Monty Hall With A Twist
On a game show, there is a popular game that is always carried out in the same way. A contestant is given the choice of three doors: behind one door is a car and behind the other two doors are goats. The contestant picks a door, say #1, and the host, who knows what's behind each door, always opens another door, say #3, to reveal a goat. Then, the contestant is given the option to switch to the other unopened door, door #2 in this case.
A long-term fan of the game show has noticed a hint in the staging of the game by the game show host. Thus, this fan can correctly guess the door with the car behind 50% of the time before any door selection is made.
Now, this fan has been selected as a contestant for the game. Using the best possible strategy, what is the probability (as a percentage) that this fan will end up with the car prize?
The answer is 75.
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2 solutions
Most people make too much of the traditional Monty Hall Problem (MHP)
In reality the host is actually offering a two-for-one switch. As a stage performer he uses deception to confuse observers and to disguise the true terms of the game. His deception is in two parts - first he reveals a goat that a casual observer can easily predict because there is always at least one goat between any two doors - the second part of the deception is to make the switch offer to just one door.
The contestant enjoys a two-third chance by switching from the sum of two one-third chance doors - the goat door and the door not selected. All doors carry the chance of 1/3 the prize irrespective of content - a revealed goat does not negate the 1/3 chance of the car. For instance - all accept that each closed door has a 1/3 chance of the car and yet two of those doors will hide a goat. The 1/3 chance of the car is permanently fixed to each door.
As for the use of Bayes theorem to explain the MHP then either Mr Bayes was wrong - which I seriously doubt - or the application of it is flawed because it results in assigning a 2/3 chance of the car to a single door? It takes two cars for a single door to have a 2/3 chance of a car - think goats.
It's the switching contestant who has the 2/3 chance of the prize not a door.
Here is the simple MHP equation:
A minus B equals C
where A = three 1/3 chances B= One 1/3 chance - (the original chosen door) C = Two 1/3 chances (the other two doors)
The application of Bayes is overkill...
There are only two possible outcomes when selecting a door: the door with the car behind it and the door with a goat.
From the 50% chance of guessing the right door we know that the contestant can select the right door 1 out of 2 times.
If he guesses the wrong door, he has the chance to switch and again there is a 50% chance of selecting the right door when he switches.
P ( w i n ) = P ( g u e s s ) + ( 1 − P ( g u e s s ) ) × P ( s w i t c h )
P ( w i n ) = 2 1 + 2 1 × P ( s w i t c h )
⇒ P ( w i n ) = 2 1 + 2 1 × 2 1 = 4 3
⇒ % P ( w i n ) = 7 5
N o t e :
We can generalise this as:
% P ( w i n ) = 5 0 × ( 1 + P ( g u e s s ) )
Good guy Monty Hall, he gave two guessing opportunities to contestants with atleast 50% chance of winning in both and more than 50% overall chance of winning.
The contestant would deliberately select a door that he thinks does not have the car and then always switch when he gets the option.
There is a 50% change that the contestant knows where the car is and a 50 % chance it will be behind one of the other 2 doors. We have no reason to prefer one of the other doors over the other. The probability the car is behind either one of the other two doors 25% each.
For example; The contestant thinks the car is behind door #1 (50% chance), door #2 (25% chance), door #3 (25% chance). The contestant selects door #2. At this point there is a 25% chance the car is behind the door he selected (#2) and a 75% chance it is behind one of the other doors (#1 or #3). The game show host showing that one of the doors in the #1, #3 group is not the winner tells us nothing that was not already known. The contestant always knew that in any event at least one of the #1, #3 group was not a winner. It is not possible for both doors #1 and #3 to be winners there is only one winning door. The contestant can completely ignore that the host has shown that one of these doors is not a winner. The constant is given the chance to switch from a door with a 25% chance of success to a group with a 75% change of success he should switch.