# Thread: Simple puzzles to stump people

1. Originally Posted by Santtu
As much as I'd like to join you and give credit to people not knowing mathematics, it's impossible to do this way. The situation described earlier already reflected reality accurately. With you presenting entirely other framework, one with your premises - of course the results may change. It's ok to present premises, ok, of course. I can't agree that the multi-round Monty hall problem would "naturally" and realistically extend to the rules and dynamics you said. As I've read from the subject, the host (and the game show) really actually keeps the rules the same. Now why is that? Now opening such possibilities we go very far outside mathematics, so this is not certain anymore. But, in short, producers may just decide on a program format by some criteria and to change that format by other criteria. What's been observed (as I understand from reading some sources) that the host does not try to decieve the guest, and there is not a meta-game involved in trying to interpret the everchanging rules.

Apparently the program has been presented with great enough similarity each time, so that it has gained the audience's trust that the rules will stay the same.

Now that in my interpretation the facts dont support your postulation, I welcome you to speculate. However such non-supported speculation may not be entirely relevant to the scenario we discussed before, not relevant enough as to disprove it.

SOrry for long post.

Remember that you postulated the whole thing in trying to make something that you wanted?
Let me reiterate what I have already said. If you consider that this game is one of a series of games (i.e. this game is played daily), then the outcome is different then if you consider that the game only happens once. If you consider this scenario as one of a series then it becomes a question of economics more than purely math. Economically you should assume that the show wants to maximize profits and therefore minimize costs. If it is possible for the chance of receiving a car to be 1/2, then the show will make the probability 1/2.

As I stated before it is possible to make the probability of winning by switching to be 1/2 if Monty sometimes opens up all the doors without giving the contestant a chance to switch. And if you've watched "Let's Make a Deal" you'll see that he does in fact do this. Sometimes the contestant is given a chance to switch and sometimes he reveals what they get after just one pick. The show very much has an air of unpredictability. You don't really know what Monty is going to do next.

The only real way to know if the producers are playing some metagame or not is to actually observe how often the game is played in this fashion and record if the observed probability is closer to 2/3 or 1/2 (or perhaps it is close to neither). If the observed probability is close to 2/3 then we can assume there is no metagame, otherwise we must assume that the producers are somehow affecting the probabilities of the game.

2. Originally Posted by The_Liquid_Laser
My point is basically this. A person who is not mathematically trained will intuitively say that the probability is 1/2. A person who is properly mathematically trained will say the probability is 2/3. However the math problem is actually an artificial construct. In reality the probability is closer to 1/2. [B]Logic is more useful when applied to abstract problems, but intuition is actually more acurate when it comes to what actually happens in reality. Sometimes learning can "untrain" what would normally be good common sense.
Tell that to those who have been wrongfully convicted in criminal trials on "common sense", i.e. various forms of the base rate fallacy. (Link)

3. Originally Posted by The_Liquid_Laser
Let me reiterate what I have already said. If you consider that this game is one of a series of games (i.e. this game is played daily), then the outcome is different then if you consider that the game only happens once.
I agree with different winning strategies for onetime and repeated games. I guess I confused what you ment to be the strategy for the onetime vs. repeated game. We would have to know what effect different outcomes would have on the audience, i.e. would the popularity decline or improve depending of certain winning ratioes, etc. We can either try to model themselves or accept that the producers go by their own model.

I perhaps wrongly assumed that this is a co-operative game, when it's actually a zero-sum game. It's only co-operative outside the stated game mechanics, i.e. how favourable an impression the host makes vs how much money it costs to run the show.

Some non-zero sum games have proven best outcomes for co-operative play when it's uncertain (in a certain probability range) whether there will be a new round. The stability of such solutions may or may not fullfill various game-theoretic criteria.

Interestingly, if a round is known to be a last round, with some game types and payoff matrixes, the winning strategy is to go for personal profit.

Originally Posted by The_Liquid_Laser
As I stated before it is possible to make the probability of winning by switching to be 1/2 if Monty sometimes opens up all the doors without giving the contestant a chance to switch. And if you've watched "Let's Make a Deal" you'll see that he does in fact do this.
I haven't watched the show and this fact didn't come up in my searches. This does put a different spin on the game. I have taken the description of "Monty Hall Problem" as an accurate description of the game and I feel like a fool now :P

Originally Posted by The_Liquid_Laser
The only real way to know if the producers are playing some metagame or not is to actually observe how often the game is played in this fashion and record if the observed probability is closer to 2/3 or 1/2 (or perhaps it is close to neither). If the observed probability is close to 2/3 then we can assume there is no metagame, otherwise we must assume that the producers are somehow affecting the probabilities of the game.
True.

