Quote (Uguu @ Nov 18 2014 01:43am)
there's 2 players.
50% chance to win and lose here brah
since there's 2 choice for who wins
You roll a die. What's the probability that if I roll the same die, I will roll higher than you?
By your logic, its 50%. Because I either roll higher than you or I don't.
That is however absurd. Because it can't possibly be 50%. You have a win condition from the start. Of an N sided die, if you rolled N, then I can't possibly beat it. I have a 1/N chance of losing before I even roll. It will always be lower than 50%.
P(first > second) + P(first < second) + P(first = second) = 1
P(first > second) = P(first < second), due to the principle of symmetry
P(first = second) = 1/N
Therefore
P(second > first) + P(second < first) + 1/N = 1
P(second > first) + P(second > first) + 1/N = 1
2(P(second > first) = 1 - (1/N)
P(second > first) = [ 1 - (1/N) ] / 2
lim N->+infinity f(N) = [1 - (1/N)] / 2 = 1/2
This means it will NEVER be 50%. So you will always have an advantage. Going first. Now extrapolate that to you trying to beat theprevious roll each time as the subset of win rolls decreases as the number of rolls increases. This means each time you roll the dice you decrease your chance of winning. In this case the player who rolls second has a disadvantage, because he doesn't even need to roll the die to lose. All that he needs to loseis for player 1 to roll the highest value on his first roll, and he will win, because there is no way player 2 can beat that roll. Therefore, the person who rolls the dice the least has the greater probability of winning. But it is not 50%, because player 1 can win from his first roll.
This post was edited by Minkomonster on Nov 18 2014 01:25am