An amazing feature of Imandra is the ability to reason about functional inputs and generate functional instances and counterexamples. For example we can introduce a variant type and a second-order function that evaluates its argument on each variant and computes the sum of the results.

In :
type t = A | B | C
let sum = (f) => f(A) + f(B) + f(C)

Out:
type t = A | B | C
val sum : (t -> Z.t) -> Z.t = <fun>


Now, let's ask Imandra to come up with an instance of a function f that returns positive values and that produces 42 when calling sum(f):

In :
instance( (f) => f(A) > 0 && f(B) > 0 && sum(f) == 42 )

Out:
- : (t -> Z.t) -> bool = <fun>
module CX : sig val f : t -> Z.t end

Instance (after 0 steps, 0.013s):
let f : t -> int = fun (x_0 : t) -> if x_0 = C then 40 else 1

Instance
proof attempt
ground_instances:0
definitions:0
inductions:0
search_time:
0.013s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 3 arith tableau max rows: 1 arith tableau max columns: 4 arith pivots: 1 rlimit count: 350 mk clause: 3 datatype occurs check: 3 mk bool var: 5 arith assert upper: 1 propagations: 2 datatype accessor ax: 3 arith num rows: 1 num allocs: 6.23861e+06 final checks: 1 added eqs: 1 del clause: 3 arith eq adapter: 1 time: 0.005 memory: 15.41 max memory: 15.84
Expand
• start[0.013s]
let (_x_0 : int) = sko_f_0 A in
let (_x_1 : int) = sko_f_0 B in
(_x_0 > 0) && ((_x_1 > 0) && (_x_0 + _x_1 + sko_f_0 C = 42))
• #### simplify

 into: let (_x_0 : int) = sko_f_0 A in let (_x_1 : int) = sko_f_0 B in not (_x_0 <= 0) && not (_x_1 <= 0) && (_x_0 + _x_1 + sko_f_0 C = 42) expansions: [] rewrite_steps: forward_chaining:
• Sat (Some let f : t -> int = fun (x_0 : t) -> if x_0 = C then (Z.of_nativeint (40n)) else (Z.of_nativeint (1n)) )

A natural application of this ability is to apply it to formal reasoning about random variables and probability distributions over them. In the following, we'll consider an application of this kind of reasoning to analyse the famous Monty Hall problem.

