Math Fun

You got it Sil. Here’s a bit more to chew on maybe:
[tab]As you said, James Hadn’t considered the possibility of Rout Min. What James actually proved is that, if Rout, then Min necessarily. Because Rout AND Mout produces Pout AND Pin, as he showed.

The misunderstanding James had was to think that A → B and A → ¬B is a contradictory scenario. B and ¬B is contradictory, but A → B and A → ¬B is not contradictory at all (not unless you’re also saying A). You can reduce logical implication thusly:
A → B is the same as saying B v ¬A
So, (A → B and A → ¬B) is reducible to (B v ¬A) ^ (¬B v ¬A). The way to make that conjunction true, is of course to make ¬A true. No contradiction at all, it just requires A not to be true.

So, A → B and A → ¬B isn’t a contradiction, it’s just a disproof of A.[/tab]

[tab]Some more that I was just thinking:
James’ thought process was obviously that A → B and A → ¬B is a contradiction.
To be a true contradiction, it would have to be (A → B) and ¬(A → B), and so he may have thought that ¬(A → B) = (A → ¬B)
Let’s look at what ¬(A → B) really equals:

As we said, A → B = B v ¬A
¬(A → B) = ¬(B v ¬A) = ¬B ^ A

A → ¬B = ¬B v A

¬B v A =/= ¬B ^ A (although they do look similar)
So ¬(A → B) =/= (A → ¬B)[/tab]

Sleeping Beauty volunteers to undergo the following experiment. On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.

Each interview consists of one question, “What is your credence now for the proposition that our coin landed heads?”

[hint: A viable answer is to remove the abstract term ‘credence’ and turn the situation into something more…tangible. If you so wish.]

[tab]When she’s interviewed, there’s a .333[bar] chance that it’s Monday and the coin has landed heads, a .333[bar] chance that it’s Monday and the coin has landed tails, and a .333[bar] chance that it’s Tuesday and the coin has landed tails. This means that her credence for the proposition that the coin landed heads is logically one third full each interview.[/tab]

For the following:

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

Insert any mathematical operators (except ones that involve writing a number) into the left hand side of each of the above equations, such that they become correct.
No manipulation of any number shapes or outside-of-the-box thinking is required to do this. Use of brackets/parentheses is encouraged for clarity.

Still missing:
8 8 8=6
9 9 9=6

The rest:
tab!=6
(1+1+1)!=6
2+2+2=6
(3x3)-3=6
(4-(4/4))!=6
5+(5/5)=6
6+6-6=6
7-(7/7)=6[/tab]

I think I need to incorporate another operation…

Good effort! :slight_smile:

8s are the hardest, btw.

[tab]Good job thinking of using factorials - that was the least obvious one to me.

You’re right about needing to incorporate another operation to finish things off.[/tab]
Don’t open the next tab if you want to see if you can do it all without any help…
[tab]Here’s a clue, squirt. Another operation would neaten up the 4s a little while you’re at it.[/tab]

I figured out a cheat-y way that solves all of them:
[tab]( ( d/dx(8) )! + ( d/dx(8) )! + ( d/dx(8) )! )!

That works no matter what number you use, because the derivative of any constant is 0, and the factorial of 0 is 1. It’s a bit of a cheat because d/dx is arguably adding another number (dx).

It also makes the problem boring.

Still no legit solution to 8 and 9.[/tab]

Haha!

Wow, it’s amazing the creativity that difficult problems can drive us to.
The function you’re looking for really is fairly elementary.
Another hint, should you need it:
[tab]Start with the 9s. Think about what you can do to a number like that, especially in light of a number I’ve already mentioned in my previous hint.[/tab]

Hmmm…
[tab]I think you’re suggesting taking the square root, which I wrote off because I saw it as incorporating another number: 2.

If that’s what you’re suggesting, 9 is easy: ( 9/sqrt(9) ) +sqrt(9) = 6
and 8 - sqrt( sqrt( 8+8 ) = 6
But why not cube root + cube root + cube root?
And if we’re doing that, why not x root of c + c + c, where x is whatever irrational number root results in 6?
Or log base y where y is a number that when raised to the 6th power equals c + c + c?

I don’t know why I treated the square root as adding a number, but not the factorial. As I think about it, I find the line between adding another number and just adding an operator to be fuzzy.
After all, you can always define a new operator. So we could define the unary operator B such that Bc=(c x 0) + 6.
B(c + c + c) = 6 for all c.[/tab]

[tab]I consider square root to have a hidden number as well, Carleas. It’s raising to the power of 1/2. Just because we happen to have a symbol that doens’t look like a number that means ‘raise to the power of 1/2’ doesn’t mean it’s not using a number. I mean, I could make a symbol that means Raise to the power of 0 and multiply by 2…right? I could just invent that symbol. Maybe it could be the dollar sign. 8£ + 8£ + 8£ = 6. BOOM! Realistically, that’s just as valid as using square root.[/tab]

The point to the puzzle was to stick to conventional mathematics as it currently is with its current symbols. Adding new creative “operators” isn’t acceptable.

