My boyfriend insists this can't be solved without integrals.

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>not enjoying bottoming
Also, putting arbitrary constraints on a problem instead of designing a better problem shouldn't be allowed.

>not enjoying bottoming
faux faggot spotted

if you assume that the situation is very poisson-y then you can model it as a queue with 10 servers, all currently busy and one customer currently in the queue.

look up M/M/1 queues and you'll probably find all the formulas you could possibly want already worked out by engineers

>>8493817
This seems like a thinly-veiled homework thread (in which case you should at least post in the /sqt/ on this board). I'll share my answer anyways.
Consider [math]n[/math] uniform random variables [math]X_i \in \{1,\hdots,k\}[/math].
We want
[eqn]\mathbb{E}\left[\min_{1 \leq i \leq n}\{X_i\}\right].[/eqn]
Well,
[eqn]\mathbb{P}\left(\min_{1 \leq i \leq n}\{X_i\} = 1\right) = 1 - \left(\frac{k-1}{k}\right)^n,[/eqn]
[eqn]\mathbb{P}\left(\min_{1 \leq i \leq n}\{X_i\} = 2\right) = \frac{(k-1)^n - (k-2)^n}{k^n},[/eqn]
and in general,
[eqn]\mathbb{P}\left(\min_{1 \leq i \leq n}\{X_i\} = j\right) = \frac{(k-j+1)^n - (k-j)^n}{k^n}.[/eqn]
We have [math]k[/math] possible outcomes, so we assign a weight to these probabilities, sum them up, and divide by [math]k[/math].
That is,
[eqn]\mathbb{E}\left[\min_{1 \leq i \leq n}\{X_i\}\right] = \frac{1}{k}\sum_{j=1}^{k}j\cdot\frac{(k-j+1)^n - (k-j)^n}{k^n}.[/eqn]
Plug in 10 and 75 for [math]n[/math] and [math]k[/math], respectively, and you should have your answer.

>>8493893
Sorry, don't divide by [math]k[/math]!

I dont remember how to do this shit but you can use the average time to the parameters of the poisson distribution of the tables becoming available

>>8493893
>>8493909
Ignore this post - it answers a different question than the one posed by the OP. Namely,
"Given 10 tables, all occupied and each with some integer waiting time between 1 and 75, what is the average waiting time?"
Incidentally, the answer simplifies to
[eqn]\mathbb{E}\left[\min_{1 \leq i \leq n}\{X_i\}\right] = \sum_{j=0}^{k-1}\frac{(k-j)^n}{k^n}[/eqn]

>>8493817
>for someone in your position

I'm a white male, so according to my university, I don't have to wait for anything.

Your boyfriend is right because the average is a (Lebesgue) integral, better pepper that angus OP.

The waiting time is the minimum of 10 IID exponential random variables, so its survival function (1 - CDF) is the exponential survival function raised to the power of 10, see e.g. http://stats.stackexchange.com/questions/18438/does-a-univariate-random-variables-mean-always-equal-the-integral-of-its-quanti

Maybe you could get around it by disguising the integration as a Riemann sum + limiting process, but that's too much effort for a /sci/ post, especially when it's not my ass that's at stake here.

About 6.4 minutes

Just solve it numerically. Code up a simulation and get the average from that. No integrals necessary.

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>>8494108
500k simulations.