Non-Poker Probability Questions

[quinn]

There have been some questions that have been seriously bothering me lately, and I was hoping some math or stat experts could answer them. Please justify your answers mathematically if you can.

1.
Say you are taking a multiple-choice test. There are three possible answers, (A, B, C). You have no clue what the answer is, so you take a guess. Let's say you guess "A". Then, you are told that the answer is NOT C. Now you are given the opportunity to either stick with A, or switch to B. Do you switch? Why? Or does it not matter whether you switch or not?

2.
You roll two dice without looking at them when they land. However, someone else saw one of the dice. They tell you that it landed "1". Now, what are the chances that the other die is a "1" as well?

3. A woman has two babies, and one is a boy. What are the chances that both are boys?

[jimdmcevoy]

1. A and B are equally likely to be correct.

2. 1/6

3. This depends how you got the information. For instance, if you asked the woman, "Do you have at least one boy?" and she said yes, then the chance that the other baby is a boy is 1/3. However, if you asked the woman "What sex is your yougest of the two babies?" and she said male, then there is a 1/2 chance the other baby is a boy.

[jimdmcevoy]

I just realized I gave no explanation for anything I said.

Well this is my reasoning:

I think of all the possibilities and and their corresponding probabilities. Then with the new information I figure out which possibilities are no longer possible.

Now you have to raise the probabilities of the other possibilities to get a total probability of 1 when you add them all together, but you must keep all the probabilites proportional to each other in the same way as they were before you had the new information.

My explanation sucks bad, hopefully someone else can explain it much clearer.

[gaming_mouse]

[ QUOTE ]
Say you are taking a multiple-choice test. There are three possible answers, (A, B, C). You have no clue what the answer is, so you take a guess. Let's say you guess "A". Then, you are told that the answer is NOT C. Now you are given the opportunity to either stick with A, or switch to B. Do you switch? Why? Or does it not matter whether you switch or not?

[/ QUOTE ]

This is the Monte Hall problem. It is always best to switch. Switching gives a 2/3 chance of being correct, while sticking gives only 1/3. For an explanation, search for "Monte Hall" in this forum. It has been explained many, many times.

[ QUOTE ]

You roll two dice without looking at them when they land. However, someone else saw one of the dice. They tell you that it landed "1". Now, what are the chances that the other die is a "1" as well?

[/ QUOTE ]

If the person only saw 1 of the die, then the answer is 1/6, b/c the dice are independent. However, the puzzle you are probably thinking of is when the person sees both dice, but says, "one of the dice landed on 1."

Now let:

A=chance at least 1 die lands on 1.
B=chance that both dice land on 1.

We want P(B|A) = P(A&B)/P(A)

P(A&B) = 1/36
P(A) = 1 -(5/6)^2 = 11/36

The answer now is 1/11.

[ QUOTE ]
3. A woman has two babies, and one is a boy. What are the chances that both are boys?

[/ QUOTE ]

This is a well known riddle, similar to the question I just solved in 2. I'll do it this time without formulas. There are 4 equally likely pairings of sexes for the children:

BB
BG
GB
GG

The info given rules out GG. The chance that both are boys is therefore 1 in 3.

gm

[rwesty]

1. Switch, Monte Hall.
2. 1/6, Independent.
3. 0, Question states that only one is a boy.

[jimdmcevoy]

Woops, seems I answered the first one wrong, maybe.

Did you guess A and not tell anyone and then you were told that C is incorrect? I had this mental picture of you taking a test in class and the teacher saying to the whole class that C is incorrect after you have chosen A...

[quinn]

Well, thanks for the answers.

Now my question is this: Why is my intuition failing me?
It tells me this:
A woman has two babies. One of them is a boy and we don't know the gender of the other one. So what? You have another random kid. It's going to be 1/2.
..
But that's wrong. I must be leaving something out in my line of thought.

[gaming_mouse]

[ QUOTE ]
Now my question is this: Why is my intuition failing me?
It tells me this:
A woman has two babies. One of them is a boy and we don't know the gender of the other one. So what? You have another random kid. It's going to be 1/2.

[/ QUOTE ]

Of course that is most people's intuition, which is why this is a well-known puzzle.

The thing wrong with your reasoning is that there is no "other" kid. You don't know the gender of one, and not of the other. You only know that of the two, at least one is a boy.

gm

[pzhon]

[ QUOTE ]

1.
Say you are taking a multiple-choice test. There are three possible answers, (A, B, C). You have no clue what the answer is, so you take a guess. Let's say you guess "A". Then, you are told that the answer is NOT C. Now you are given the opportunity to either stick with A, or switch to B. Do you switch? Why? Or does it not matter whether you switch or not?


[/ QUOTE ]
Whether to switch or not depends on the procedure followed by the person who told you that C is not correct. This is also true of the Monthy Hall problem.

If the person will only give you the chance to switch when you are right, then you should never switch.

If the person will always reveal an incorrect answer and give you the chance to switch whether you are right or wrong, you should switch to increase your probability from 1/3 to 2/3.

If it will always be announced whether answer C is correct or not, then it doesn't matter whether you switch or not.

See the rec.puzzles archive answer (http://rec-puzzles.org/sol.pl/decision/monty.hall) for the Monty Hall problem.