# A certain 'Monty Hall' variant was played. How many doors were there, and, what was the value of the prize?

Recall the standard Monty Hall scenario where a presenter hides one prize with high value $$h$$ behind one closed door (e.g., in the classical version: the one high prize is a car), and, equal prizes with equal low value $$l$$ behind all other closed doors (e.g., in the classical version: all other low prizes are goats).

The total amount of initial closed doors is $$c \ge 3$$ (e.g., in the classical version: $$3$$). The presenter explains (the standard Monty Hall rules): The player will be asked to choose one closed door as hiding the prize to receive (but it stays closed), and, next, the presenter (who knows what's behind each door) will open $$1 \le o \le c-2$$ doors different from the player's choice (e.g., in the classical version: $$1$$) which he can choose, and also effectively chooses, to all reveal an equal low value prize.

Then, the presenter will offer the player an alternative: "You get a second chance: either you prefer, and stick to, your first choice, or, you prefer to change your mind and chose another, remaining, closed, door as hiding the prize you receive.".

Everything happens according to the rules, and, the player, who, like many people, heard about the Monty Hall puzzle, informs the presenter to switch and prefer to chose another remaining closed door. "Good choice ... " says the presenter, "compared to alternative, you increased your average prize value with multiplicative factor $$1.24$$ and with additive term $$2400000$$ dollars.". The player plays as preferred (and, doing so, indeed can expect exactly higher average prize value as presenter mentioned).

Someone noticed:

• The (one) high prize value ($$h$$) was the three times the (all other equal) low prize value ($$l$$).

That is to say: $$h=3l$$

• The amount ($$o$$) of doors opened by the presenter was even, and, the initial amount ($$c$$) of closed doors was equal to three times amount of opened doors divided by two.

That is to say: $$c=3o/2$$

How many doors were there and what was the value of the one high valued prize?

• Should price instead be prize throughout? Dec 31, 2021 at 0:30
• Yes it should, I apologize for poor English and will adjust where I can ! Dec 31, 2021 at 0:40
• @RobPratt wow I didn't notice it, lol (and seems like nobody bother to comment on it for almost 3 days) Dec 31, 2021 at 8:28

The expected values of the staying and switching strategies are $$E_{stay}=(1/c)h+(1-1/c)l$$ $$E_{switch}=(1/c)l+(1-1/c)((c-o-2)l+h)/(c-o-1)$$ Now $$E_{switch}/E_{stay}=1.24$$ is a rational function in $$o$$ only; solving gives only one integral solution $$o=12$$, so there are $$c=18$$ doors. Then solving $$E_{switch}-E_{stay}=\frac{4l}{15}=2.4×10^6$$ gives $$l=9×10^6$$, so $$h=27×10^6$$.

• You found solution! Congrats! Dec 28, 2021 at 20:57
• FYI: price expectation 'stay' and price expectation 'switch' happen to also be simple integers in this particular case. Have a nice day. Dec 28, 2021 at 21:09
• Dear 'Parcly Taxel' again: my very sincere and humbly expressed congratulations! Would you perhaps mind to explain (hidden: if only for me, and, not spoiling for everyone) how you solved 'rational function in $o$' to give 'integral solution for $o$'? Best 2022! Jan 4, 2022 at 5:14
• @FirstNameLastName Any good CAS will do the heavy lifting for you. Jan 4, 2022 at 6:13

This isn't an official answer to the question, but rather its intent is to compliment @Parcly Taxel's existing answer.

Some abbreviations:

c for closed
o for open
h for high
l for low
p for probability
a for average

k for keep
s for switch

Some probabilities:

probability of high prize when keeping choice $$pKH(c) = 1 / c$$ probability of low prize when keeping choice $$pKL(c) = 1 - pKH(c)$$ probability of high prize when switching choice $$pSH(c,o) = pKL(c) / (c - (o + 1))$$ probability of low prize when switching choice $$pSL(c,o) = 1 - pSH(c,o)$$

