A math riddle for you. Feel free to submit your own.

[septimus]

@iamthewalrus_3

Ok, I’ve been thinking about the 124-card deck, and here’s something that’s interesting: 5! (the number of ways to arrange 5 cards) is 120, which is exactly how many cards are unknown in a 124-card deck case of this problem. Of course, the fifth card in our arrangement is hidden, but since we had it to begin with, maybe there’s a way to choose the last card so that it’s the only fixed point?.

I think you’ve demonstrated that a deck of size 124 is the absolute maximum for which an encoding guarantee is possible. It may seem surprising that we can succeed with 124.

I will not post a solution in the thread; for one thing I don’t know a succinct way to present a solution. I just now whipped together a 100-line program of dense C that implements and proves a solution (the encoder literally tries all 5! orderings until it finds one that works :smack:). I will e-mail that code to any C programmer on request or, when the Sultan CODE tag is fixed, post it here.

[Little_Nemo]

This is more of a math paradox. It’s based on a YouTube video.

There’s a “game”. The rules of the game are simple.

A person walks into a room and rolls a pair of standard dice. If the result is snake eyes (a pair of ones) the person is immediately killed and the game is over. If the person rolls anything else they can leave the room unharmed.

If the first person didn’t roll snake eyes, a new group is sent into the room. This time there are ten people in the group. One person rolls the pair of dice for the entire group with the same results; a pair of ones means everyone in the room is immediately killed and the game ends, any other result means everyone gets to leave unharmed and another group is sent in.

Every group is ten times larger than the previous group. Every group faces one dice roll, which effect everyone in the group. The game continues until one group is killed and then ends. Nobody ever goes back into the room a second time. Assume, for the sake of this post, that an infinite amount of people are available and the room can hold all of them.

So if a person plays this game, what are the odds that they will survive?

One way to look at it is the probability of the dice roll. The odds of rolling a pair of ones is 36-1. Everyone in the game faces a single dice roll with the same odds. So the chances of being killed are approximately 2.8% and the chances of survival are approximately 97.2%.

Another way to look at it is to consider the ratios. If the first group (of one person) rolls snake eyes, they are killed and the game ends with one person killed and no survivors (a 100% casualty rate). If the second group (of ten people) rolls snake eyes, then the one person from the first group survived and the ten people in the second group are killed - so ten out of eleven people in the game are killed (a 90.9% casualty rate). If the third group is killed, you have a hundred people die out of a total of one hundred and eleven (a 90% casualty rate). If the fourth group is killed, you have a 1000 out of a total of 1111. If it’s the fifth group, you have 10,000 killed out of 11,111. The sixth group, 100,000 killed out of 111,111. And so on. In other words, approximately ninety percent of the people in the game will be killed and approximately ten percent will survive.

So how do you have a game where all of the players individually have a 97.2% of survival but collectively only 10% of them survive?

[Dead_Cat]

@Reindeer_Flotilla

Here’s one I liked: A mom is 21 years older than her son. In 6 years, she will be 5 times as old. Where is the father now?

What’s clever about this riddle is doing the math actually produces the answer.

Nice. I initially thought there was an error, or a mistake in my algebra, then realised that presumably the intended answer is

The father is in bed with the mom.

[septimus]

@Little_Nemo

… Every group is ten times larger than the previous group. Every group faces one dice roll, which effect everyone in the group. The game continues until one group is killed and then ends. Nobody ever goes back into the room a second time. Assume, for the sake of this post, that an infinite amount of people are available and the room can hold all of them.

So if a person plays this game, what are the odds that they will survive?

… So how do you have a game where all of the players individually have a 97.2% of survival but collectively only 10% of them survive?

VERY cute. I’ll Spoiler my “solution”, though it isn’t very interesting.

[SPOILER]It may be tempting to try to map this to something like Simpson’s Paradox, but that can’t be done — it’s too simple to be Simpsoniacal!

It’s just another Paradox of Infinity. Make a change to force things to be finite and the contradiction disappears.

For example, change the rules so that there are only a googol-plex (10[sup]10[sup]100[/sup][/sup]) of humans available and that if snake-eyes are never rolled in a googol (10[sup]100[/sup]) shakes then everyone survives. The odds of that happening are miniscule but UIAM it is enough to drop the 90% figure back down to 2.8%![/SPOILER]

[Dead_Cat]

@Little_Nemo

This is more of a math paradox. It’s based on a YouTube video.

There’s a “game”. The rules of the game are simple.

A person walks into a room and rolls a pair of standard dice. If the result is snake eyes (a pair of ones) the person is immediately killed and the game is over. If the person rolls anything else they can leave the room unharmed.

If the first person didn’t roll snake eyes, a new group is sent into the room. This time there are ten people in the group. One person rolls the pair of dice for the entire group with the same results; a pair of ones means everyone in the room is immediately killed and the game ends, any other result means everyone gets to leave unharmed and another group is sent in.