# Reasoning about Monty Hall probabilities¶

Let’s ask Imandra if 1/3 of a 1,000,000$is the best outcome for any strategy the player can employ. To do this we’ll use Imandra verify directive to verify that, for any valid strategy, the scenario with equally probable prize placement gives an expected reward less or equal than 1000000/3. In : verify ( (strategy) => Real.({ valid_strategy(strategy) ==> ( expected_reward ( make_scenario_pmf(pDoorsEqual,strategy) ) <= 1000000.0 / 3.0 ) }))  Out: - : (door -> choice -> real) -> bool = <fun> module CX : sig val strategy : door -> choice -> real end  Counterexample (after 0 steps, 0.013s): let strategy : door -> choice -> real = fun (x_0 : door) -> fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then 3/1000000 else if (x_0 = DoorB) && (x_1 = Stay) then 999997/1000000 else 0  Refuted proof attempt ground_instances:0 definitions:0 inductions:0 search_time: 0.013s details: Expand smt_stats:  num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 11 arith pivots: 2 rlimit count: 1971 mk clause: 3 datatype occurs check: 5 mk bool var: 10 arith assert upper: 1 arith row summations: 2 propagations: 2 datatype accessor ax: 5 arith num rows: 2 num allocs: 1.45941e+07 final checks: 1 added eqs: 1 del clause: 3 arith eq adapter: 1 time: 0.006 memory: 17.7 max memory: 18.09 Expand • start[0.013s] let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in let (_x_6 : real) = 1 / 3 in let (_x_7 : real) = _x_6 * _x_0 in let (_x_8 : real) = if Stay = Swap then 1000000 else 0 in let (_x_9 : bool) = DoorB = DoorA in let (_x_10 : bool) = DoorC = DoorA in let (_x_11 : real) = _x_6 * _x_2 in let (_x_12 : bool) = DoorA = DoorB in let (_x_13 : bool) = DoorC = DoorB in let (_x_14 : real) = _x_6 * _x_4 in let (_x_15 : bool) = DoorA = DoorC in let (_x_16 : bool) = DoorB = DoorC in let (_x_17 : real) = _x_6 * _x_1 in let (_x_18 : real) = if Swap = Stay then 1000000 else 0 in let (_x_19 : real) = _x_6 * _x_3 in let (_x_20 : real) = _x_6 * _x_5 in (_x_0 >= 0) && ((_x_1 >= 0) && ((_x_2 >= 0) && ((_x_3 >= 0) && ((_x_4 >= 0) && ((_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)))))) ==> _x_7 * 1000000 +. _x_7 * (if _x_9 then 1000000 else _x_8) +. _x_7 * (if _x_10 then 1000000 else _x_8) +. (_x_11 * (if _x_12 then 1000000 else _x_8) +. _x_11 * 1000000 +. _x_11 * (if _x_13 then 1000000 else _x_8)) +. (_x_14 * (if _x_15 then 1000000 else _x_8) +. _x_14 * (if _x_16 then 1000000 else _x_8) +. _x_14 * 1000000) +. (_x_17 * _x_18 +. _x_17 * (if _x_9 then _x_18 else 1000000) +. _x_17 * (if _x_10 then _x_18 else 1000000) +. (_x_19 * (if _x_12 then _x_18 else 1000000) +. _x_19 * _x_18 +. _x_19 * (if _x_13 then _x_18 else 1000000)) +. (_x_20 * (if _x_15 then _x_18 else 1000000) +. _x_20 * (if _x_16 then _x_18 else 1000000) +. _x_20 * _x_18)) <= 1000000 / 3 • #### simplify  into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in not ((_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)) || (1000000/3 * _x_0 +. 1000000/3 * _x_2 +. 1000000/3 * _x_4 +. 2000000/3 * _x_1 +. 2000000/3 * _x_3 +. 2000000/3 * _x_5 <= 1000000/3) expansions: [] rewrite_steps: forward_chaining: • Sat (Some let strategy : door -> choice -> real = fun (x_0 : door) -> fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then (Q.of_string "3/1000000") else if (x_0 = DoorB) && (x_1 = Stay) then (Q.of_string "999997/1000000") else (Q.of_nativeint (0n)) ) Imandra refuted our statement and provided us with a counterexample strategy. Imandra suggests that the expected reward will be greater than 1000000/3 if the player chooses DoorA and then Swaps his choice in 0.0003% of the cases. The counterexample strategy is defined in the CX module, so let's see what is the expected reward with this strategy: In : expected_reward( make_scenario_pmf(pDoorsEqual, CX.strategy) )  Out: - : real = 1000003/3  The counterexample strategy that Imandra produced increased our expected winnings by 1$. This counterexample strategy employs swapping instead of staying. What would be our expected reward if we always swap?

In :
let random_then_swap_strategy = (door : door,choice) =>
switch(door,choice){ | (_,Swap) => Real.(1.0 / 3.0) | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsEqual,random_then_swap_strategy)
)

Out:
val random_then_swap_strategy : door -> choice -> real = <fun>
- : real = 2000000/3


By always swapping we’ve increased our chances of winning up to 2/3. Is this the best possible expected reward one might get in this game?

