Heaven
The posted solution fails to consider the two day scenario where the person chooses the 1day door twice, followed by the right door. This has a probability of 1/3*1/3*1/3=1/27.
Lars
Thursday, September 18, 2003
I agree with the solution but am not quit sure about the derivation. The average number of guesses it takes to pick the right door would be 3. This means that, on average, doors 2 and 3 would have been visited two times on average: the possibilities are 22, 23,32,33. I summed all of these and divided by four to get the same average of three:( (1+1)+(1+2)+(2+1)+(2+2))/4 = 12/4=3.
James kloberdance
Friday, November 14, 2003
the answer may not be right
Since there r 6 arrangements of doors
(Heaven means to wait 0 days)
Arrangements: Door A Door B Door C
a 0 1 2
b 0 2 1
c 1 0 2
d 1 2 0
e 2 0 1
f 2 1 0
A poor guy might try 5 times to enter door A to reach heaven. ( e.g. he meets cdefa )
Assume always entering Door A
there r 2 cases to try 1 time, and totally wait 0 days
(when meeting the arrangements a or b)
there r 4*2 = 8 cases in trying 2 times. (e.g. ca, cb or eb)
averagely wait 1 days in 4 cases, and 2 days in the other 4 cases. totally wait 1*4 + 2*4 =12 days
there r 4*3*2= 24 cases in trying 3 times. (e.g. cda )
averagely wait 2 days in 4 cases; 4 days in 4 cases and 3 days in 16 cases. totally 2*4 + 4*4 + 3*16 =72 days
there r 4*3*2*2= 48 cases in trying 4 times.
wait 4 days in 24 cases; and 5 days in 24 cases.
then totally 4*24 + 5*24=216 days
there r 4! *2 = 48 cases in trying 5 times, and totally wait 48*6 = 288 days
average days = total days/cases = (288+216+72+12+0)/(48+48+24+8+2) =4.523 days
PeterPan
Thursday, November 27, 2003
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