[tab]Yes, I was suggesting that you use the square root, since it can be denoted by √ which does not require using a number. Contrast this with other roots such as the cubed root that requires using the number “3” however you present it. The square root may also be represented by raising a quantity to the power “1/2”, which involves using a number, but using the √ symbol remains within the designated rules. The ! symbol to denote “factorial” is fine for the same reason, even if its expansion would involve adding other numbers.

Perhaps I didn’t explain that rule sufficiently.

Also, only the integers provided at the beginning of the problem are the only ones allowed - so no “x root of c + c + c, where x is whatever irrational number root results in 6” or “log base y where y is a number that when raised to the 6th power equals c + c + c”. And no use-defined operators either! Lol. Only conventional mathematical ones - I should have been more explicit on that one too :stuck_out_tongue: Nice idea though. I’ve enjoyed all of your contributions to this problem, even though some of them did not fall inside the rules I was trying to communicate.

( 9/sqrt(9) ) +sqrt(9) = 6 and 8 - sqrt( sqrt( 8+8 ) = 6 are both lovely.

I would even say they’re better than the solutions I was expecting, which were:
(√9 x √9) - √9 = 6 and (√(8/8 + 8 ))! = 6

Though I do prefer √4 + √4 + √4 = 6 for the 4s.[/tab]
So Carleas has solved them all =D>

A shame nobody else contributed.

EDIT:
I got distracted writing this reply, as usual, so by the time I previewed it there were more contributions.
JSS, that’s correct.
FJ:
[tab]By that argument, all conventional mathematical operators would involve extra numbers if fully expanded. My bad for not explicitly stating exclusively conventional mathematical operators in the rules. Carleas has already experimented with user-defined operators in much the same way as you suggest, though maybe you resisted opening the tabs which show this.

Not to say both of your arguments are invalid - they’re good points. I just didn’t mean to imply that you could use them when I described the problem. Both square roots and factorials were supposed to be fine, along with addition, subtraction, multiplication and division. No numbers have to be used when using certain conventional mathematical operators for each of these.[/tab]

I came to this problem late but I will say I did like the way Carleas called differentials a cheat, I don’t think they were since they don’t use numbers per se but assume general solutions to the differences between one value and another with a related dependant value and rule to evaluate it. In fact every single one of the values could be solved by differentials, substitution rules etc, addition of one number is irrelevant if you use the right maths, in fact you can basically solve an value of summation in a differential without an operator other than d/dx, it’s kinda too easy though if you know A’ level plus maths, and you can see why it would not be allowed. Kudos to the solutions that weren’t “cheats”. :slight_smile:

I’ve got a ‘puzzle’ so to speak, but it’s pretty open ended.

I’ve got a data set: I’ve tested drug A and drug B on both women and men. For the sake of simplicity, we’ll just say that each subject either died or healed (just a binary measurement of the effect of the drug on each individual).

The percentage of men who were healed by drug A was higher than the percentage of men who were healed by drug B.
The percentage of women who were healed by drug A was higher than the percentage of women who were healed by drug B.

But, the percentage of people overall who were healed by drug B was higher than the percentage of people healed by drug A.

How is this possible? What does a data set have to look like to produce this result?

And no, it’s not a trick question – the answer is not anything like “Many of the participants were hermaphrodites or transvestites.” Just keep it simple: all participants are men or women.

Bonus points if you know the name of this mathematical anomaly.

I spent a while on this before I gave up and googled it, and learned that I had added some assumptions that were getting in my way.

This first entabment will only discuss assumptions, but I tab them because talking about them can sometimes suggest the answer; purists may want to avoid:
[tab]I had assumed that the treatment group sizes were the same, so that the number of individuals given each drug were the same, and thus that the number of men given drug A was the same as the number of men given drug B, and ditto for women. The trial sizes and compositions can be different.[/tab]

This next entabment will discuss the answer:
[tab]I’d never heard of Simpson’s paradox before this, so thanks for pointing it out. They have some interesting real-world examples in that article. Particularly interesting was the example of gender-biased admissions.

I’m having difficulty succinctly expressing what must be true of the data set for this to occur… I know that the compositions need to be reversed, so for example, more women participate in the drug A trial and more men participate in the drug B trial. And I intuitively grasp the “vector interpretation” given in the wikipedia article, especially the image:

But I need to think more on this to distil what’s going on. Every time I try to express the criteria that are necessary, it sounds like a tautological restatement of the question itself. I know the two trial groups must be of different sizes and ratios of men to women, and I think it is expressible as some relationship between the success group of one drug for one gender to the failure group of the other drug for the other gender.[/tab]

[tab]Carleas: That’s the one

I would explain it like this:
It’s about weighting (think weighted averages).

Women have a higher healing rate than men. A has a higher healing rate than B.
If the healing rate of Women taking B is higher than the healing rate of Men taking A, and A is disproportionately taken by men while B is disproportionately taken by women, this (with some more mathematical constraints) can allow for the Simpson’s Paradox to occur.

I’ve taken a screenshot of a journal article about it that you might find enlightening:

[/tab]

A new one circulating the net, similar to other problems that have been discussed here.

[tab]August 14

Edit; lol no, that can’t be it… has to be either June 18 or July 16

Edit2: July 16[/tab]

Yep.

[tab]I’m curious why you said it has to be June 18 or July 16. I eliminated May and June first and went from there, I’d be interested to know your reasoning.[/tab]