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsEqual,strategy)
) <= 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances:0
definitions:0
inductions:0
search_time:
0.015s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 11 arith pivots: 2 rlimit count: 1953 mk clause: 3 mk bool var: 10 arith assert upper: 1 arith row summations: 2 propagations: 2 conflicts: 1 datatype accessor ax: 5 arith conflicts: 1 arith num rows: 2 num allocs: 8.25349e+06 added eqs: 1 arith eq adapter: 1 time: 0.006 memory: 18 max memory: 18.42
Expand
• start[0.015s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 1 / 3 in
let (_x_7 : real) = _x_6 * _x_0 in
let (_x_8 : real) = if Stay = Swap then 1000000 else 0 in
let (_x_9 : bool) = DoorB = DoorA in
let (_x_10 : bool) = DoorC = DoorA in
let (_x_11 : real) = _x_6 * _x_2 in
let (_x_12 : bool) = DoorA = DoorB in
let (_x_13 : bool) = DoorC = DoorB in
let (_x_14 : real) = _x_6 * _x_4 in
let (_x_15 : bool) = DoorA = DoorC in
let (_x_16 : bool) = DoorB = DoorC in
let (_x_17 : real) = _x_6 * _x_1 in
let (_x_18 : real) = if Swap = Stay then 1000000 else 0 in
let (_x_19 : real) = _x_6 * _x_3 in
let (_x_20 : real) = _x_6 * _x_5 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
==> _x_7 * 1000000 +. _x_7 * (if _x_9 then 1000000 else _x_8)
+. _x_7 * (if _x_10 then 1000000 else _x_8)
+. (_x_11 * (if _x_12 then 1000000 else _x_8) +. _x_11 * 1000000
+. _x_11 * (if _x_13 then 1000000 else _x_8))
+. (_x_14 * (if _x_15 then 1000000 else _x_8)
+. _x_14 * (if _x_16 then 1000000 else _x_8) +. _x_14 * 1000000)
+. (_x_17 * _x_18 +. _x_17 * (if _x_9 then _x_18 else 1000000)
+. _x_17 * (if _x_10 then _x_18 else 1000000)
+. (_x_19 * (if _x_12 then _x_18 else 1000000) +. _x_19 * _x_18
+. _x_19 * (if _x_13 then _x_18 else 1000000))
+. (_x_20 * (if _x_15 then _x_18 else 1000000)
+. _x_20 * (if _x_16 then _x_18 else 1000000) +. _x_20 * _x_18))
<= 2000000 / 3
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in not ((_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)) || (1000000/3 * _x_0 +. 1000000/3 * _x_2 +. 1000000/3 * _x_4 +. 2000000/3 * _x_1 +. 2000000/3 * _x_3 +. 2000000/3 * _x_5 <= 2000000/3) expansions: [] rewrite_steps: forward_chaining:
• Unsat

We’ve proven that the best possible expected reward in the Monty Hall game is 2/3 of a \$1,000,000 in an (arguably, rather counterintuitive) result that one has to always swap his first choice to double his chances of winning.

Suppose now that we don’t know the winning strategy, but we do know that it is possible to win in 2/3 of the cases. Can we ask Imandra to synthesise the optimal strategy? For this purpose we can use the instance directive:

In :
instance ( (strategy) => Real.({
valid_strategy(strategy) &&
( expected_reward (
make_scenario_pmf(pDoorsEqual,strategy)
) == 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> real end

Instance (after 0 steps, 0.012s):
let strategy : door -> choice -> real =
fun (x_0 : door) ->
fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then 1 else 0

Instance
proof attempt
ground_instances:0
definitions:0
inductions:0
search_time:
0.012s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 11 arith pivots: 2 rlimit count: 1935 mk clause: 6 datatype occurs check: 5 mk bool var: 12 arith assert upper: 2 arith row summations: 2 propagations: 4 datatype accessor ax: 5 arith num rows: 2 num allocs: 1.51719e+07 final checks: 1 added eqs: 2 del clause: 6 arith eq adapter: 2 time: 0.006 memory: 18.39 max memory: 18.79
Expand
• start[0.012s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 1 / 3 in
let (_x_7 : real) = _x_6 * _x_0 in
let (_x_8 : real) = if Stay = Swap then 1000000 else 0 in
let (_x_9 : bool) = DoorB = DoorA in
let (_x_10 : bool) = DoorC = DoorA in
let (_x_11 : real) = _x_6 * _x_2 in
let (_x_12 : bool) = DoorA = DoorB in
let (_x_13 : bool) = DoorC = DoorB in
let (_x_14 : real) = _x_6 * _x_4 in
let (_x_15 : bool) = DoorA = DoorC in
let (_x_16 : bool) = DoorB = DoorC in
let (_x_17 : real) = _x_6 * _x_1 in
let (_x_18 : real) = if Swap = Stay then 1000000 else 0 in
let (_x_19 : real) = _x_6 * _x_3 in
let (_x_20 : real) = _x_6 * _x_5 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
&& (_x_7 * 1000000 +. _x_7 * (if _x_9 then 1000000 else _x_8)
+. _x_7 * (if _x_10 then 1000000 else _x_8)
+. (_x_11 * (if _x_12 then 1000000 else _x_8) +. _x_11 * 1000000
+. _x_11 * (if _x_13 then 1000000 else _x_8))
+. (_x_14 * (if _x_15 then 1000000 else _x_8)
+. _x_14 * (if _x_16 then 1000000 else _x_8) +. _x_14 * 1000000)
+. (_x_17 * _x_18 +. _x_17 * (if _x_9 then _x_18 else 1000000)
+. _x_17 * (if _x_10 then _x_18 else 1000000)
+. (_x_19 * (if _x_12 then _x_18 else 1000000) +. _x_19 * _x_18
+. _x_19 * (if _x_13 then _x_18 else 1000000))
+. (_x_20 * (if _x_15 then _x_18 else 1000000)
+. _x_20 * (if _x_16 then _x_18 else 1000000) +. _x_20 * _x_18))
= 2000000 / 3)
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in (_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1) && (1000000/3 * _x_0 +. 1000000/3 * _x_2 +. 1000000/3 * _x_4 +. 2000000/3 * _x_1 +. 2000000/3 * _x_3 +. 2000000/3 * _x_5 = 2000000/3) expansions: [] rewrite_steps: forward_chaining:
• Sat (Some let strategy : door -> choice -> real = fun (x_0 : door) -> fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then (Q.of_nativeint (1n)) else (Q.of_nativeint (0n)) )

