Counter-intuitive statistics puzzles

  • Thread starter Thread starter NATURE.M
  • Start date Start date
  • Tags Tags
    Statistics
AI Thread Summary
The discussion revolves around the Monty Hall Problem and other counter-intuitive statistics puzzles. A key example presented is a drug test scenario with 99% sensitivity and specificity, illustrating that the probability of a positive result indicating actual drug use is significantly lower than expected, depending on the prevalence of drug users in the population. Another puzzle involves the gender distribution of children in a hypothetical society where families continue having children until they have a boy, raising questions about the eventual ratios of boys to girls. Additional puzzles include determining the probability of one child being a girl given that one child is a boy, and the impact of the order of matches in a tennis award scenario where winning two consecutive matches is required. Participants express interest in exploring these puzzles further, highlighting their complexity and the deceptive nature of initial intuitions.
NATURE.M
Messages
298
Reaction score
0
Recently, I came upon the Monty Hall Problem, and found it to be quite interesting. Any suggestions of similar counter-intuitive statistics puzzles would be much appreciated.
 
Physics news on Phys.org
wiki said:
Suppose a drug test is 99% sensitive and 99% specific. That is, the test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. If a randomly selected individual tests positive, what is the probability he or she is a user?

The answer is not close to 99% as one might think. In fact, it depends on how many other people are drug user? If only 0.5% of people in the group was a user, then even if someone tests positive, the chance of that person being a user is only around 33%.
Details here
 
Puzzle 1:
In the land of Tyrannia, parents want to have boys rather than girls. So the rules of Tyrannia state that everybody who gets a girl, can have another child, until they have a boy. So for example, a couple can get a boy right away, and is then not allowed another child. But a couple can get 10 girls and then a boy, and then is not allowed another child.

What is the eventual percentage of boys and girls?

Puzzle 2:
A couple has two children. One of the children is a boy. What is the probability that the other is a girl.

A couple has two children. The oldest of the children is a boy. What is the probability that the other is a girl.

Are these two probabilities the same?

Other things to review: http://en.wikipedia.org/wiki/Simpson's_paradox
 
How many random people do you need to gather in room in order for the probability that two have the same birthday to be 90% or higher?
 
In order to get a certain tennis award, you need to play three matches. You can play against an easy player or against a very tough player. But you can't play the same player in a row.

So you have the following two options:
Match 1: Play against the easy player
Match 2: Play against the hard player
Match 3: Play against the easy player

or

Match 1: Play against the hard player
Match 2: Play against the easy player
Match 3: Play against the hard player

To get the award, you need to win two matches in a row. Which schedule do you choose?
 
Last edited:
You can play the following game:

You first pay a certain sum as entry fee, namely x dollars.

Then you flip a coin. You keep flipping a coin until you hit head. You count the amount of tails you got, call this ##n##. You get paid ##2^n## dollars.

For example, if you throw 3 tails in a row, then you get paid ##8## dollars.

How large should the entry fee in order for you to play this game? What if you play this game several times?
 
Thanks a lot for the suggestions micromass and I_am_learning. The puzzles defiantly seem challenging upon first glance. I'll go through each one in more depth tomorrow, as I'm awfully tired at the moment.
 
Going forward nature, you must provide links to the puzzles.
 
micromass said:
In order to get a certain tennis award, you need to play three matches. You can play against an easy player or against a very tough player. But you can't play the same player in a row.

So you have the following two options:
Match 1: Play against the easy player
Match 2: Play against the hard player
Match 3: Play against the easy player

or

Match 1: Play against the hard player
Match 2: Play against the easy player
Match 3: Play against the hard player

To get the award, you need to win two matches in a row. Which schedule do you choose?

2 makes sense. Given that you're "very likely" to beat the easy player, you're given two chances to beat the hard player vs only 1 chance.
 
  • #10
johnqwertyful said:
2 makes sense. Given that you're "very likely" to beat the easy player, you're given two chances to beat the hard player vs only 1 chance.

Yeah after thinking over that puzzle thoroughly, the answer seemed rather trivial, although the appearance can be somewhat deceiving.
 
Back
Top