It Took Them 606 Takes To Perfect This Ad. Can You Do Better?

I hope to illustrate the principle with a toy model and also to relay a cautionary tale about overcomplication in exploring a problem.

It seems that people agree about the answer: doing the harder things first saves time because you avoid doing tasks you know are most likely to be repeated. This answer is intuitive without any math... so of course I wanted to build a mathematical model to capture the effect.

I started by enumerating the possibilities for the case where $n$ is 2. Here we have two events with probabilities $p_1 < p_2$ . Obviously, the shortest time the movie can take is two steps and has only one possible ordering: film stunt one successfully then film stunt two successfully. For three steps, there is also one possible ordering: fail at filming stunt one, then succeed at filming stunt one, and finally succeed at filming stunt two. For four steps and above, the possibilities begin to splinter off into interesting patterns.

I found it a bit much to enumerate the possibilities straightaway so I decided on a diagramming technique. There are four basic events that can happen: succeed at filming stunt one ( $1^+$ ), fail at filming stunt one ( $1^-$ ), succeed at filming stunt two ( $2^+$ ), or fail at filming stunt two ( $2^-$ ). For the first couple lengths of filming, the possible events are enumerated as follows:

$\begin{aligned} &\mbox{time} &\mbox{sequences} \\ &1 &\emptyset \\ &2 &1^+2^+ \\ &3 &1^-1^+2^+ \\ &4 &1^-1^-1^+2^+ + 1^+2^-1^+2^+ \\ &\ldots &\ldots \end{aligned}$

With a bit of staring, it is clear that there is some further structure to the events. First, every successful take ends with the sequence $1^+2^+$ . Moreover, there are two basic intermediate possibilities that can repeat before ending in the success state $\boxed{1^+2^+}$ . They are $\boxed{1^-}$ and $\boxed{1^+2^-}$ .

These intermediate states can be combined in any possible order, all of which constitute a valid string of events for a filming session. Inspired by this, and also by the fact that I was scribbling it all on napkins (a sure sign of a good theory in the making), I thought the solution to the problem lay just a bit further ahead in summing the diagrammed series generated by this representation:

$\begin{aligned} \langle t\rangle &= \sum\limits_t t p(t) \\ &= \boxed{1^+2^+}\left(2 + 3\boxed{1^{-}}+4\left(\boxed{1^-}^2 + \boxed{1^+2^-}\right) + 5\left(\boxed{1^-}^3+2\boxed{1^-}\times\boxed{1^+2^-}\right)+\ldots\right) \end{aligned}$

Sadly for me, the way to sum such a series in closed form escapes me. This was massively depressing, as is any failure to exploit diagramming techniques.

Fortunately, there is a simpler way. I'll calculate for the case of three stunts because it makes the generalization obvious.

Consider the fact that, as other commenters have pointed out, the expected number of attempts to succeed at something with probability $p$ is given by $p^{-1}$ . As filming the stunts 1, 2, and 3 in succession has probability $\displaystyle p_1p_2p_3$ , we expect to try it $\displaystyle \frac{1}{p_1p_2p_3}$ times before succeeding.

Said in another way, we expect to fail $\displaystyle S_{fail} = \frac{1}{p_1p_2p_3}-1$ times and succeed on the $\displaystyle \frac{1}{p_1p_2p_3}\mbox{th}$ attempt.

Of the $S_{fail}$ attempts some of them will end on the first stunt with probability $\left(1-p_1\right)$ , taking one scene. Less will fail on the second stunt with probability $p_1\left(1-p_2\right)$ , taking two scenes. A few will fail on the third stunt with probability $p_1p_2\left(1-p_3\right)$ , taking up three scenes. Finally, the successful take will require three scenes.

We can now evaluate the average duration (in scenes) for an unsuccessful attempt $\langle t\rangle = \sum\limits_{t=1}^3 t p(t)$

$\begin{aligned} \sum\limits_{t=1}^3 t p(t) = \frac{\left(1-p_1\right) + 2 p_1\left(1-p_2\right) + 3 p_1p_2\left(1-p_3\right)}{\left(1-p_1\right) + p_1\left(1-p_2\right) + p_1p_2\left(1-p_3\right)} \\ &= \frac{1 + p_1 + p_1p_2 - 3p_1p_2p_3}{1-p_1p_2p_3} \end{aligned}$

As there are $S_{fail}$ unsuccessful attempts, we expect them to take a total of $S_{fail}\langle t\rangle$ scenes to shoot.

$\begin{aligned} N_{fail}\langle t \rangle &= \frac{1 + p_1 + p_1p_2 - 3p_1p_2p_3}{1-p_1p_2p_3}\left(\frac{1}{p_1p_2p_3}-1\right) \\ &= \frac{1 + p_1 + p_1p_2 - 3p_1p_2p_3}{p_1p_2p_3} \end{aligned}$

This gives us the total number of scenes required to film the failed attempts. For the total number of scenes to film a success, we add 3, the number of scenes shot during the success.

Somewhat incredibly, this makes the total number of scenes equal to

$S_{total}(p_1,p_2,p_3) = \frac{1+p_1+p_1p_2}{p_1p_2p_3}$

This immediately suggests a general result, the number of scenes required to film $n$ events in succession is given by:

$S_{total}^n = \displaystyle \frac{1 + \sum_{i=1}^{n-1}\prod\limits_{j=1}^i p_j }{\prod\limits_{j=1}^n p_j }$

We can check this result with a few test cases. I wrote a routine to simulate 50,000 attempts at filming the three events $1,2,3$ in succession with $p_1 = 0.11, p_2 = 0.31, p_3 = 0.51$ as well as in the reverse order ( $3,2,1)$ . Below is an empirical probability distribution of the number of required scenes.

Bootstrapping the output of the simulation yield estimates for the average duration as well as the error on that measure: $S_{short} = 66.0\pm0.28$ scenes, and $S_{long} = 95.65 \pm 0.42$ scenes.

This compares favorably with the predictions $S_{total}(0.11, 0.31, 0.51) = 65.78$ and $S_{total}(0.51, 0.31, 0.11) = 95.92$ .

We can also use our results to predict the result Aditya should find with his probabilities $\left(0.9,0.1,0.2,0.6\right)$ . In the optimal order we predict $209.6$ to his $208.2$ scenes and in the reverse we predict $471.9$ to his $488.0$ scenes. We would predict that if Aditya runs the simulation on more test cases, the results of our formula and his sample statistics should converge.

The given answer is that stunts with a high probability of success should be last, but I disagree. Given n stunts with probabilities p1, p2, ..., pn, the probability of success is p1 * p2 * ... * pn. Multiplication is commutative so it doesn't matter how they are ordered.

Now, I appreciate that if the ones with low probability are first, that they can stop the take sooner, but a failed take is a failed take, whether the failure is on the first stunt, the last stunt, or anywhere in between.

Am I wrong?