Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help: Probability of Equal Chance

  1. Nov 24, 2009 #1
    Here is the story:

    There are totally 20 balls. Only one ball is red and the rest of 19 balls are blue. Everybody wants a red one. 20 persons take turn to draw from the box.

    Point 1: the probability of drawing a read one has nothing to do with the order. It means the first person or the last person has the equal chance (1/20) to get it.

    Point 2: the first person has 1/20 chance; the second person has 1/19 if the first is blue; the third person has 1/18 if the second is still blue... and so on.

    Point 3: the first person has bigger chance to get a red one.

    For me, I definitely agree to point 1 and I know it is an ancient math problem. Anybody tell me the link or where it is from. I was arguing with my coworkers for the whole day for nothing. They laughed at me for my poor math.

    Otherwise, tell me my mistake. Thank you for your time and sorry for my poor language.
  2. jcsd
  3. Nov 24, 2009 #2


    User Avatar
    Gold Member

    What mistake? That the first person has the highest chance of picking a red ball?

    Reread your Point 2, concentrating on the part where you say "if the first is blue". You have failed to factor-in the total probability by ignoring this as a possible outcome.
  4. Nov 24, 2009 #3
    Which point do you agree to?
  5. Nov 24, 2009 #4
    This is the real story.

    The IT guy sent an email to everyone yesterday and said there are totally 1 laptop and 19 computers for free. The first person replys to his email will get the first opportunity to draw. So, I was the first and IT guy told me that I have a bigger chance to win a laptop. I was arguing I have the equal chance with the last person to get the laptp. What he is planning is to put 20 paper(only one writes laptop) in the box and I am the first to draw from the box.

    My boss told me that I was stupid as I will have least chance to get the laptop. He concluded the one in the middle (10th or 11th) will have the biggest chance to get laptop.

    Now there are three stories. IT guy says, biggest chance to have a laptop; my boss says, least chance to get a laptop; me says, equal chance (1/20) like everybody else behind me.
    Last edited: Nov 24, 2009
  6. Nov 24, 2009 #5


    User Avatar
    Gold Member

    You are correct. Everyone has an equal shot.

    Think about this: say you draw first and then everyone else draws and none of you look at your pieces of paper.

    As it stands, there are 20 people and 20 pieces of paper. Does it make any difference which piece of paper came out of the box first?
  7. Nov 24, 2009 #6
    I know it is from an ancient Greek math, but I can not recall where it is from. Could you please tell me exactly where it is from.

    The point is even I tell the rest of them (19 persons) I get no laptop and it will not change the chance for any one of them (19 persons) to get a laptop. Is that right?

    Let me say: I get 1/20 chance; the second gets (19/20) * (1/19); the third gets (18/20)*(1/18)... and so on
  8. Nov 24, 2009 #7


    User Avatar
    Gold Member


    I don't know where it's from. Sorry.

    But beware: it is not the same as the very well-known Monty Hall problem. Anyone who points you at the Monty Hall problem is mistaken. (And if no one does, then just ignore this.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook