Middling math question (via PM)
Received via PM:
If you were betting 1 unit at decimal odds of D on both sides of a pure middle then:
So in other words, you'd be risking D-2 units to win 2D-2 units, which corresponds to decimal odds of 1 + (2D-2)/(2-D) = D/(2-D).
Since breakeven probability for given decimal odds is the reciprocal of those odds, the breakeven middle probability would be 2/D - 1.
So, for a middle offered at odds of -110 (decimal = 210 / 110 ≈ 1.9091) on either side, the breakeven middle probability would be 2/1.9091 - 1 = 1/21.
For a middle offered at odds of -120 (decimal = 220 / 120 ≈ 1.8333) on either side, the breakeven middle probability would be 2/1.8333 - 1 = 1/11.
For a middle offered at at odds of -130 (decimal = 230 / 130 ≈ 1.7692) on either side, the breakeven middle probability would be 2/1.7692 - 1 = 3/23.
For a middle offered at at odds of -140 (decimal = 240 / 140 ≈ 1.7143) on either side, the breakeven middle probability would be 2/1.7143 - 1 = 1/6.
In general, for US-style fave odds of L (L ≤ -100 so decimal = (L-100)/L) on either side, the breakeven middle probability would be 2L/(L-100) - 1 = (L+100)/(L-100).
This is kind of a math question, but not really.
I recently read this regarding middling...
"If the betting odds are -110 for each side of the middle, you only have to hit 1/21 of them to break even. So 5% middle hits is a profitable expected value, but that means 19 out of 20 tries you lose money (-4.545%) to get the one that hits (+90.91%)."
Does this mean that if the vig is increased to -120 on each side, you would then have to win atleast 1 of every 11 tries? and if it was -130, would you then have to win 1 of every 6? and so on?
I recently read this regarding middling...
"If the betting odds are -110 for each side of the middle, you only have to hit 1/21 of them to break even. So 5% middle hits is a profitable expected value, but that means 19 out of 20 tries you lose money (-4.545%) to get the one that hits (+90.91%)."
Does this mean that if the vig is increased to -120 on each side, you would then have to win atleast 1 of every 11 tries? and if it was -130, would you then have to win 1 of every 6? and so on?
- If you hit your middle you'd win D-1 units on each bet for a total 2D-2 units.
- If you failed to hit your middle you'd win win D-1 units on one side and lose 1 unit on the other side for a net result of D-2 units (which will be negative for odds shorter than even).
So in other words, you'd be risking D-2 units to win 2D-2 units, which corresponds to decimal odds of 1 + (2D-2)/(2-D) = D/(2-D).
Since breakeven probability for given decimal odds is the reciprocal of those odds, the breakeven middle probability would be 2/D - 1.
So, for a middle offered at odds of -110 (decimal = 210 / 110 ≈ 1.9091) on either side, the breakeven middle probability would be 2/1.9091 - 1 = 1/21.
For a middle offered at odds of -120 (decimal = 220 / 120 ≈ 1.8333) on either side, the breakeven middle probability would be 2/1.8333 - 1 = 1/11.
For a middle offered at at odds of -130 (decimal = 230 / 130 ≈ 1.7692) on either side, the breakeven middle probability would be 2/1.7692 - 1 = 3/23.
For a middle offered at at odds of -140 (decimal = 240 / 140 ≈ 1.7143) on either side, the breakeven middle probability would be 2/1.7143 - 1 = 1/6.
In general, for US-style fave odds of L (L ≤ -100 so decimal = (L-100)/L) on either side, the breakeven middle probability would be 2L/(L-100) - 1 = (L+100)/(L-100).