Let’s Make A Deal! Programming Praxis.
Monty Hall 문제로 잘 알려져 있는 확률 문제를 실제로 시뮬레이션 해보는 프로그램.
아래는 OCaml 코드
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open Random;;
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open Array;;
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open Printf;;
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let rec shuffle arr n =
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if n = 0 then arr
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else begin
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let a = Random.int (length arr) in
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let b = Random.int (length arr) in
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let va = Array.get arr a in
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Array.set arr a (Array.get arr b);
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Array.set arr b va;
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shuffle arr (n-1)
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end
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exception Found
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let find_unchosen_goat arr choice =
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let ug = ref 0 in
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try
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Array.iteri
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(fun i _ ->
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if i != choice && Array.get arr i != 1 then begin ug := i; raise Found end
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else ug := i)
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arr;
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0
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with _ -> !ug
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let find_change_choice arr old_choice unchosen_goat =
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let cc = ref 0 in
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try
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Array.iteri
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(fun i _ ->
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if i != old_choice && i != unchosen_goat then begin cc := i; raise Found end
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else cc := i)
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arr;
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0
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with _ -> !cc
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let check_choice arr choice = Array.get arr choice = 1
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let deal change =
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let doors = shuffle [|0;0;1;|] 10 in
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let choice = Random.int 3 in
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let unchosen_goat = find_unchosen_goat doors choice in
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let last_choice =
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if change then find_change_choice doors choice unchosen_goat
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else choice
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in
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check_choice doors last_choice
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(** Written by myself – simulate the whole game *)
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let simulate ngames =
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let switch_game_result = Array.map (fun _ -> deal true) (Array.init ngames (fun i -> i)) in
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let switch_wins = Array.fold_left (fun wins result -> if result then wins+1 else wins) 0 switch_game_result in
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let stay_game_result = Array.map (fun _ -> deal false) (Array.init ngames (fun i -> i)) in
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let stay_wins = Array.fold_left (fun wins result -> if result then wins+1 else wins) 0 stay_game_result in
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Printf.printf "Winning rate for switch choice : %f\n" ((float_of_int switch_wins) /. (|>float_of_int ngames));
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Printf.printf "Winning rate for stay choice : %f\n" ((float_of_int stay_wins) /. (|>float_of_int ngames))
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(** Translation of suggested solution – simplified simulation*)
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let monty n =
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Random.self_init ();
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let rec monty n switch stay =
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let auto,pick = Random.int 3, Random.int 3 in
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if n = 0 then (switch,stay)
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else
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if auto = pick then monty (n-1) switch (stay+1)
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else monty (n-1) (switch+1) stay
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in
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monty n 0 0
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(** More simplified *)
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let monty2 n =
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Random.self_init ();
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let rec monty2 n switch =
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let pick = Random.int 3 in
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if n = 0 then switch
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else
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if pick = 0 then monty2 (n-1) switch
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else monty2 (n-1) (switch+1)
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in
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monty2 n 0
결과를 보면 선택을 바꾼 경우 이길 확률이 2/3가 된다는 것을 볼 수 있다.
# monty2 1000;; - : int = 674 # monty2 10000;; - : int = 6657 # monty2 100000;; - : int = 66647
