# Probability of an event whose chance of occurring doubles once after the first event in a sequence

To explain my problem I'll use the following example. There is a bag with 9 blue balls and one red ball. On the first event, a ball is taken out of the bag (90% chance it is blue and 10% chance it is red). On the second event, the bag is replaced with another bag that contains 4 blue balls and 1 red ball. A ball is taken out from the second bag and placed back inside. All future events will continue to use the second bag which has an 80% chance of blue ball and 20% chance of red ball.

I want to find the event at which the chance of getting at least one red ball is greater than 50%.

If only the second bag (with 5 balls) was used, I could use the equation:

$1 - (\frac45)^n = 0.5$

And then solve for "n" and round up where "n" is the number of events, $\frac45$ is the chance of getting a blue ball, and 0.5 represents 50% chance of getting a red ball.

I can't figure out how to change my equation in order to account for the additional chance from the first event (which used the bag with 10 balls). Essentially the chance of red ball doubles after the first event.

• What I mean is the chance of getting at least 1 red ball after a long sequence of events. So for every event new, there is a higher possibility of getting at least one red ball than the previous event – Jorge Luque Jun 16 '16 at 6:05
• So your chance of no reds in $n$ trials is $\frac{9}{10}\left(\frac{4}{5}\right)^{n-1}$. This is less than 0.5 for $n\ge4$. – almagest Jun 16 '16 at 6:25
• @almagest: The question is about the chance of at least $1$ red ball in $n$ trials. – barak manos Jun 16 '16 at 6:50
• @barakmanos Just so. – almagest Jun 16 '16 at 6:56

Calculate $1$ minus the probability of the complementary event:

$1-\left(1-\frac{1}{10}\right)^{1}\cdot\left(1-\frac{1}{5}\right)^{n-1}>\frac{1}{2}\iff$

$1-\left(\frac{9}{10}\right)^{1}\cdot\left(\frac{4}{5}\right)^{n-1}>\frac{1}{2}\iff$

$\left(\frac{9}{10}\right)^{1}\cdot\left(\frac{4}{5}\right)^{n-1}<\frac{1}{2}\iff$

$\left(\frac{4}{5}\right)^{n-1}<\frac{5}{9}\iff$

$\left(\frac{4}{5}\right)^{n}<\frac{4}{9}\iff$

$n>\log_{\frac{4}{5}}\frac{4}{9}\implies$

$n>3$

So $4$ trials or more will get you a probability higher than $50\%$.

• Why are you calculating $1-\frac{9}{10}-\left(\frac{4}{5}\right)^n$? – almagest Jun 16 '16 at 6:54
• @almagest: Ooops should be $\times$, thanks... – barak manos Jun 16 '16 at 6:55