Comments posted to this topic are about the item Monty Hall Paradox
Mr or Mrs. 500
That is an interesting puzzle. Here is another solution using a CTE instead of a loop.
set nocount on
-- Create a table to store the results.
create table Doors
,WinLose as case when ChosenDoor = WinningDoor then 0 else 1 end -- Win = 1, Lose = 0
-- Use a CTE to generate 10,000 "contestants" and their chosen/winning doors.
-- If the chosen door equals the winning door the contestant will switch when given the
-- option and lose. Otherwise they will switch to the winning door and win.
as (select i
from (select 1 union select 2 union select 3 union select 4 union select 5 union
select 6 union select 7 union select 8 union select 9 union select 0) AS X(i))
insert into Doors(ID, ChosenDoor, WinningDoor)
select (D3.i * 1000 + D2.i * 100 + D1.i * 10 + D0.i) AS seq
,abs(cast(cast(newid() as varbinary) as int)) % 3 + 1 ChosenDoor
,abs(cast(cast(newid() as varbinary) as int)) % 3 + 1 WinningDoor
from Digits as D0, Digits as D1, Digits as D2, Digits as D3
-- Select the total number of winners (1) and losers (0).
group by WinLose
drop table Doors
set nocount off
Very cool alternative. Thanks for sharing! 🙂
I think this is bogus. The reason that I think it is bogus is that on the second bet you are excluding the door you picked from your first bet ONLY if it is not the right door. So you are adding new information.
What I lack in youth, I make up for in immaturity!
I think this is bogus. The reason that I think it is bogus is that on the second bet you are excluding the door you picked from your second bet ONLY if it is not the right door. So you are adding new information.
Thanks for sharing you're wonderful insight.
I am surprised anyone even found this old post. LOL.
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