Hello, I was playing poker and I was 31% to win the game when we were all-in but I lost. However, this loss was insignificant for I had 3x as many chips than him therefore we became even in chips. Later, when the chips were even, I was 42% to win the game and I won therefore winning me the game. My question is ... what were my total odds of winning this game? Ultimately, my odds of winning this game has to be higher than 42% because I was 42% at one point for the win. However, I did have an additional chance of winning at 31%. What is the formula to figure this out? I would think my odds of winning would be something like 42% + ?%. Let me know please, thanks! Also, let me know what the odds would be if it was 80% and 50% =)
2006-09-18 16:02:26 · 2 answers · asked by Ertai2 4 in Science & Mathematics ➔ Mathematics
What you do for unconditional probability like this is start subtracting one chance from 1, then the next one OF the remainder FROM the remainder, and so on. So 1-0.42=0.56, 0.56*(1-0.31)=0.56*0.69=0.3864, 1-0.3864=0.6136. Your chances of winning, at the beginning, were 61.36%. In your second scenario, they would of course be 90%.
2006-09-18 16:06:53 · answer #1 · answered by Anonymous · 0⤊ 0⤋
100 percent
2006-09-18 23:03:47 · answer #2 · answered by Anonymous · 0⤊ 1⤋