How many licks?

01Define what counts as reaching the center

Here, “reaching the center” means first exposing the Tootsie Roll filling by licking, with no biting. It does not mean reaching the geometric midpoint or removing the entire shell.

Represent erosion using a finite collection of fixed paths toward the filling. For path $j\in\{1,\ldots,m\}$, let $h_j>0$ be the initial hard-candy thickness. Let $d_{ij}\ge0$ be the potential advance along that same path during lick $i$. These advances may vary, be dependent, or be zero on paths a lick does not affect. The model does not allow the shell to regrow.

We use $\inf\varnothing=\infty$. Thus the count is infinite if no finite sequence of licks exposes the filling.

Proof. After $n$ licks, the remaining thickness along path $j$ is $\max\{h_j-\sum_{i=1}^{n}d_{ij},0\}$. At least one path reaches the filling exactly when its cumulative advance is at least $h_j$. Taking the first integer with that property gives the displayed formula. $\square$

This is an identity for a model of erosion along fixed paths. A real three-dimensional dissolution model would need additional geometry and physics. The plot above uses one fixed path.

02Identical licks

For one path of thickness $h>0$, suppose every lick has the same potential advance $\mu>0$. The final physical removal is clipped at the remaining shell thickness.

Proof. The least integer $n$ for which $n\mu\ge h$ is the least integer at least $h/\mu$, which is its ceiling. $\square$

Shell thickness alone cannot determine a count: changing the advance per lick changes the result. Neither parameter is supplied by the original question.

03Allow the licks to vary

Let $D_1,D_2,\ldots$ be independent, identically distributed potential advances along one fixed path. Write $\mu=\mathbb E[D_i]>0$ and $c=\operatorname{SD}(D_i)/\mu>0$. Choose a gamma distribution with shape $k=1/c^2$ and scale $\theta=\mu c^2$:

This choice has $\mathbb E[D_i]=k\theta=\mu$ and $\operatorname{Var}(D_i)=k\theta^2=\mu^2c^2$. Gamma advances are a modeling assumption, not an experimentally established law of licking.

Let $S_n=\sum_{i=1}^{n}D_i$ and $N=\inf\{n\ge1:S_n\ge h\}$. Clipping the final physical removal does not change when this first crossing happens.

Proposition. The cumulative probability of reaching the filling by lick $n$ is $F_N(0)=0$ and, for every integer $n\ge1$,

The regularized upper incomplete gamma function is

Proof. Nonnegative advances imply $\{N\le n\}=\{S_n\ge h\}$. To find the law of $S_n$, the gamma density gives, for $s\ge0$,

Independence makes the Laplace transform of the sum equal to the product:

This is the Laplace transform of $\operatorname{Gamma}(nk,\theta)$. Uniqueness of Laplace transforms for probability distributions on $[0,\infty)$ therefore gives $S_n\sim\operatorname{Gamma}(nk,\theta)$. Integrating its density from $h$ to infinity and substituting $t=u/\theta$ yields

Substitute $k=1/c^2$ and $\theta=\mu c^2$ to obtain the proposition. $\square$

The count is finite with probability one. By the strong law of large numbers, $S_n/n\to\mu>0$ almost surely. Consequently $S_n\to\infty$ almost surely, so a fixed finite thickness $h$ is eventually crossed.

04Read the plotted probabilities

The 3D view displays one seeded sequence of advances. After lick $n$, its marked path has depth $\min\{S_n,h\}$ and remaining thickness $\max\{h-S_n,0\}$. The first contact with the filling occurs at $N$. The cutaway is a separate inspection aid; it never changes the lick count. The surrounding depression is schematic and need not follow a physical dissolution law.

A completed simulated path can coexist with a probability below 100% on the chart: that path has finished, while the chart describes all possible paths under the same parameters. Changing the camera or cutaway does not change the sampled advances.

The solid staircase is $F_N(n)$: the chance that filling has been exposed after $n$ completed licks. The dashed staircase uses identical advances of size $\mu$. The horizontal axis focuses on the transition region; each step represents an integer count.

The probability of taking exactly $n$ licks and the $p$-quantile are

The displayed median is $q_{0.5}$. The shaded interval runs from $q_{0.1}$ to $q_{0.9}$ and contains at least 80% of the model’s probability. It need not contain exactly 80%, because the count is discrete. It is a prediction interval under assumed parameters, not a confidence interval estimated from observations.

For the illustrative starting values $h=5\,\mathrm{mm}$, $\mu=0.016\,\mathrm{mm}$ per lick, and $c=0.5$, the 10th percentile, median, and 90th percentile are respectively 302, 313, and 324 licks. The chart converts micrometers to millimeters before evaluating the formula.

The distribution depends on thickness and mean advance through their ratio $h/\mu$. Scaling both by the same positive factor leaves the distribution unchanged.

05Boundary cases and limits

Zero variation. At $c=0$, define each advance to be exactly $\mu$ and use $N=\lceil h/\mu\rceil$. Do not substitute zero into the gamma formula.

There is a subtle discontinuity if $h/\mu=m$ is an integer. As $c\to0^+$, the gamma model satisfies

Indeed, $S_m$ has mean $h$ and variance $m\mu^2c^2$, and its standardized gamma distribution tends to the standard normal as its shape $m/c^2$ grows. Thus $\Pr(S_m\ge h)\to1/2$. For fixed integers $n<m$ and $n>m$, concentration around $n\mu$ gives crossing probabilities tending to 0 and 1, respectively. At exactly $c=0$, however, equality reaches the filling at lick $m$. If $h/\mu$ is not an integer, $N\to\lceil h/\mu\rceil$ in probability as $c\to0^+$.

An independent special-case check. When $c=1$, advances are exponential and $N-1\sim\operatorname{Poisson}(h/\mu)$. Therefore $\mathbb E[N]=h/\mu+1$ and $\operatorname{Var}(N)=h/\mu$. This also shows why dividing thickness by mean advance need not give the exact expected count.