We've got an example of an optimal strategy - this one doesn't randomize over doors, but suggests to always choose the DoorA and then swap.

# Biased Monty Hall¶

Now let's apply what we've learned about the "standard" Monty Hall paradox to a more complicated game setup - suppose that we somehow know that the host is biased towards the DoorA -- the probability that the host will hide the prize behind the first door is 3 times higher than the individual probabilities for two other doors. Should one still always swap? Or, maybe, it makes more sense to always choose the more likely DoorA and stay?

We encode our biased host in the biased PMF over doors (also doing a sanity check that it is a valid probability distribution):

In :
let pDoorsBiased = (door : door) => Real.(
switch (door){
| DoorA => 3.0 / 5.0
| DoorB | DoorC => 1.0 / 5.0
}
);

valid_door_pmf(pDoorsBiased)

Out:
val pDoorsBiased : door -> real = <fun>
- : bool = true


In the previous section, we've proven that the maximal possible expected reward in the "standard" Monty Hall problem is 2/3 of a million. Again, we ask Imandra whether we can improve on this in the biased case:

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsBiased,strategy)
) <= 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> real end

Counterexample (after 0 steps, 0.013s):
let strategy : door -> choice -> real =
fun (x_0 : door) ->
fun (x_1 : choice) ->
if (x_0 = DoorA) && (x_1 = Stay) then 399997/600000
else if (x_0 = DoorC) && (x_1 = Swap) then 200003/600000 else 0