Some averages (expectations):

average prize when keeping choice $$aKP(c,h,l) = h * pKH(c) + l * pKL(c)$$ average prize when switching choice $$aSP(c,o,h,l) = h * pSH(c,o) + l * pSL(c,o)$$

for any $$3<=c$$ and any $$1<=o<=c-2$$
for any $$1<=h$$ and any $$0<=l
we have: $$k = aKP(c,h,l)$$ $$s = aSP(c,o,h,l)$$ expanded: $$k = h*(1/c)+l*(1-(1/c))$$ $$s = h*(1-(1/c))/(c-(o+1))+l*(1-((1-(1/c))/(c-(o+1))))$$ this allows to compute and inspect: $$s/k$$ and $$s-k$$ and $$c/o$$ and $$h/l$$

In output (see program) one spots realistic cases with rational numbers having none or very few digits after decimal point for all variables.
I chose among those cases for following ratios: $$c=3*o/2$$ $$h=3*l$$ then we abbreviate:

d for doors
p for prize

and we have:

$$o=2*d$$ and $$c=3*d$$
$$l=p$$ and $$h=3*p$$

and (see program) one computes, given (I ignore factor 1000000):

$$d=6$$
$$p=9$$

outcomes:

$$o=12$$
$$l=9$$
$$c=18$$
$$h=27$$

$$k=10$$
$$s=12.4$$

$$s/k=1.24$$
$$s-k=2.4$$

Conversely to find solution:

given:
$$s/k=1.24$$ and $$s-k=2.4$$
one finds:
$$k=10$$ and $$s=12.4$$
and one (this is a bit tedious) rewrites: $$k=p*(3d+2)/3d$$ $$s=p*(3d^2+3d-2)/(3d(d-1))$$ giving ($$p$$ cancels): $$2.4=s/k=(3d^2+3d-2)/(3d^2-d-2)$$ and quadratic equation: $$9d^2-53d-6=0$$ having solution:
$$d=6$$
All other values are now easy to find next...

code:


using Printf

# probability of high prize when keeping choice
# pKH(c) = 1 / c
function probKeepHigh(closed)
return 1 // closed
end

# probability of low prize when keeping choice
# pKL(c) = 1 - 1 / c
function probKeepLow(closed)
return 1 - probKeepHigh(closed)
end

# probability of high prize when changing choice
# pCH(c,o) = pKL(c) / (c - (o + 1))
function probChangeHigh(closed, open)
return probKeepLow(closed) // (closed - (open + 1))
end

# probability of low prize when changing choice
# pCL(c,o) = 1 - pCH(c,o)
function probChangeLow(closed, open)
return 1 - probChangeHigh(closed, open)
end

# average prize when keeping choice
# aKP(c,h,l) = h * pKH(c) + l * pKL(c)
function avgKeepPrize(closed, high, low)
return high * probKeepHigh(closed) + low * probKeepLow(closed)
end

# average prize when changing choice
# aCP(c,o,h,l) = h * pCH(c,o) + l * pCL(c,o)
function avgChangePrize(closed, open, high, low)
return high * probChangeHigh(closed, open) + low * probChangeLow(closed, open)
end

for closed in (3 : 20)
for open in (1 : closed - 2)

for high in (1 : 30)
for low in (0 : high - 1)

# keep and change choice prize averages
keep = avgKeepPrize(closed, high, low)
change = avgChangePrize(closed, open, high, low)

@printf("closed ") ; print(closed)
@printf(" open ") ; print(open)
@printf(" high ") ; print(high)
@printf(" low ") ; print(low)
@printf("\n")

@printf("keep ") ;  print(keep)
@printf(" change ") ; print(change)
@printf(" (change / keep) ") ; print(change // keep)
@printf(" (change - keep) ") ; print(change - keep)
@printf("\n")

@printf("%.11f %.11f %.11f %.11f\n",
keep, change, change // keep, change - keep)
@printf("\n")

end
end

end
end


• fwiw: The Monty Hall story is known for triggering controversy, but, when told unambiguously, it is only an exercise in probability theory. Attempting to convince myself I got the values and formulas right for more general cases, I spotted some nice figures leading to the puzzle. Have a nice day! Dec 29, 2021 at 16:31