4. While not a replacement at all for what Ygolo posted, here's the logic behind the "children" questions.

The tree for the 3/4 problem (at least one girl, at least one boy, but can be used for similar problems);

This is how we think of the problem. However, it's flawed when conditions are added (here is the 2/3 problem

The red line denotes what isn't possible given the rules of the problem. That makes only three possibilities that are allowed, with two being accurate.

The way the mind tends to think of the tree, however, is like this;

Which is why we instinctively think of it as 1/2. However, the assumption of the boy first is incorrect.

Here's another way of looking at it, if this doesn't make sense. Lets say that you pick 1000 couples that have two kids. Of that, there should be an even split of 4 (BB, BG, GB, GG). But then you say - well, I know this couple that has a boy. So you go from;

to

In other words, you have changed your sample group.

(Forgive the drawings, I only have excel to work with here )

5. What I don't understand are why BG and GB are considered separate possibilities. They are both one boy and one girl, just the order is different.

6. Originally Posted by athenian200
What I don't understand are why BG and GB are considered separate possibilities. They are both one boy and one girl, just the order is different.
Think of it as this tree;

The x at the top is "Sex" between the couple - from this can either come a boy, or a girl. Up to here it makes sense that there are two different outcomes.

Now, inside each outcome, repeat sex - another B or G (another fork). That makes the BG the result of one outcome and GB the result of another. Both can be viewed as "one boy and one girl", but both are not "a boy, then a girl". When calculating how many possible outcomes, you have to measure all of the different combinations something can happen (a boy, then a girl + a girl, then a boy = a boy and girl).

Does that make sense?

7. Originally Posted by athenian200
What I don't understand are why BG and GB are considered separate possibilities. They are both one boy and one girl, just the order is different.
There are lots of choices. If you choose BG and GB to be the same, you won't have a uniform probability distribution. You're derivation would go a little different.

Keep in mind that you are deriving this distribution based on what is given in the problem (The probability of having a boy being 50% and a girl 50%. Hermaphrodites and others are 0%, if you were to consider them)

8. Originally Posted by ptgatsby
Think of it as this tree;

The x at the top is "Sex" between the couple - from this can either come a boy, or a girl. Up to here it makes sense that there are two different outcomes.

Now, inside each outcome, repeat sex - another B or G (another fork). That makes the BG the result of one outcome and GB the result of another. Both can be viewed as "one boy and one girl", but both are not "a boy, then a girl". When calculating how many possible outcomes, you have to measure all of the different combinations something can happen (a boy, then a girl + a girl, then a boy = a boy and girl).

Does that make sense?
Yes, actually. It's sounds like you're just looking at it purely mathematically. I don't see any tangible difference between the two, but... can you actually measure this in real life the way you could the Monty Hall thing?

9. Originally Posted by athenian200
Yes, actually. It's sounds like you're just looking at it purely mathematically. I don't see any tangible difference between the two, but... can you actually measure this in real life the way you could the Monty Hall thing?
Well... that is a real life situation, really Forget all about the problems, this is the literal representation of what could happen. Like ygolo says, this is about distribution.

Tell me this:

What is the chance of having a boy and a girl in a family? Is it the same as having two girls or two boys?

10. Originally Posted by athenian200
Yes, actually. It's sounds like you're just looking at it purely mathematically. I don't see any tangible difference between the two, but... can you actually measure this in real life the way you could the Monty Hall thing?
I am not sure what the real life probabilities are for having a boy or a girl, but if it is 50&#37;/50% roughly and if the gender of the first an second child are truly independent (part of my experience tells me the some family are prone to girls and other to boys, but without looking at demographics it's just a blind guess), then it will come out that way in real life.

But you can "simulate" the problem as follows. Find a "fair" coin (flip the coin a bunch of times to see if the odds are 50%/50%). Now let tails represent a boy, heads represent a girl.

The experiment is as follows: Flip the coin twice, record the outcomes. The possible outcomes are HH, HT, TH, TT. (Or if you want you could map the outcomes to "both heads", "both tails", or "on head one tail".)

Repeat the experiment many many times (say a 100 times). Compute the percent of times you get HH, HT, TH, and TT each separately. You should get that the percent of times that you get each of the outcomes is about 25%. If you repeat the experiment a 1 thousand times, you should be even closer to 25% each.

EDIT: I forgot you knew how to code. You could probably run a long computer simulation using a random number generator if you wanted.

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