The Monty Hall problem...

Skip down to see the new problem, this section serves to restate the original Monty Hall problem and its solution.

To recap: There exists a game show with a host ("Monty Hall") and a contestant. There are 3 closed doors and it is known that behind one of the doors there is a prize (represented by a car) and behind each of the two others is a single goat. The contestant is asked to choose a door, without opening it. The host then opens one of the other two doors and a goat is always revealed (either by design or by accident) - the door containing the car is never opened, at this stage. The host then offers the contestant the chance to stick with his original choice or switch to the other closed door. The door the contestant finally settles on is opened and if the car is behind that door, she gets to keep it.

The question is: should the contestant stick or switch?

It turns out that the correct answer to this problem is that the contestant should switch because 2/3 of the time, the car will be behind the other door.

There are many explanations about why this is true, but the one I like is the following. When the contestant makes the original choice, the original selection has 1/3 a chance of being the correct one (all other things being equal). This means that the group of the other 2 doors had a 2/3 chance of containing the door with the car. This remains true when one of the doors in that group is opened - that group of 2 doors originally had a chance of 2/3 of containing the car and it still does, but now we know that one of those 2 doors contains a goat and we know which of those 2 doors that is. So the total probability of the group of 2 doors hiding a car is unchanged, but it can't be the one with the goat, which means that the remaining door must have a 2/3 chance of being the one with the car - hence the contestant should switch.

...with the twist

So that's the original Monty Hall problem. Here it is again with a twist.

We are going take the original problem and make some additional stipulations:

• the host knows which door conceals the car
• if the host has a choice of action, then his contract with the TV network states that he must choose the choice that minimises the chance that the contestant will win the prize
• after the host opens the first door containing a goat, the host MAY choose to explain the logic of the solution to the original Monty Hall problem to the contestant. His choice must be guided by the second stipulation
• the host is unaware if the contestant already knows the correct solution to the Monty Hall problem
• the contestant is aware of these stipulations, even if she does not know what the Monty Hall problem is or what its correct solution is.

In the case that the host chooses not to explain the logic of Monty Hall problem to a contestant who happens to know the correct solution to the Monty Hall problem anyway, should the contestant stick or switch?

Process A selects a daughter at random from all the daughters of 2-child families, then pairs that daughter with her mother. It can be shown that observations generated by this process describe girl-girl families with a probability of 1/2.

Process B selects a mother at random from all 2-child families with at least one girl and then pairs the mother with one of her daughters. In boy-girl families, this is the only daughter. In girl-girl families a coin-flip is used to select one of the two daughters. It can be shown that observations generated by this process describe girl-girl families with a probability of 1/3.

None of this is controversial - different randomisation processes result in observations with different probability distributions.

Consider now Process C. Process C has two inputs, one from Process A and one from Process B. The function of Process C is to match observations from Process A with observations from Process B and, if a match is detected, to emit a new observation. Otherwise, nothing is emitted.

Q: What is the probability that observations emitted by Process C describe a girl-girl family?

Each of Jerry's juggling balls is made of two skins. Some juggling balls are red and blue - consisting of a red skin and a blue skin; some are pure red - consisting of two red skins. Jerry doesn't sell blue balls - some say it was because those words ("Blue Balls!") were the last two words that Florida yelled as she stormed out of his apartment and then his life, never to be seen again. He still doesn't know what she meant by it.

Half the blue skins are stamped with the name "Feynman", half with the name "Schrôdinger". Only 1 in every 1000 red skins are stamped with "Florida!".

(Jerry didn't stamp every red skin with "Florida!" because he was trying to let go. Indeed he would have stamped only 1 in every 1,000,000 because, after all, she was one in a million, but he didn't think he was going to sell that many balls)

Jerry's juggling balls are sold in sets of three. The balls come in a cloth bag with a string tie in the neck. Every set of three contains a Feynman, a Schrôdinger and a red ball (consisting of two red skins).

Sonia receives a set of Jerry's juggling balls for her birthday from her older brother, Scott, who is an amateur statistician.

Neither Scott nor Sonia have seen the juggling balls and neither knows, or has any reason to believe, that any of the balls have a red-skin stamped "Florida!".

Scott asks Sonia to put a blindfold on and then asks her to select a ball from the bag. Sonia reaches into the bag and pulls out a ball. Scott tells Sonia that the ball has at least one red skin.

Q1. What, at this point, is the probability that the ball is a red ball?

Scott then tells Sonia that the phrase "Florida!" is printed on the ball.

Q2. What, now, is the probability that the ball is a red ball?

Sonia is excited to have received a set of Jerry's Juggling Balls with a rare juggling ball stamped with "Florida!".

During Sonia's birthday dinner, Scott tells his family some big news. He has met a girl who has a single sibling.

Q3. What is the probability that Scott's girl friend's sibling is also a female?

Scott then reveals that his girl friend's name is Florida.

Q4. What is the probability that Florida's sibling is also a female?

Sonia says: "That's funny one of my teacher's two children is a girl named Florida!"

Q5. What is the probability that Sonia's teacher's other child is a girl?

Sonia then corrects herself: "Actually, that was my old teacher. My current teacher also has a daughter and one other child, but neither of them is called Florida"

Q6. What is the probability that Sonia's current teacher's other child is a girl?

When the families arrive all the boys, all the fathers and the mothers of the boy-boy families go to watch the boys run around outside.

This leaves 100 girls and their 75 mothers inside the hall to play games. They then decide to run two raffles - one for the mums, one for the girls. The rules are the same for each: write your name and either BG or GG on a piece of paper depending on whether your family is a BG or GG family and then place the paper into a bucket - a red one for the mums and a blue one for the girls.

Q1. What is the chance that that a raffle drawn from the red (mum's) bucket will be won by a GG ticket (a GG family)?

A) 50%
B) 33%
C) Something else

Q2. What is the chance that that a raffle drawn from the blue (daughter's) bucket will be won by a GG ticket (a member of a GG family)?

A) 50%
B) 33%
C) Something else

What this example illustrates is that one needs to be careful about identifying what one is counting when judging probabilities in "A Girl Named Florida" problem and ones like it - are you trying to count groups (families) or members of groups (children)?

So, in this example, and considering only the 75 families in the hall, there is a 50% chance that a girl will be member of a family which has 33% likelihood of being a two girl family and this is not a contradiction.