# RosettaCode Monty Hall

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## Reference

Statement of the Monty Hall problem on RosettaCode: here.

Deadline for adding to RosettaCode page: 31 Aug 2012; submitter:

## Eiffel code

Here's a candidate implementation. This compiles and runs, producing output similar to this:

Staying wins 333504 times.

Switching wins 666496 times.

class MONTY_HALL create make feature {NONE} -- Initialization make local games_count: INTEGER do create random_generator.make games_count := 1000000 across 1 |..| games_count as game loop play end print ("Staying wins " + staying_wins.out + " times.%N") print ("Switching wins " + (games_count - staying_wins).out + " times.%N") end feature -- Commands play local doors: ARRAYED_LIST [BOOLEAN] chosen, shown: INTEGER do create doors.make_filled (Door_count) -- False is a goat, True is a car doors [next_random_door] := True -- Put a car behind a random door chosen := next_random_door -- Pick a door, any door -- Monty selects a door which is neither the winner nor the choice from shown := next_random_door until shown /= chosen and not doors [shown] loop shown := next_random_door end if doors [chosen] then -- If you would have won by staying, count it staying_wins := staying_wins + 1 end ensure staying_wins_valid: staying_wins = old staying_wins or staying_wins = old staying_wins + 1 end feature {NONE} -- Implementation Door_count: INTEGER = 3 -- The total number of doors. staying_wins: INTEGER -- The number of times that the strategy of staying would win. random_generator: RANDOM -- A random number generator for selecting doors. next_random_door: INTEGER -- A door chosen at random. do random_generator.forth Result := random_generator.item \\ Door_count + 1 ensure valid_door: Result >= 1 and Result <= Door_count end end

## Comments

Note that the implementations in many other languages maintain a separate variable to count the number of switch wins. They calculate whether switching wins at each step via some funky logic that relies on zero-based array indexing, which would be inconvenient in Eiffel. But we don't need to do that at all anyway, because calculating it at each step is completely redundant: we can just do a final subtraction at the end, right?