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The die can land in ways, and so the number of elements in the equiprobable sample space is , each element consisting of a sequence of H's and T's followed by a score from to .
The distribution of the number of heads is Binomial, and the probability of getting heads (where ) is:
The probability that the score on the die, , is less than is given by:
So to calculate the probability that, for any given throw of the coins, the score on the die is less than the number of heads, you'll need to multiply the two respective probabilities together, and then add all the resulting products.
As a summation formula, this is .
According to my calculations (using a spreadsheet to do the arithmetic) this comes out as to 4 d.p.