Refuted
proof attempt
ground_instances:0
definitions:0
inductions:0
search_time:
0.013s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 13 arith pivots: 3 rlimit count: 2307 mk clause: 3 datatype occurs check: 5 mk bool var: 10 arith assert upper: 1 arith row summations: 3 propagations: 2 datatype accessor ax: 5 arith num rows: 2 num allocs: 2.4864e+07 final checks: 1 added eqs: 1 del clause: 3 arith eq adapter: 1 time: 0.006 memory: 18.97 max memory: 19.37
Expand
• start[0.013s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 3 / 5 in
let (_x_7 : real) = 1 / 5 in
let (_x_8 : real) = _x_7 * _x_0 in
let (_x_9 : real) = if Stay = Swap then 1000000 else 0 in
let (_x_10 : bool) = DoorB = DoorA in
let (_x_11 : bool) = DoorC = DoorA in
let (_x_12 : bool) = DoorA = DoorB in
let (_x_13 : real) = _x_7 * _x_2 in
let (_x_14 : bool) = DoorC = DoorB in
let (_x_15 : bool) = DoorA = DoorC in
let (_x_16 : real) = _x_7 * _x_4 in
let (_x_17 : bool) = DoorB = DoorC in
let (_x_18 : real) = if Swap = Stay then 1000000 else 0 in
let (_x_19 : real) = _x_7 * _x_1 in
let (_x_20 : real) = _x_7 * _x_3 in
let (_x_21 : real) = _x_7 * _x_5 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
==> _x_6 * _x_0 * 1000000 +. _x_8 * (if _x_10 then 1000000 else _x_9)
+. _x_8 * (if _x_11 then 1000000 else _x_9)
+. (_x_6 * _x_2 * (if _x_12 then 1000000 else _x_9) +. _x_13 * 1000000
+. _x_13 * (if _x_14 then 1000000 else _x_9))
+. (_x_6 * _x_4 * (if _x_15 then 1000000 else _x_9)
+. _x_16 * (if _x_17 then 1000000 else _x_9) +. _x_16 * 1000000)
+. (_x_6 * _x_1 * _x_18 +. _x_19 * (if _x_10 then _x_18 else 1000000)
+. _x_19 * (if _x_11 then _x_18 else 1000000)
+. (_x_6 * _x_3 * (if _x_12 then _x_18 else 1000000)
+. _x_20 * _x_18 +. _x_20 * (if _x_14 then _x_18 else 1000000))
+. (_x_6 * _x_5 * (if _x_15 then _x_18 else 1000000)
+. _x_21 * (if _x_17 then _x_18 else 1000000) +. _x_21 * _x_18))
<= 2000000 / 3
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in not ((_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)) || (600000 * _x_0 +. 200000 * _x_2 +. 200000 * _x_4 +. 400000 * _x_1 +. 800000 * _x_3 +. 800000 * _x_5 <= 2000000/3) expansions: [] rewrite_steps: forward_chaining:
• Sat (Some let strategy : door -> choice -> real = fun (x_0 : door) -> fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Stay) then (Q.of_string "399997/600000") else if (x_0 = DoorC) && (x_1 = Swap) then (Q.of_string "200003/600000") else (Q.of_nativeint (0n)) )

Imandra suggests that we can do better than 2/3 if we use a mixed strategy of either choosing the DoorA and staying or choosing the DoorC and swapping. Let's test both of these pure strategies and see what would be the expected rewards for each:

In :
let aStayStrategy = (door,choice) =>
switch(door,choice){ | (DoorA,Stay) => 1.0 | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsBiased,aStayStrategy)
);
let cSwapStrategy = (door,choice) =>
switch(door,choice){ | (DoorC,Swap) => 1.0 | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsBiased,cSwapStrategy)
);

Out:
val aStayStrategy : door -> choice -> real = <fun>
- : real = 600000
val cSwapStrategy : door -> choice -> real = <fun>
- : real = 800000


Choosing the DoorC and then swapping increases the chances of winning up to 80%! Again, is this the best outcome one can get?

