In a certain sense, the answer depends on how lazy you want to be.
If you are willing to do some very mild bookkeeping, then it is easy to calculate on any particular day the exact probability of select a half-tablet based on your previous draws. This is a form of conditional probability calculation. If you don't like to do any bookkeeping at all, then a marginal probability calculation can be obtained. The former gives a very simple expression, while the latter is more complicated. (This is the price you pay for additional laziness.)
A simple observation
Note that at any point in time, if we draw a whole tablet, then (i) the total number of tablets stays the same, (ii) the number of whole tablets decreases by one and (iii) the number of half-tablets increases by one.
Otherwise, if we draw a half-tablet, then the total number of tablets and the number of half-tablets both decrease by one, and the number of whole tablets stays the same.
Hence, knowing only the number of whole tablets chosen over the total number of days, we can calculate exactly how many half-tablets and whole tablets are left in the jar. Specifically, let $X_t$ be the number of whole tablets drawn up to time $t$ (inclusive). Then, the number of remaining whole tablets is $W_t = n - X_t$ and the number of half-tablets is $H_t = X_t - (t-X_t) = 2 X_t - t$. (The last quantity can be seen to arise from the $X_t$ half-tablets created by drawing each of the $X_t$ whole tablets and replacing them with a half-tablet minus the $t-X_t$ half-tablets we must have consumed.)
Conditional probability
Thus, by keeping track only of what day we are on and a running total of how many whole tablets we've drawn, we can find the conditional probability of drawing a half-tablet on day $t+1$ as
$$\renewcommand{\Pr}{\mathbb P}
\Pr( \text{draw half-tablet on day $t+1$} \mid (t,X_t) ) = \frac{H_t}{H_t+W_t} = \frac{2 X_t - t}{n - t + X_t} .
$$
So, by using very simple bookkeeping, we can calculate the probability of drawing a half-tablet on each day for each individual realization of this process. In particular, this allows us to determine exactly the first day on which we have greater than 50% chance of drawing a half-tablet.
Unconditional (marginal) probability
If we are too lazy to keep track of how many whole tablets we've drawn (or simply don't have a pen), then we can completely ignore the past and rely on marginal probabilities of drawing a half-tablet. From the above observation, it's apparent that $H_t$ follows a time-imhomogenous Markov process satisfying
$$
H_{t+1} =
\begin{cases}
H_t - 1 & \mathrm{w.p.}\; H_t / (H_t + W_t) \\
H_t + 1 & \mathrm{w.p.}\; W_t / (H_t + W_t)
\end{cases} \>.
$$
We can, therefore, use a simple recursion to calculate $\mathbb P(H_t = h)$ for each valid $h$. Some $R$ code to implement the recursion is found below.
Results for $n=100$
When we start with $n = 100$ tablets, then the first time that we have over a 50% chance of getting a half-tablet is at $t = 90$, i.e., after 90 tablets have already been drawn from the bottle (or in other words, on the 91st day). The probability is
$$\mathbb P(\text{half-tablet on 91st draw}) \approx 0.502571 \>.$$
Below is a graph of the probability of getting a half-tablet as a function of $t$.
Here is some very simple (and very inefficient!) $R$ code to calculate the probability of being in each state as a function of $t$. At the end is the calculation of the probability of drawing a half-tablet at each point in time.
# Eat-your-vitamins stochastic process
n <- 100
P <- matrix(0,n+1,2*n+1)
P[1,1] <- 1
for( t in 0:(2*n-1) )
{
j <- t + 1
offset <- 1 * (t %% 2 != 0)
M <- min(t,2*n-t)
for( h in seq(offset,M,by=2) )
{
i <- h + 1
w <- n - (h+t)/2
if( h > 0 )
P[i-1,j+1] <- P[i-1,j+1] + P[i,j] * h / (h+w)
if( w > 0 )
P[i+1,j+1] <- P[i+1,j+1] + P[i,j] * w / (h+w)
}
}
# Probability of getting a half-tablet as a function of time.
H <- matrix(0:n,nrow(P),ncol(P))
W <- n - H/2 - t(matrix(0:(ncol(P)-1),ncol(P),nrow(P)))/2
pr.half <- colSums( P*H/(H+W) )
