Riddle Minute

[nickspoon]

I also think it's 1, but I may be way off. Huffman encoding against the alphabet "ABCDE" would result in A, B, C, D, and E being encoded as 10, 0111, 0101, 011, and 1, and since you told us what A through D are, E must be 1.

If that's the case, then that's interesting because it's not what I was thinking of. In my system, F would be 1101. The value of each character is independent of its context.

Also, I would be surprised if I got this right, but is the answer to your joker riddle ~0.161?

[Doc Sigma]

I also think it's 1, but I may be way off. Huffman encoding against the alphabet "ABCDE" would result in A, B, C, D, and E being encoded as 10, 0111, 0101, 011, and 1, and since you told us what A through D are, E must be 1.

If that's the case, then that's interesting because it's not what I was thinking of. In my system, F would be 1101. The value of each character is independent of its context.

Also, I would be surprised if I got this right, but is the answer to your joker riddle ~0.161?

I'll need to rethink yours, then -- in my solution, F would not exist. Heh.

As for the joker riddle, nope -- considering the answer you got, you're thinking too hard. (Which is the beauty of this one... it kinda makes you think too hard at first.)

[Archaemic]

Uhhh, fairly certain that Huffman coding doesn't work that way.
E would be 11 or 00 with Huffman coding if the sample set were just ABCDE and the other values were as mentioned.

[Doc Sigma]

Uhhh, fairly certain that Huffman coding doesn't work that way.
E would be 11 or 00 with Huffman coding if the sample set were just ABCDE and the other values were as mentioned.

Why wouldn't there be any one-bit encodings? E would be 1. Or 0. Hmm. Hang on. *draws up the tree*

Wait, never mind, with a five-character alphabet, the most expensive string would be three bits long, not 4. So it's not Huffman encoding.

[Dr. Sticks]

for the card riddle, do you stop as soon as you deal the joker BECAUSE it was the 53rd card dealt? in that case, the odds are 1:1

[nickspoon]

The position of the joker can be any one in 53, so each of the following has a 1/53 chance of occurring (situation;probability of four aces):
1st card is Joker;0
2nd card is Joker;0
3rd card is Joker;0
4th card is Joker;0

Those are simple. Now, the probability of all of the first four cards being aces is one in 270725 (52C4).
5th card is Joker;1/270725
The probability of four aces in the first five cards is 48 in 52C5 (because there are 48 possibilities for the fifth card).
6th card is Joker;48/2598960
First six cards, 48C2 in 52C6.
7th card is Joker;1128/20358520

And so on and so forth. The general formula for the probability getting four aces in X cards is therefore:
48C(X-4)/52CX, 53>X>4.

Thus, P(four aces before a joker) = (48C0/52C4 + 48C1/52C5 + 48C2/52C6 + ... + 48C48/52C52) / 53

...which leads to a probability of precisely a fifth.

[Doc Sigma]

The position of the joker can be any one in 53, so each of the following has a 1/53 chance of occurring (situation;probability of four aces):
1st card is Joker;0
2nd card is Joker;0
3rd card is Joker;0
4th card is Joker;0

Those are simple. Now, the probability of all of the first four cards being aces is one in 270725 (52C4).
5th card is Joker;1/270725
The probability of four aces in the first five cards is 48 in 52C5 (because there are 48 possibilities for the fifth card).
6th card is Joker;48/2598960
First six cards, 48C2 in 52C6.
7th card is Joker;1128/20358520

And so on and so forth. The general formula for the probability getting four aces in X cards is therefore:
48C(X-4)/52CX, 53>X>4.

Thus, P(four aces before a joker) = (48C0/52C4 + 48C1/52C5 + 48C2/52C6 + ... + 48C48/52C52) / 53

...which leads to a probability of precisely a fifth.

YES!!! Except you did it the really, really hard way. Hint: The cards other than the joker and the aces do not matter at all.

[LewisTheTank]

My guess:

E=1

Reasoning?

EDIT: 2222nd post, awesome.

My calculator told me. The number 2 (or '0010') was very useful. It (my calculator) never lies, but HAS been wrong before. (Thanks to my input...)

[LewisTheTank]

A - 1
B - 11
C - 21
D - 1211
E - 111221
F - 312211
G - ?

G=13112221
(And just to show off a little):
H=1113213211
(This was another one my dad loved to keep me in practice with. I loved that jackass, but this IS a good brain-teaser!)

[Doc Sigma]

A - 1
B - 11
C - 21
D - 1211
E - 111221
F - 312211
G - ?

G=13112221
(And just to show off a little):
H=1113213211
(This was another one my dad loved to keep me in practice with. I loved that jackass, but this IS a good brain-teaser!)

Excellent! I've always loved this one.

[nickspoon]

My calculator told me. The number 2 (or '0010') was very useful. It (my calculator) never lies, but HAS been wrong before. (Thanks to my input...)

I don't know what rule you're thinking of here, but the pattern that I'm looking for cannot be found on a calculator.

And Sigma, the formula reduces to (X(X-1)(X-2)(X-3))/6497400, which would expand to a quartic and is probably not the simpler solution you're looking for.

[Doc Sigma]

I GOT IT!!! I'll give some more answers, this should prove that I know it, WITHOUT giving away the answer. Because this is a cool one that other people should try, too!

A: 10
B: 0111
C: 0101
D: 011
E: 1
F: 1101
G: 001
H: 1111
...
T: 0
...
Z: 0011

Numbers, too!

1: 10000
2: 11000
3: 11100
...
0: 00000

[nickspoon]

I GOT IT!!!

That you did! Very well done.

Also, I loved your riddle mainly because of how a ridiculous amount of maths amounts to such a simple answer, even if said maths wasn't the best method

[Doc Sigma]

Basically, the other 48 cards do not matter. Dealing four aces followed by the joker... does it really matter which other cards got dealt in the process? Logically, no. It might as well just be five cards, the four aces and the joker. If you don't buy the logic, then do the math... which you already did.

My well has run dry at the moment... will post another riddle later.

[nickspoon]

The other cards don't matter, yes, which is the reason for the 48C(X-4) - that's the number of combinations of the cards besides the aces.

Anyway, next riddle, nice easy one:

A town is run by an evil mayor. He has his eyes on the miller's daughter, and offers the miller a deal. He will give the miller's daughter a bag with two pebbles in it, one white and one black. The daughter will pull one stone from the bag with the townsfolk as an audience; if she draws the white stone, she goes free, and if she draws the black stone, she must marry the mayor. The miller has no choice but to agree; the mayor is a powerful man. However, he doesn't trust the mayor not to cheat. How can the miller's daughter act to ensure a fair game?