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsBiased,strategy)
) <= 800000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances:0
definitions:0
inductions:0
search_time:
0.021s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 13 arith pivots: 3 rlimit count: 2287 mk clause: 3 mk bool var: 10 arith assert upper: 1 arith row summations: 3 propagations: 2 conflicts: 1 datatype accessor ax: 5 arith conflicts: 1 arith num rows: 2 num allocs: 3.73802e+07 added eqs: 1 arith eq adapter: 1 time: 0.01 memory: 19.69 max memory: 20.1
Expand
• start[0.021s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 3 / 5 in
let (_x_7 : real) = 1 / 5 in
let (_x_8 : real) = _x_7 * _x_0 in
let (_x_9 : real) = if Stay = Swap then 1000000 else 0 in
let (_x_10 : bool) = DoorB = DoorA in
let (_x_11 : bool) = DoorC = DoorA in
let (_x_12 : bool) = DoorA = DoorB in
let (_x_13 : real) = _x_7 * _x_2 in
let (_x_14 : bool) = DoorC = DoorB in
let (_x_15 : bool) = DoorA = DoorC in
let (_x_16 : real) = _x_7 * _x_4 in
let (_x_17 : bool) = DoorB = DoorC in
let (_x_18 : real) = if Swap = Stay then 1000000 else 0 in
let (_x_19 : real) = _x_7 * _x_1 in
let (_x_20 : real) = _x_7 * _x_3 in
let (_x_21 : real) = _x_7 * _x_5 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
==> _x_6 * _x_0 * 1000000 +. _x_8 * (if _x_10 then 1000000 else _x_9)
+. _x_8 * (if _x_11 then 1000000 else _x_9)
+. (_x_6 * _x_2 * (if _x_12 then 1000000 else _x_9) +. _x_13 * 1000000
+. _x_13 * (if _x_14 then 1000000 else _x_9))
+. (_x_6 * _x_4 * (if _x_15 then 1000000 else _x_9)
+. _x_16 * (if _x_17 then 1000000 else _x_9) +. _x_16 * 1000000)
+. (_x_6 * _x_1 * _x_18 +. _x_19 * (if _x_10 then _x_18 else 1000000)
+. _x_19 * (if _x_11 then _x_18 else 1000000)
+. (_x_6 * _x_3 * (if _x_12 then _x_18 else 1000000)
+. _x_20 * _x_18 +. _x_20 * (if _x_14 then _x_18 else 1000000))
+. (_x_6 * _x_5 * (if _x_15 then _x_18 else 1000000)
+. _x_21 * (if _x_17 then _x_18 else 1000000) +. _x_21 * _x_18))
<= 800000
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in not ((_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)) || (600000 * _x_0 +. 200000 * _x_2 +. 200000 * _x_4 +. 400000 * _x_1 +. 800000 * _x_3 +. 800000 * _x_5 <= 800000) expansions: [] rewrite_steps: forward_chaining:
• Unsat

And we've proven that the DoorC-Swap is the best strategy the player can employ in this setup. Intuitively the reason for that is that it is better to choose the least likely door, forcing the host to reveal more information about the prize location in the more likely doors.

# Even more complex example¶

Using this approach one can go further and analyse even more complicated Monty-Hall-like games (for example, the prize value might depend on the door and/or player actions).

In :
let pDoorsComplex = (door : door) => Real.(
switch (door){
| DoorA => 5.0 / 10.0
| DoorB => 3.0 / 10.0
| DoorC => 2.0 / 10.0
}
);

let reward_complex = (scenario) => {
let prize = switch(scenario.prize) {
| DoorA => 1000000.
| DoorB => 2000000.
| DoorC => 4000000.
};
if( scenario.prize == scenario.first_guess ) {
if(scenario.choice == Stay) prize else 0.0;
} else {
if(scenario.choice == Swap) prize else 0.0;
}
};

let expected_reward_complex : (scenario => real) => real =
(scenario_pmf) => Real.({
let pr = (s) => scenario_pmf(s) * reward_complex(s);
let avg = (choice) => {
let avg = (first_guess) => {
pr { prize: DoorA, first_guess, choice} +
pr { prize: DoorB, first_guess, choice} +
pr { prize: DoorC, first_guess, choice}
};
avg(DoorA) + avg(DoorB) + avg(DoorC)
};
avg(Stay) + avg(Swap)
})

Out:
val pDoorsComplex : door -> real = <fun>
val reward_complex : scenario -> real = <fun>
val expected_reward_complex : (scenario -> real) -> real = <fun>

