Like everyone who encountered this in their intro stats class, I love this problem. And it bothers everyone the first time.
Are there other problems like the Monty Hall problem, perhaps in other disciplines?
Are there other problems like the Monty Hall problem, perhaps in other disciplines?
http://consc.net/papers/envelope.html
And another that I've figured out the "right" answer to but think there might be more to it is this:
Suppose I set up a two-player game as follows:
1. One player is the "odd" player, the other is
the "even" player.
2. Each player may choose a single number per turn,
either one or two.
3. The result of each turn is the sum of the two players'
choices. For example, if the even player picks 1 and the
odd player picks 2, the result of that turn is 3.
4. If the result is odd, the even player pays the odd
player that amount, and vice-versa. In our example
result of 3, the even player would pay the odd player 3
dollars.
I guess that the odd player can always choose 1. In that case he can either win 3 or lose 2. Making that the favorable choice for the odd player.
Even player could always choose 2 in which case he can either win 4 or lose 3. Making this the favorable choice for the even player. However, if both players play their favorable choice, odd player always wins: 2 + 1 = 3.
I'm not sure if I got this right, but that's my thinking so far...
Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Have you ever cheated on your spouse?"
Now, what is the probability the person is a cheater given they answered yes?
Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/
var runs = 1000,
honest_admissions = 0,
automatic_admissions = 0;
function flip() {
return Math.random() < 0.5 ? 'h' : 't';
}
for (var i = 0; i < runs; i++) {
if (flip() == 't') {
honest_admissions++;
} else if (flip() == 'h') {
automatic_admissions++;
}
}
document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.')
I think this is indeterminate. You have to know the ratio of total yesses to total participants.
The real world use-case I was recalling is simpler: flip a coin and answer honestly if it's heads and "yes" if it's tails. Any individual response is non-incriminating, but if N is large enough where you can assume N/2 got heads, you can know the number of "honest cheaters."
I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse.
If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%).
If I flip tails, I answer 2Y (X) or 2N (1-X).
So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X)
And combined, Y = 0.25+0.5X
What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)