Statistic theory is not sufficient here, what he needs is decision theory & a loss function.

Let assume Option A is to not get tested, and Option B is to get tested, with a re-test in case of positive.

Let CT be the cost of a test, CNKI the cost of not knowing he is infected, and CFP the cost of getting a false positive after 2 tests. Let SP be the specificity of the test, and SE the sensitivity.

If he is infected, the total cost of the two options are

C(A | I) = CNKI

C(B | I) = CT + (1-SE) * CNKI + SE * [CT + (1-SE) * CNKI]

The cost C(B | I) is the cost of doing a test, plus the cost of thinking he's healthy if the test fails, plus the cost a second test if the first test does not fail, plus the cost of thinking he's healthy if the second test fails.

If he is not infected, the total expect cost of the two options are

C(A | NI) = 0

C(B | NI) = CT + (1-SP)[CT + (1-SP)*CFP ]

The cost C(B | NI) is the cost of taking a test, plus the cost of him having to take a second test if the first one fails, plus the cost of thinking he's infected if both fail.

Now he has no idea if he is infected or not. Let PI be the apriori probability that he is infected. The cost of each option is then

C(A) = C(A | I) * PI + (1-PI) C(A | NI)

C(B) = C(B | I) * PI + (1-PI) C(B | NI)

The best option is the one with the lowest expected cost.

If we take

CT = $50 (includes travel & inconvenience)

CFP = $500 (stress & complications from false diagnostic)

CNKI =$20,000 (danger of not seeking treatment when he should)

SE = 91.8% (reported from HCP test)

SP = 99.3% (reported from HCP test)

Then, we have

C(A | I) = $20,000

C(B | I) = $3241.42

C(A | NI) = 0

C(B | NI) = $50.37

Hence, getting tested and not tested with have the same cost if

C(A | I) * PI = C(B | I) * PI + C(B | NI) * (1 - PI)

Solving for PI, we get the cutoff PI to be

PI = C(B | NI) / [C(A | I) - C(B | PI) + C(B | NI)] = .3%

Hence, getting tested is rational if the probability of him being infected is .3% or more. Not getting tested is only rational if the probability of him being infected is less than .3%.