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward_complex (
make_scenario_pmf(pDoorsComplex,strategy)
) <= 1400000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances:0
definitions:0
inductions:0
search_time:
0.013s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 15 arith pivots: 2 rlimit count: 2473 mk clause: 3 mk bool var: 10 arith assert upper: 1 arith row summations: 2 propagations: 2 conflicts: 1 datatype accessor ax: 5 arith conflicts: 1 arith num rows: 2 num allocs: 5.33225e+07 added eqs: 1 arith eq adapter: 1 time: 0.006 memory: 20.6 max memory: 21
Expand
• start[0.013s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 5 / 10 in
let (_x_7 : real) = 3 / 10 in
let (_x_8 : bool) = DoorB = DoorA in
let (_x_9 : real) = if _x_8 then _x_6 else _x_7 in
let (_x_10 : real) = if _x_8 then 1000000 else 2000000 in
let (_x_11 : bool) = Stay = Swap in
let (_x_12 : real) = if _x_11 then _x_10 else 0 in
let (_x_13 : bool) = DoorC = DoorB in
let (_x_14 : bool) = DoorC = DoorA in
let (_x_15 : real) = if _x_14 then _x_6 else if _x_13 then _x_7 else 2 / 10
in
let (_x_16 : real)
= if _x_14 then 1000000 else if _x_13 then 2000000 else 4000000
in
let (_x_17 : real) = if _x_11 then _x_16 else 0 in
let (_x_18 : real) = if _x_11 then 1000000 else 0 in
let (_x_19 : bool) = DoorA = DoorB in
let (_x_20 : bool) = DoorA = DoorC in
let (_x_21 : bool) = DoorB = DoorC in
let (_x_22 : bool) = Swap = Stay in
let (_x_23 : real) = if _x_22 then 1000000 else 0 in
let (_x_24 : real) = if _x_22 then _x_10 else 0 in
let (_x_25 : real) = if _x_22 then _x_16 else 0 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
==> _x_6 * _x_0 * 1000000 +. _x_9 * _x_0 * (if _x_8 then _x_10 else _x_12)
+. _x_15 * _x_0 * (if _x_14 then _x_16 else _x_17)
+. (_x_6 * _x_2 * (if _x_19 then 1000000 else _x_18)
+. _x_9 * _x_2 * _x_10
+. _x_15 * _x_2 * (if _x_13 then _x_16 else _x_17))
+. (_x_6 * _x_4 * (if _x_20 then 1000000 else _x_18)
+. _x_9 * _x_4 * (if _x_21 then _x_10 else _x_12)
+. _x_15 * _x_4 * _x_16)
+. (_x_6 * _x_1 * _x_23
+. _x_9 * _x_1 * (if _x_8 then _x_24 else _x_10)
+. _x_15 * _x_1 * (if _x_14 then _x_25 else _x_16)
+. (_x_6 * _x_3 * (if _x_19 then _x_23 else 1000000)
+. _x_9 * _x_3 * _x_24
+. _x_15 * _x_3 * (if _x_13 then _x_25 else _x_16))
+. (_x_6 * _x_5 * (if _x_20 then _x_23 else 1000000)
+. _x_9 * _x_5 * (if _x_21 then _x_24 else _x_10)
+. _x_15 * _x_5 * _x_25))
<= 1400000
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in not ((_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1)) || (500000 * _x_0 +. 600000 * _x_2 +. 800000 * _x_4 +. 1400000 * _x_1 +. 1300000 * _x_3 +. 1100000 * _x_5 <= 1400000) expansions: [] rewrite_steps: forward_chaining:
• Unsat
In :
instance ( (strategy) => Real.({
valid_strategy(strategy) &&
( expected_reward_complex (
make_scenario_pmf(pDoorsComplex,strategy)
) == 1400000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> real end

Instance (after 0 steps, 0.013s):
let strategy : door -> choice -> real =
fun (x_0 : door) ->
fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then 1 else 0

