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A geometric distribution is used to model the number of failures before a success in repeated, independent Bernoulli trials. In our case, a 'success' is defined as hitting the balloon and a 'failure' is missing it. Given the success probability \(p = 0.1\) (10%), our geometric probability function is \(P(X = k) = (1 - p)^{k - 1} \cdot p\) where \(X\) is the number of trials until the first success and \(k\) is a particular trial.

In a geometric distribution, the expected value or mean, which is the average number of trials expected for the first success is given by \(E[X] = 1/p\). Hence, the expected number of dart throws to hit the balloon is \(1/0.1 = 10\).

To calculate your expected winnings, we need to subtract the cost of playing the game (that would be lost even if you win) from the cash prize. For each throw, you are spending $$\$ 5$$. So for 10 throws, you would spend $$\$ 50$$. However, you are only willing to spend up to $$\$ 20$$, which is 4 throws. Therefore if you do not hit the balloon within 4 throws, you stop and your winnings would be $$\$ 0$$. Given a 10% chance of hitting the balloon, the odds are against hitting it within 4 throws. Hence the expected winnings would be very close to $$\$ 0$$. If the first success happens within those 4 throws, i.e., if \(k <= 4\), the expected winnings are $$\$ 60$$ after subtracting the cost of game from the cash prize, otherwise your winnings are $$\$ 0$$. Therefore, your expected winnings combine the probability you will win within 4 throws with the amount left over from the prize after deducting the cost of game and the probability of not winning within those 4 throws. The exact winnings can be calculated with the formula \(E[Winnings] = \sum_{k=1}^{4}(100 - 5k) \cdot ((1 - p)^{k - 1} \cdot p)\).