> 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%.
- 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:
Or it is possible to win on the first day only
WLLLL � LLL (2.1)
Or to win on the second day only
Or to win on the third day only
Etc, etc, etc
Or to win the last day only
Also it is not impossible to win twice:
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.
Now let back on your question. Let compute probability of event LLLLLLL�.LLLL. you don't want � all
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.
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) +
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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