Instance
proof attempt
ground_instances:0
definitions:0
inductions:0
search_time:
0.013s
details:
Expand
smt_stats:
 num checks: 1 arith assert lower: 8 arith tableau max rows: 2 arith tableau max columns: 15 arith pivots: 2 rlimit count: 2455 mk clause: 6 datatype occurs check: 5 mk bool var: 12 arith assert upper: 2 arith row summations: 2 propagations: 4 datatype accessor ax: 5 arith num rows: 2 num allocs: 7.06205e+07 final checks: 1 added eqs: 2 del clause: 6 arith eq adapter: 2 time: 0.006 memory: 21.01 max memory: 21.42
Expand
• start[0.013s]
let (_x_0 : real) = sko_strategy_0 DoorA Stay in
let (_x_1 : real) = sko_strategy_0 DoorA Swap in
let (_x_2 : real) = sko_strategy_0 DoorB Stay in
let (_x_3 : real) = sko_strategy_0 DoorB Swap in
let (_x_4 : real) = sko_strategy_0 DoorC Stay in
let (_x_5 : real) = sko_strategy_0 DoorC Swap in
let (_x_6 : real) = 5 / 10 in
let (_x_7 : real) = 3 / 10 in
let (_x_8 : bool) = DoorB = DoorA in
let (_x_9 : real) = if _x_8 then _x_6 else _x_7 in
let (_x_10 : real) = if _x_8 then 1000000 else 2000000 in
let (_x_11 : bool) = Stay = Swap in
let (_x_12 : real) = if _x_11 then _x_10 else 0 in
let (_x_13 : bool) = DoorC = DoorB in
let (_x_14 : bool) = DoorC = DoorA in
let (_x_15 : real) = if _x_14 then _x_6 else if _x_13 then _x_7 else 2 / 10
in
let (_x_16 : real)
= if _x_14 then 1000000 else if _x_13 then 2000000 else 4000000
in
let (_x_17 : real) = if _x_11 then _x_16 else 0 in
let (_x_18 : real) = if _x_11 then 1000000 else 0 in
let (_x_19 : bool) = DoorA = DoorB in
let (_x_20 : bool) = DoorA = DoorC in
let (_x_21 : bool) = DoorB = DoorC in
let (_x_22 : bool) = Swap = Stay in
let (_x_23 : real) = if _x_22 then 1000000 else 0 in
let (_x_24 : real) = if _x_22 then _x_10 else 0 in
let (_x_25 : real) = if _x_22 then _x_16 else 0 in
(_x_0 >= 0)
&& ((_x_1 >= 0)
&& ((_x_2 >= 0)
&& ((_x_3 >= 0)
&& ((_x_4 >= 0)
&& ((_x_5 >= 0)
&& (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1))))))
&& (_x_6 * _x_0 * 1000000 +. _x_9 * _x_0 * (if _x_8 then _x_10 else _x_12)
+. _x_15 * _x_0 * (if _x_14 then _x_16 else _x_17)
+. (_x_6 * _x_2 * (if _x_19 then 1000000 else _x_18)
+. _x_9 * _x_2 * _x_10
+. _x_15 * _x_2 * (if _x_13 then _x_16 else _x_17))
+. (_x_6 * _x_4 * (if _x_20 then 1000000 else _x_18)
+. _x_9 * _x_4 * (if _x_21 then _x_10 else _x_12)
+. _x_15 * _x_4 * _x_16)
+. (_x_6 * _x_1 * _x_23
+. _x_9 * _x_1 * (if _x_8 then _x_24 else _x_10)
+. _x_15 * _x_1 * (if _x_14 then _x_25 else _x_16)
+. (_x_6 * _x_3 * (if _x_19 then _x_23 else 1000000)
+. _x_9 * _x_3 * _x_24
+. _x_15 * _x_3 * (if _x_13 then _x_25 else _x_16))
+. (_x_6 * _x_5 * (if _x_20 then _x_23 else 1000000)
+. _x_9 * _x_5 * (if _x_21 then _x_24 else _x_10)
+. _x_15 * _x_5 * _x_25))
= 1400000)
• #### simplify

 into: let (_x_0 : real) = sko_strategy_0 DoorA Stay in let (_x_1 : real) = sko_strategy_0 DoorA Swap in let (_x_2 : real) = sko_strategy_0 DoorB Stay in let (_x_3 : real) = sko_strategy_0 DoorB Swap in let (_x_4 : real) = sko_strategy_0 DoorC Stay in let (_x_5 : real) = sko_strategy_0 DoorC Swap in (_x_0 >= 0) && (_x_1 >= 0) && (_x_2 >= 0) && (_x_3 >= 0) && (_x_4 >= 0) && (_x_5 >= 0) && (_x_0 +. _x_1 +. _x_2 +. _x_3 +. _x_4 +. _x_5 = 1) && (500000 * _x_0 +. 600000 * _x_2 +. 800000 * _x_4 +. 1400000 * _x_1 +. 1300000 * _x_3 +. 1100000 * _x_5 = 1400000) expansions: [] rewrite_steps: forward_chaining:
• Sat (Some let strategy : door -> choice -> real = fun (x_0 : door) -> fun (x_1 : choice) -> if (x_0 = DoorA) && (x_1 = Swap) then (Q.of_nativeint (1n)) else (Q.of_nativeint (0n)) )

More generally, this illustrates how we can use Imandra’s ability to reason about functional values and real arithmetic to formally prove and refute statements about probability distributions. These abilities allow us to us to formally analyse many kinds of scenarios requiring the ability to reason about random variables.