I am a computer programmer and have a question reguarding probability.
If a player is sitting at a table of 10 people and has of 48% chance of winning. What would his chance of winning be if there were 7 people at the table, instead of 10.
Any forumals and or links to other sites reguarding calculations simular to this would be greatly appriciated.
I know it isn't as easy as doing 10/.48 = 7/x
I was thinking that there are 9 other players who combined have a 52%, or 5.78% per player, chance of beating me. If 3 players leave, at 5.78% per player, then do I have a 17.34% (3 * 5.78%) more of a chance of winning... equaling 65.34%?
each one of 9 players either win the expert or not.
set P(one novice loses)=p, it makes sense to say that for 9 players their ps doesn't interfere each other's and are equal(equally novices).
for 6 novices, the expert winning chance would be p[sup]6[/sup]=0.48[sup]6/9[/sup]=61.3%
Last edited by George,Y (2006-04-11 03:25:56)
Any way you can repeat in newbie terms. I some what under stand the concept, but im still a little confused.
NVM I re-read it and understand. May I also elaberate on this as well.
toss a dime, the chance of getting head up is 1/2. toss two dimes, the chance of getting two heads up is 1/2 (1/2) =1/4 , the 1/2 in the bracket is the chance of getting head up for the second dime, giving the first dime heads up. it's the same as the chance of getting head up without the first dime. This is called probalisticly independent.
the expert win = he beats all novices = all novices fail to win him = 1st novice fails, same time 2nd one fails, same time(1st and 2nd both fail) 3rd one fails ...
assume whether a novice beat the expert or fail is not influenced by other novice or novices at all. and the chance is p
1st 2nd 3rd 4th 5th 6th 7th 8th 9th novice
p p p p p p p p p =0.48 chance of everyone fails