We can compute this using the formula for conditional probability.
In this case what A is is the probability the coin toss is valid.
A is valid, and we know that the probability of A is equal to 0.99998.
What B is is the probability that it's heads.
So the probability of B is the probability of H, which is 0.49999.
Now we just have to plug these into the formula.
But what we need to use the formula is the probability of A intersect B.
And what A is the probability of A intersect B. What A is is valid.
Instead of heads and tails, what B is is heads.
If we intersect heads and tails with heads, we get tails.
We know the probability of tails is 0.49999.
That means using the formula we have the probability of tails, which is 0.49999,
which is the probability of A intersect B.
We're dividing that by the probability of A,
which is a valid event which is 0.99998.
We get 0.5.
I should note that this is not the case for real coin tosses.
There is no physical coin ever manufactured that has exactly
chances of landing on both sides.
In fact, with real coin tosses, at least with American currency,
there is a much higher percentage--much higher meaning close to 51%
rather than 50%--that the coin lands on the same side that it started on.
When we talk about mathematical coin tosses,
we're going to assume that there is no edge case
and that it's equally likely
that we have a uniform distribution and there are only 2 outcomes.
When we talk about mathematical coin tosses,
we're going to assume that we have a uniform distribution,
and there are only two outcomes.