Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Help Me ! » Chance of Win Based on amount Players » 2006-04-11 03:34:17

NVM I re-read it and understand.  May I also elaberate on this as well.

#2 Re: Help Me ! » Chance of Win Based on amount Players » 2006-04-11 03:27:41

Any way you can repeat in newbie terms.  I some what under stand the concept, but im still a little confused.

#3 Help Me ! » Chance of Win Based on amount Players » 2006-04-11 02:57:11

tmaiden
Replies: 4

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.

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%?

Board footer

Powered by FluxBB