Arbitrage - backing multiple selections help

[Dogoa]

I hope that somebody can help me with this little math problem.
Been trying to understand how to calculate average odds of multiple selections.

If i have four backing selections :

odds 8,00 = 100/8 = 12,5%
odds 6,6 = 100/6,6 = 15,15%
odds 19,5 = 100/19,5 = 5,12%
odds 100 = 100/100 = 1%

Average % = (12,5+15,15+5,12+1)/4 = 8,44%

Average odds = 100/8,44 = 11,84

So backing (dutching) these four selections is the same as one back selection at odds 11,84.

And if my lay selections is a odds 3,25 and that cover all outcomes of the event, then i can make a profit.

Can somebody pleace tell me if i made my calculations correct.

Many thanks Dogoa

[milfor]

No, the division by 4 is wrong.

You have dutched events with the implied probabilities 12,5 + 15,15 + 5,13 + 1 = 33,75%

The corresponding odds would be 100 / 33,75 = 2,96. So laying at 3,25 is not a good idea.

[milfor]

Sorry for the typing error: 33,78%

[Dogoa]

Many thanks for the help.