Every group is ten times larger than the previous group. Every group faces one dice roll, which effect everyone in the group. The game continues until one group is killed and then ends. Nobody ever goes back into the room a second time. Assume, for the sake of this post, that an infinite amount of people are available and the room can hold all of them.

So if a person plays this game, what are the odds that they will survive?

One way to look at it is the probability of the dice roll. The odds of rolling a pair of ones is 36-1. Everyone in the game faces a single dice roll with the same odds. So the chances of being killed are approximately 2.8% and the chances of survival are approximately 97.2%.

Another way to look at it is to consider the ratios. If the first group (of one person) rolls snake eyes, they are killed and the game ends with one person killed and no survivors (a 100% casualty rate). If the second group (of ten people) rolls snake eyes, then the one person from the first group survived and the ten people in the second group are killed - so ten out of eleven people in the game are killed (a 90.9% casualty rate). If the third group is killed, you have a hundred people die out of a total of one hundred and eleven (a 90% casualty rate). If the fourth group is killed, you have a 1000 out of a total of 1111. If it’s the fifth group, you have 10,000 killed out of 11,111. The sixth group, 100,000 killed out of 111,111. And so on. In other words, approximately ninety percent of the people in the game will be killed and approximately ten percent will survive.

So how do you have a game where all of the players individually have a 97.2% of survival but collectively only 10% of them survive?

And, as a corollary, your odds of death are exactly the same whichever group you are in based on the first way of looking at it, but on the second way of looking at it you are much more likely to die being in a bigger group than a smaller group.

I guess the answer is along the lines of The first survival figure is based on the independent events of the dice rolls, whereas the second is dependent on all of the dice rolls

[iamthewalrus_3]

@septimus

I think you’ve demonstrated that a deck of size 124 is the absolute maximum for which an encoding guarantee is possible. It may seem surprising that we can succeed with 124.

I will not post a solution in the thread; for one thing I don’t know a succinct way to present a solution. I just now whipped together a 100-line program of dense C that implements and proves a solution (the encoder literally tries all 5! orderings until it finds one that works :smack:). I will e-mail that code to any C programmer on request or, when the Sultan CODE tag is fixed, post it here.

Ha ha. Yeah, I’ve been thinking about it, and I might have a solution, but I haven’t convinced myself that it actually works yet. I’d probably have to write some code to try it out.

If anyone who doesn’t know the answer is interested in puzzling it out with me, I’ll post what I have.

[iamthewalrus_3]

@Little_Nemo

So how do you have a game where all of the players individually have a 97.2% of survival but collectively only 10% of them survive?

Isn’t this just the difference between dependent and independent events?

The 97.2% chance of survival is based on the probability of an independent die roll. The 10% total survival can be adjusted any way you want by varying the size of the group because all those deaths are dependent on a single event.

[begbert2]

@Little_Nemo

So if a person plays this game, what are the odds that they will survive?

[spoiler]If a person plays the game, that means they actually end up in the room while a die is rolled, which means that regardless of how many people have been in the room before them or how many people are in the room with them, their chances of survival are 35/36.

The odds of whether a person will die if they’re merely queued up to play the game is a different question entirely - and their odds of survival are essentially 100%. The queue has been stated to be infinitely long, the odds are extremely high that a randomly selected person from the queue will be in a later group, and the odds are low that a later group will be invited into the room at all, since any earlier group losing will end the game. Certainly they have better than a 35/36 chance of surviving because their survival rate only drops that low once you’ve ruled out the possibility that a prior group lost.[/spoiler]

[Lord_Mondegreen]

@excavating_for_a_mind

Suits are typically ranked Clubs Diamonds Hearts Spades.

Thanks, I didn’t know that. I think I’ll teach my younger daughter this trick so we can amaze my wife and older daughter.

[Reindeer_Flotilla]

@Dead_Cat

Nice. I initially thought there was an error, or a mistake in my algebra, then realised that presumably the intended answer is

The father is in bed with the mom.

You presume 100% correctly. Well played!

[septimus]

@excavating_for_a_mind

Suits are typically ranked Clubs Diamonds Hearts Spades.

This order (which is alphabetic order) is in ascending order of usual rank.

That is: Spades is the highest suit, then hearts, diamonds; and clubs is the lowest-ranked of all. That suit ordering is NOT universal, but it is the ordering in contract bridge and, I think, is the default ordering in the U.S.A.

[Musicat]

I imagine every doper knows this one, but just in case…

A hunter sets out from base camp, looking for a bear. He travels one mile south.

No bear, so he travels another mile east.

Still doesn’t see a bear. He continues a mile north, finds and shoots a bear, and notices that he has returned to base camp.

What color is the bear?

