Backing out a margin-free probability

MediumMarket Making & Betting Games~3m

A three-outcome market (home / draw / away) is quoted at decimal odds $2.50$ , $3.40$ , $3.00$ . The raw implied probabilities sum to more than 1 because the book carries a margin. Removing that margin proportionally (scale every implied probability by the same factor so they sum to 1), what is the fair, margin-free probability of the home outcome, to four decimals?

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Raw $1/\text{odds}$ sums above 1 - that sum is the factor to divide out.
Fair probability of home $= (1/2.50) / \text{sum}$ .

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0.3893 (± 0.001)

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Backing out a margin-free probability

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