A (hopefully) simple probability question

two people each flip a coin at the same time. each time a person gets tails, they lose 1 point. they continue until both players get heads on the same flip. what are the odds that:

- the game continues to the 2nd, 3rd, 4th etc rounds (stop when the odds get very low)
- one player loses 1, 2, 3 etc more points than the other (again stop when odds are very low)

possible outcomes:
P1___|___P2
H________H
H________T
T________H
T________T

probability of each outcome is .5 * .5 = .25

probability of "passing" the round = 1 - .25 = .75
probability of "passing" n rounds = (.75)^n

probability of losing a point = .5

dunno lol.

lol, anyone wanna put this into laymans terms? twas hoping for something like

round 2 = X%
round 3 = X%

one player at +1 points = X%
""" +3 " = X%

etc :)

The probability that it stops on round n is pretty simple...

it's [3^(n-1)]/4^n as to stop on round n it has to not stop on all n-1 rounds preceeding it, each with a probability of 3/4, and then the probability it stops on round n is 1/4.

This is all assuming fair coins and indepence of trials and all that shit.



The second answer is quite complex, you could probably set up a recurrence relation or maybe use branching processes, I haven't really got the time to do it right now.

Read up on Discrete random variables.

I wants lots and lots of some delectable pot!

Marijuana MUST be legalized.