If you were to start out with a 2% chance of winning a prize at the casino but each time you lost you were granted an additional 2% chance of winning, what would be the odds that you lost enough to reach a 70% chance of winning.
So say you start gambling with a 2% chance to win and you lose, so now you have a 4% chance to win. now with that 4% chance of winning you lose again, so now you have a 6% chance. but you lose again so now you have an 8% chance to win. and you lose again so now you have a 10% chance, and so fourth.
What would be the odds that you lost so many times that you were able to reach a 70% chance of winning.
So first I hope you are talking about an increase in percentage points and second that the winning probability is set back to 2% once you win before hitting the 70% winning chance. I guess your question is how likely it to loose n times in a row to directly hit the 70% winning probability right?
In such a case you need to loose 34 times in a row to reach a 70% chance of winning since $$2 + 2\cdot n \stackrel{!}{=} 70 \Leftrightarrow n = 34$$
So the odds to actually reach this scenario is $$P_1(loose) \cdot P_2(loose) \cdots \cdot P_{34}(loose) = 0.98 \cdot 0.96 \cdot \cdots \cdot 0.32 = 0.0000000024%$$
Therefore the probability that this happens is 0.00000024%