what you have to pay when you lose and what you collect when you win. To the uninitiated, it may not seem like much, but it keeps thousands of sports books and bookmakers in business for decades and has provided an almost insurmountable obstacle to sports betting profitability ever since its institution.

So what percentage of winners constitutes your break-even point?

This is worth discussing, because it happens to be the subject of much confusion. And it hasn't been correctly addressed in many places. What is the break-even point for the football or basketball bettor, with the 11/10 taken into consideration, and assuming the same amount bet per game? Invariably, people say 55%, since 55 is 10% more than 50. Others say 60%, even 62%, for whatever reason.

Assuming the same amount is indeed bet each time out (not real world, but the only way you can really figure an equation like this)……….

The correct answer is 52.38%. There is a very easy formula you can use for this. It goes like this:

BREAK-EVEN % = Price / (1 + Price)

In this case, the player is LAYING 11/10, or $1.10 to $1. Therefore, the PRICE is 1.10. So,

1.10 / (1 + 1.10) = 1.10 / 2.10 = .5238095

That’s 52.38%, rounded off. That’s the break-even point. Anything above that will bring you a profit. Don’t believe me? Let’s do the math, the same way someone like me might do it otherwise:

Let’s say you were going to make 100 bets at $100 apiece. When you figure in the 11/10, you’d be putting $11,000 into action IF you were to lose every play. But let’s get more realistic. If you were playing $100 per and went 50-50 on those plays, it would break down this way:

Win: 50 x $100 = $5000
Lose: 50 x $110 = $5500
In this situation, you would LOSE $500, or FIVE units

How about if we hit 55% of our DECISIONS (Remember, you'll push - or tie - some games where no money changes hands, so we won't count these)?

Win: 55 x $100 = $5500
Lose: 45 x $110 = $4950

Actually, 55% is quite profitable. You would be AHEAD $550 here. That represents a $5.50 average profit on every play.

At 53%, you have $5300 on the WINNING side, and $5170 on the LOSING side, for $130 profit. At 52%, it’s $5200 won and $5280 lost, for an $80 net loss. So it’s somewhere inbetween, right? Likely shaded more toward 52% than 53%.

Well, with a little interpolation, you can get there:

At 52.38%,

Win: 52.38 x $100 = $5238.00
Lose: 47.62 x $110 = $5238.20

Voila! We’ve essentially found our “break-even” point.

Now go get 'em.

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