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

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).

hence 0.48=p[sup]9[/sup]

for 6 novices, the expert winning chance would be p[sup]6[/sup]=0.48[sup]6/9[/sup]=61.3%

]]>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.

Thanks!

My understanding:

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%?