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Heaven

The posted solution fails to consider the two day scenario where the person chooses the 1-day 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 2-2, 2-3,3-2,3-3. 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 c-d-e-f-a )


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. c-a, c-b or e-b)
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. c-d-a )
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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