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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 M

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?

DanielStraight 13y ago
This could be the most bothersome problem of all: http://www.airplaneonatreadmill.com/ http://www.airplaneonatreadmill.com/
ksmiley 13y ago
I know of one other mathematical fact that matches the Monty Hall problem in the loudness of its opponents: 0.999(...) exactly equals one. Despite a number of proofs existing, you can still find ten large pages of arguments on Wikipedia's talk page. http://en.wikipedia.org/wiki/Talk:0.999.../Arguments http://en.wikipedia.org/wiki/Talk:0.999.../Arguments
tomgagnier 13y ago
This problem is known as "Restricted Choice" in the card game bridge. It occurs when deciding whether to play for the drop of a missing honor card or finesse. See http://en.wikipedia.org/wiki/Principle_of_restricted_choice http://en.wikipedia.org/wiki/Principle_of_restricted_choice for more.
drblast 13y ago
The two envelopes paradox is a good one: http://consc.net/papers/envelope.html 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. Question: Is this game fair? If not, is it possible to give either player a per-turn stipend to make the game fair? If so, how much is that amount?
- tomca32 13y ago
  I'll bite. 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...
  - drblast 13y ago
    Here's a hint; is there a strategy one player can play such that no matter what the other player does, the first will win money on average?
jcampbell1 13y ago
I like this question: 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?
- deleted 13y ago
  [deleted]
- bernardom 13y ago
  Wait, is there an interesting answer to this? 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.)
- mildavw 13y ago
  Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly. Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/ 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.') 2/3 of yeses are honest and 1/3 didn't have to answer the question!
  - mildavw 13y ago
    Oops. That's completely wrong. It's the case where every honest answer is yes! 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."
basch 13y ago
It bothers everyone because the teacher never _stresses_ the fact that the host doesn't open a door at random. It's purposely deceptive.

DanielStraight 13y ago

This could be the most bothersome problem of all: http://www.airplaneonatreadmill.com/ http://www.airplaneonatreadmill.com/

ksmiley 13y ago

I know of one other mathematical fact that matches the Monty Hall problem in the loudness of its opponents: 0.999(...) exactly equals one. Despite a number of proofs existing, you can still find ten large pages of arguments on Wikipedia's talk page. http://en.wikipedia.org/wiki/Talk:0.999.../Arguments http://en.wikipedia.org/wiki/Talk:0.999.../Arguments

tomgagnier 13y ago

This problem is known as "Restricted Choice" in the card game bridge. It occurs when deciding whether to play for the drop of a missing honor card or finesse. See http://en.wikipedia.org/wiki/Principle_of_restricted_choice http://en.wikipedia.org/wiki/Principle_of_restricted_choice for more.

drblast 13y ago

The two envelopes paradox is a good one: http://consc.net/papers/envelope.html 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. Question: Is this game fair? If not, is it possible to give either player a per-turn stipend to make the game fair? If so, how much is that amount?

tomca32 13y ago
I'll bite. 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...
- drblast 13y ago
  Here's a hint; is there a strategy one player can play such that no matter what the other player does, the first will win money on average?

jcampbell1 13y ago

I like this question: 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?

deleted 13y ago
[deleted]
bernardom 13y ago
Wait, is there an interesting answer to this? 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.)
mildavw 13y ago
Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly. Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/ 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.') 2/3 of yeses are honest and 1/3 didn't have to answer the question!
- mildavw 13y ago
  Oops. That's completely wrong. It's the case where every honest answer is yes! 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."

basch 13y ago

It bothers everyone because the teacher never _stresses_ the fact that the host doesn't open a door at random. It's purposely deceptive.