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If you have two events each with continuous probabilities of happening in any given second, how can you find the probability that one event happens N times before the other event reaches that count? This given an even probability distribution.

For example, if miner A finds 400 chunks of gold per second and miner B finds 600 chunks of gold per second, what is the likelihood miner A obtains 5 chunks of gold before miner B does?

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I have up-voted Brian Tung's answer, but I think a more leisurely account of the matter is worth something.

The question says:

If you have two events each with continuous probabilities of happening in any given second,

and then

For example, if miner A finds 400 chunks of gold per second

This is somewhat imprecisely stated in some ways. But its phrasing makes sense if we assume that the number of chunks of gold found in any interval of time is always probabilistically independent of the number found in an in another interval of time that does not overlap with that interval, and that the probability of finding more than one chunk at a given instant, given that at least one was found then is $0.$ Probably one could argue that any other interpretation is too complicated for the way the question was phrased.

In that case, the number of chunks found in any time interval has a Poisson distribution, so that \begin{align} \Pr( & \text{number of chunks} = x) = \frac{\lambda^x e^{-\lambda}}{x!}, \\[4pt] & \text{where $\lambda$ is the average number of chunks found in that amount of time.} \end{align}

We are told that

miner A finds 400 chunks of gold per second and miner B finds 600 chunks of gold per second

Here we will also assume that the numbers of chunks found by A and B are probabilistically independent of each other.

A standard theorem about Poisson processes is this:

The conditional probability distribution of the number of chunks found by miner A during a specified time, given that the total found by both miners during that time is $n,$ is the binomial distribution of the number of successes in $n$ independent trials, with probability $\dfrac{400}{400+600}$ of success on each trial.

Thus we have $$ \Pr(\text{number found by A} = x \mid \text{total number found by both} = n) \\[6pt] = \binom n x \left( \frac{400}{400+600} \right)^x \left( \frac{600}{400+600} \right)^{n-x}. \tag 1 $$ I think a number of times people have posted here asking how to prove that.

Next, the question says:

what is the likelihood miner A obtains 5 chunks of gold before miner B does?

Every baseball fan knows the following fact of arithmetic, being reminded of it each year in October: the event that team A wins four games against team B before team B wins four games against team A is the same as the event that in the first seven games between A and B, team A wins more games than team B.

The same thing applies here: The event that miner A finds five chunks before miner B finds five chunks is the same as the event that among the first nine chunks found, five or more were found by miner A.

Thus, apply formula $(1)$ above with $x=5,6,7,8,9$ and add them up. You get about $0.2665677.$

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    $\begingroup$ Upvoted in return (see my Math Meta question!). Excellent exposition. But I see you have rounded off only to eliminate one non-zero digit! :-) $\endgroup$ – Brian Tung Apr 8 '18 at 17:28
  • $\begingroup$ Thanks for the detailed exposition! This really helps understand what's going on and how to use the formula Brian came up with. $\endgroup$ – B T Apr 8 '18 at 21:08
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Given that each miner finds chunks of gold according to Poisson processes with rates in the ratio of $2:3$ (it doesn't matter what the actual rates are, only their ratio), then this problem is equivalent to a best-of-nine contest, with each round being won by $A$ with probability $2/5$, or by $B$ with probability $3/5$. The probability that $A$ "wins" the contest is then

$$ P_5(A) = \sum_{k=5}^9 \binom{9}{k} \left(\frac25\right)^k \left(\frac35\right)^{9-k} = \frac{520640}{1953125} = \frac{104128}{390625} = 0.26656768 $$


In general, with rates $\lambda_A$ and $\lambda_B$, the probability that $A$ is first to $m$ chunks of gold is

$$ P_m(A) = \frac{1}{(\lambda_A+\lambda_B)^{2m-1}}\sum_{k=m}^{2m-1} \binom{2m-1}{k} \lambda_A^k \lambda_B^{2m-1-k} $$

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  • $\begingroup$ Upvoted for a nice general formula. Thanks! $\endgroup$ – B T Apr 8 '18 at 21:07

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