House of Cards

To solve this problem, we have to consider the sources of time.

One is the total amount of time spent stacking the house of cards., which is given by $N_\text{h.o.c.}\times t_\text{h.o.c.}$ , where $t_\text{h.o.c.}$ is equal to the time it takes to build the house, and $N_\text{h.o.c.}$ is equal to the expected number of times we need to repeat this act (due to the wind).

The second source is the average amount of time spent gluing the house which is given by $\bar{t}_\text{glue}$ . Each time we get knocked down by the wind, we have to start over, building the house of cards, and then performing some fraction of the glue job before the wind comes and knocks it down again. Thus, the bar over $t$ , indicating that we're taking the mean time. The partial glue job happens a total of $\left(N_\text{h.o.c.} -1\right)$ times (the last time takes the full $t_\text{glue}$ seconds).

The last source is amount of time we spend building and reinforcing in the last attempt, the time where everything works out. Thus, we have two numbers to calculate, $N_\text{h.o.c.}$ and $\bar{t}_\text{glue}$ .

The indoor component of the house of cards construction is risk-free, it takes the same amount of time, every time, and always finishes. Thus, the variability comes in from racing the wind. Despite the wind coming an average of once every twenty seconds, we can find a window of sixty wind-free seconds if we're willing to wait long enough. How likely is this?

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Consider an ensemble of infinitely many pre-build house of cards, all being glued outdoors in infinitely many separate windy environments. Now, let $p_+(t)$ be the portion of the structures which are not blown over by wind at time $t$ . As any given system as the chance $k_\text{wind}dt$ of being knocked over per unit time, we expect the number of surviving systems to disappear like

$\dot{p_+}(t) = -k_\text{wind} \times p_+(t)$

which leads to $p_+(t) = \exp -k_\text{wind} t$ .

Thus, the chance for any given house to be standing after time $t$ in the wind is given by $p_+(t)$ . Likewise, the expected number of times we will fail before successfully gluing the house in the wind is given by $1/p_+(t)$ .

Thus, we have $N_\text{h.o.c.} = p^{-1}_+(60\text{ sec}) \approx 20.09$ .

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As for $\bar{t}_\text{glue}$ , we can once more exploit the distribution $p_+(t)$ to calculate the conditional average time for cases when the house of cards is blown over by wind, i.e. when $t < 60$ . This is given by

$\bar{t}_\text{glue} = \displaystyle \frac{\int\limits_0^{t_\text{glue}}tp_+(t) dt}{\int\limits_0^{t_\text{glue}}p_+(t) dt}$

which is given by

$\begin{aligned} \bar{t}_\text{glue} &= k_\text{wind}^{-1} - \frac{t_\text{glue} }{e^{k_\text{wind}t_\text{glue}}-1}\\ &\approx 16.85\text{ s} \end{aligned}$

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The total time taken is then

$\begin{aligned} t_\textrm{total} &= N_\textrm{h.o.c.} t_\textrm{h.o.c.} + \left(N_\textrm{h.o.c.} - 1\right) \bar{t}_\textrm{glue} + t_\textrm{glue} \\ &= e^{k_\text{wind}t_\text{glue}}\left(k_\text{wind}^{-1} + t_\text{h.o.c.} \right) - k^{-1}_\text{wind} \\ &\approx 3.45 \text{ hr} \end{aligned}$

Please correct the numbers in the figure. The last arrow should read $60$ sec.