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FW: Probability

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  • Steve Merbach
    Message 1 of 2 , Jan 3, 2001
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      > Please help settle a friendly discussion!!
      > 728 chances were sold on a raffle.
      > Each day of the calendar year one ticket is drawn and put back.
      > The probability of being picked on any given day is 1 out of 728.
      > What are the chances of winning at least once out of the 365 days.
      > I think my chances are winning are still 1 out of 728, or .137%.
      > My co-worker (the ticket seller) feels that probability of winning at
      > least once during the year are significantly higher, perhaps as high as
      > 30% to 45%.
      > HELP!!!
    • pegi@net.yu
      My dear friend, This is very simple task to compute. Let denote Pwin = 1/728 (= 1.37 of 1000) the probability to pick up the winning ticket. This is
      Message 2 of 2 , Jan 4, 2001
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        My dear friend,

        This is very simple task to compute.
        Let denote Pwin = 1/728 (= 1.37 of 1000) the probability to pick up the winning ticket. This is probability 
        of only one draw. But you have 365 draws. This significantly increase your chances. Let analyze the 
        situation. Let denote event loose as L and event win as W. During 365 days it is possible all to loose:

        LLLLLLLLL�..LLLL.        (1)

        Or it is possible to win on the first day only
                WLLLL � LLL                (2.1)
        Or to win on the second day only
                LWLLL�LLL                (2.2)
        Or to win on the third day only        
                LLWLL�LLL                (2.3)
        Etc, etc, etc 
        Or to win the last day only
                LLLLL�.LLW                (2.365)

        Also it is not impossible to win twice:

                WWLLLLL�LLLL
                WLWLLLL�LLLL
                ��������
                WLLLLLL�LLLW
                LWWLLL�LLLLL
                LWLWLL�LLLL        (3)
                �������.
                �������.
                �������.
                LLLLLLL�LLWW

         
        Also you can win three, four�times
        Theoretically you can win all 365 days and that probability is to small but still exist. Its probability is 
        (1/728)**365 where "**" means "power"   (1/728)*(1/728)*�.*(1/728). This is very, very small number 
        but it is not impossible. If you don't believe it is possible I can aware you that it is still possible but this is 
        for an future examination. 

                WWWW�.WWWW        (4)

        Now let back on your question. Let compute probability of event LLLLLLL�.LLLL. you don't want � all 
        loose .

        Because of condition that you always return back the ticket the probability of every day draw doesn't 
        change (If you do not return back the ticket the calculation is a little diferent) you have an array of 365 
        independent events and thus 

        P(LLLLL�LLL) = P(L)*P(L)*�.*P(L)     �   365 times.

        As you know that P(W) = 1/728 (=0.00137) you have to know that P(L) = 1 � P(W) = 727/728 = 0.9986.

        Then P(LLLLL�LLLL) = (0.9986)*(0.9986)*(0.9986)* �. *(0.9966)       �.   365 times.

                = 0.6054�

        Because the event LLLLL�LLL is the only one unwanted for you everything else is wanted for you and 
        have the probability 

                        1 � 0.6054 = 0.3945�

        So your friend is right.


        You can now easily compute some other events. For example what is the probability to win EXACTLY  
        once. To accomplish this task you have to take a look on lines (2.1), (2.2), � (2.365).

        P(win once) =         P(WLLLLLL�LLL) +
                        P(LWLLLLL�LLL) +
                        P(LLWLLLL�LLL) +
                        ��.
                        P(LLLLLLL�.LLW) =

                        (1/728)*(727/728)*(727/728)*�*(727/728)+
                        (727/728)*(1/728)*(727/728)*�*(727/728)+
                        (727/728)*(727/728)*(1/728)*�*(727/728)+
                        ������������������
                        (727/728)*(727/728)*(727/728)*�*(1/728) = 

                        365*(1/728)*(727/728)**364 =

                        0.30399
        This means that P(all lose and win only once) = 0.6054 + 0.3039 = 0.9093.
        This means that the probability to win at least twice is about 0.1.


        Ok, that is more than you expected. If you understand my explanation try to compute the probability to 
        win EXACTLY twice and send your answer on  >pegi@...
                        


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