I believe that is incorrect. Salagadoola means “Mechicka Booleroo”, not “Mechicka Boola”.
Right. All you get is the ordering.
Not orientation, not whispered clues, not timing, not any other trick. Just the 4 cards in order. It’s a math riddle, not a trick question.
I haven’t practiced this, but the process is simple enough that you could learn to do it quickly enough to make this a fun parlor trick.
I am curious about this. My method seems to have just enough information to solve the 52-card case.
I’ll have to think about it.
Don’t you have to specify that the animals can always see each other? If the fox can hide in a bush at some arbitrary point, the duck can’t win 100% of the time.
For the passing cards one, you also have a choice of which card to remove. There’s got to be some way of choosing which card such that the remaining cards will give you at least a bit of information. I can’t think of it just yet, but that’s the gist of it.
If anybody else is working on it I’ll put what I’ve got so far
At least 2 cards will be the same suit, lets say hearts. You put a heart first, leaving 12 possibilties. The other 3 cards can be in any of 6 combinations bringing it down to 2 possibilties. The last bit must be which heart you choose, but I can’t think of any way that you can be assured of 2 hearts that allow you to convey 50/50 information.
For the fox and duck, the solution is easy if the speed factor is less than pi+1 (like, say, 4):
Call the speed factor s (as in, the fox is s times faster than the duck). The duck swims in a small concentric circle of radius just less than 1/s. This gives the duck a slightly greater angular speed, and so it can eventually get to a point exactly opposite the fox. Once that happens, the duck turns and makes a beeline to the closest point of the shore: The fox will take pi*r/s to get there, while the duck will take (s-1)*r/s to get there, so as long as s-1 < pi, the duck wins.
For the s = 4.6 version, I expect that it starts out the same way, but that the duck changes its direction as soon as the fox commits to a direction. The duck can change the position of the target point by significantly more than it changes its own path length, so this should buy the duck at least some additional margin, but I haven’t worked out exactly how much yet.
Got it 
correct, you should switch. Although there are two remaining choices, it is not 50-50, because Monty has conditioned it by essentially giving you the best of two choices by revealing the goat door. So it is 2-1 that you will win.
OK let’s play it exactly the way I phrased it. For the low price of $100 you can play. I will put $20,000 behind one door and nothing behind the others. You chose, then I may open a door to reveal one of the nothings and offer a switch. Since you’ve already stated best strategy is to switch, that will be a forced strategy for you. Care to play?
I’d love to see the solution.
Read my other spoiler box then think
On the endless cycle A-2-3-…-K-A-2 one card must be no less than 6 steps lower than the other, so put the lower one in the pile
That is awesome. Thank you!
Well done Mr. Shine. It took me longer to figure it out the first time, but that’s the same method I came up with. There are many others, but I think that’s the simplest/most-elegant one.
Definitely interested in how you can do this with a 124-card deck, since as far as I can tell, our method uses all the bits of info available.
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.
Hmmm, not getting it.
So, the order of the three remaining cards can be coded into the distance between the two cards:
LMH = 1
LHM = 2
MLH = 3
MHL = 4
HLM = 5
HML = 6
So, if my two suited cards are the 3 and 8 of hearts, I put the 8 on the table, put the 3 first (indicating suit, and being the lower of the two suited cards). The three remaining cards are ordered HLM to indicate a distance of five and therefore 3+5 = 8.
But, the distance of five, going the other direction, would also indicate the J. How do you tell?
K364,
[spoiler]You agree on a direction to count. So, you always add the number to the card you can see.
If you need to go from a high number to a low number, you wrap around. So, if you had, say, the Q and the 2, then you’d hide the 2, show the queen, and indicate adding 3, because Q -> K -> A -> 2 is 3 steps.[/spoiler]
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?
That’s… obviously not a solution, but it seems like step 1 toward one.
Oh thanks Walrus. I got confused by the wording “so put the lower one in the pile”
The card puzzle talks about the three remaining cards being low, medium and high? What if there is a pair, or three of a kind? I guess you would need to agree on a ordering of the suits. Is that right?
Suits are typically ranked Clubs Diamonds Hearts Spades.