Calculating the chance that all $3$ members of a team win a lottery with $25\%$ chance of winning in $3$ attempts each

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A recent grad on a stem/OPT visa can stay in the US for $3$ years. Every year they can apply for the H1B lottery one time, which, simplifying for this question, has a $25\%$ chance of winning. There are $3$ people with this visa on the team.

So $3$ people get to each apply $3$ times with a $25\%$ chance of winning each time.

What are the chances after $3$ years all $3$ will have won?

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Assuming there can be multiple winners,

$$Pr(win) = 1 - Pr(not \;win)$$

Such that a 'not win' is defined by $0$ successful attempt's. Let random variable $X$ be the chance that one of the people win the lottery. Such that $X$ is distributed binomially with $p = 0.25$ and $n = 3$. Using the binomial probability formula:

$$Pr(X = x) = \binom{n}{x} \times p^x \times (1 - p)^{n-x}$$

Where $n$ is the number of attempts, $p$ is the probability of success, and $x$ is the amount of successful outcomes.

$$Pr(X = 0) = \binom{3}{0} \times 0.25^0 \times 0.75^3$$

$$Pr(X = 0) = \frac{27}{64}$$

As earlier established.

$$Pr(win) = 1 - \frac{27}{64} = \frac{37}{64}$$

So, we'vew now established that the chance of $1$ person winning in 3 years (1 attempt per year) is $37/64$. We can again treat this is binomial, where we are looking for the probability that all of them win. Such that

$$Pr(all \; win) = (\frac{37}{64})^3 = \frac{50653}{262144} $$

Which is approximately equal to $0.1932$