[Ludovic]

Probably light brown :eek:

[CookingWithGas]

@Musicat

I imagine every doper knows this one, but just in case…

A hunter sets out from base camp, looking for a bear. He travels one mile south.

No bear, so he travels another mile east.

Still doesn’t see a bear. He continues a mile north, finds and shoots a bear, and notices that he has returned to base camp.

What color is the bear?

Hmm. The classic answer iswhite, because the base camp is at the north pole.butthere are an infinite number of points where base camp could be, on a latitude circle that is 1 mile north of another latitude circle having a circumference of 1 mile, which would be close to the *south *pole. In which case there are no bears at all.

[Left_Hand_of_Dorkness]

@CookingWithGas

Hmm. The classic answer iswhite, because the base camp is at the north pole.butthere are an infinite number of points where base camp could be, on a latitude circle that is 1 mile north of another latitude circle having a circumference of 1 mile, which would be close to the *south *pole. In which case there are no bears at all.

Given that the puzzle says he shot a bear, your objection doesn’t really hold.

[Doug_K]

@CookingWithGas

Hmm. The classic answer iswhite, because the base camp is at the north pole.butthere are an infinite number of points where base camp could be, on a latitude circle that is 1 mile north of another latitude circle having a circumference of 1 mile, which would be close to the *south *pole. In which case there are no bears at all.

In the classic version the hunter goes a mile south, goes east (or west) until he spots a bear, shoots it, then goes a mile north and ends up at base camp. It isn’t specified how far east he travels. There’s only one spot on Earth that you’ll end up at the same place no matter how far east or west you traveled.

[That_Don_Guy]

@pulykamell

As a kid, one of my favorite misdirection groaner versions of this type of puzzle started with “Ok, you’re [or I’m] driving a bus. At the first stop X amount of people get on. At the next stop, half the people get off, and 3 people get on.” And then you just spin in with some types of seemingly complicated, but doable math for a couple rounds. At the end, the question is “what is the name of the bus driver?”

A version of this appears in the 2019 Old Farmer’s Almanac.

@Mr_Shine

A man offers you a car if you can choose which one of three doors it’s behind. Behind the other 2 are goats. After you choose, he opens one of the other doors and reveals a goat. He then offers you the chance to switch doors. Should you, or does it not matter?

Ah, the old Monty Hal-er, Wayne Brady problem.

And the answer depends on semantics - specifically, when it says “he opens one of the other doors and reveals a goat,” is it clear that he was going to make sure that he opened a door with a goat, or did he open a door at random and it just happened to have a goat behind it?

@markn_1

The way I’ve always heard this one is the final question is “how many stops did the bus make”? I think this is better because the answer similarly turns out to be much simpler than expected, but in this case the questionee won’t be able to give the right answer (assuming he was keeping track of the passengers rather than the stops).

There is a variation of this in one of Raymond Smullyan’s books - Alice in Puzzle-Land, I think - where, IIRC, the question was, “How many people got off at the second stop?” The person being asked the question (Alice?) says, “I wasn’t counting that!”, and the asker responds, “You must count everything, because everything counts!” (Another of the questions was the one about dividing an integer by a fraction and then adding another integer.)

@F.U.Shakespeare

A variation of a joke from “Hee Haw”:

At the age of 30, a woman gives birth to a daughter.

Ten years later, the daughter is 10, and the mother is 40, or four times as old.

Five more years later, the daughter is fifteen, and the mother is 45, or only three times as old.

Fifteen more years later, the daughter is thirty, and the mother is 60, or only twice as old.

Question: how long until they’re both the same age?

This predates Hee-Haw; it’s a classic Abbott & Costello routine, and appears in the 1941 movie Buck Privates.

[CookingWithGas]

@Left_Hand_of_Dorkness

Given that the puzzle says he shot a bear, your objection doesn’t really hold.

It’s not an objection, it’s a statement of fact.

[begbert2]

@That_Don_Guy

Ah, the old Monty Hal-er, Wayne Brady problem.

And the answer depends on semantics - specifically, when it says “he opens one of the other doors and reveals a goat,” is it clear that he was going to make sure that he opened a door with a goat, or did he open a door at random and it just happened to have a goat behind it?.

That doesn’t matter, actually - if he picks the door to open randomly it only adds two caveats:

1. If he opens a door and shows the car, then switch to it!

2. If he opens your door and shows a goat, then switch to something else!

Presuming that none of those conditions occurs, then you should still switch when he opens a door and shows a goat.
The thing that should actually worry you is if he didn’t have to open a door. If he had the option of refraining from doing so, then he might be only doing it if he knows you have the right one and wants to lure you away from it.

[CookingWithGas]

@begbert2

That doesn’t matter, actually - if he picks the door to open randomly it only adds two caveats:

1. If he opens a door and shows the car, then switch to it!

You can’t switch to a door once he opens it.