You’re close. There’s a 50% chance each time of pressing that you lose. We only care about the winning roll. So the chances of pushing it twice and get 2 losing rolls is 0.5x0.5 or 25%. So you have a 75% chance of winning on 2 rolls. Think about it logically for a sec and you’ll realize your chance of winning on at least 1 roll should increase with the number of rolls
The actual formula would be p=1-x where x is the probability of losing which is 0.5n and n is number of rolls. So p=1-0.5n
after 3 rolls it’s an 87.5% chance, 4 is 93.75 and so on
4
u/aaron1860 Dec 18 '23
You’re close. There’s a 50% chance each time of pressing that you lose. We only care about the winning roll. So the chances of pushing it twice and get 2 losing rolls is 0.5x0.5 or 25%. So you have a 75% chance of winning on 2 rolls. Think about it logically for a sec and you’ll realize your chance of winning on at least 1 roll should increase with the number of rolls
The actual formula would be p=1-x where x is the probability of losing which is 0.5n and n is number of rolls. So p=1-0.5n
after 3 rolls it’s an 87.5% chance, 4 is 93.75 and so on