# Emprworm's Riddle #17: Mess of Tokens

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I was gonna say no it's impossible because its a picture and you cant do jack with it :P ;D

There are also words, lol

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EMPRWORM'S RIDDLE #10:

Infinite Flips

As some of you know, Zamboe is contending my proofs of infinite math and probablities. All this hoopla has inspired another riddle. Its fairly easy, but knowing Zamboe, well...you never know. At any rate, I wont be getting into a 20 page argument over the answer for this one. I am hoping that someone in this forum is good with math and will validate that my answers are correct. Ok, on with the riddle:

Imagine a robot flipping a coin for all eternity. One flip after another. Lets assume this particular coin is constructed in such a way that it will never land on its side- it has a precisely 50-50 chance of landing heads or tails. The robot begins flipping the coin. If the robot flips a heads, then the game is over- he WINS, and the robot will stop and can finally go home (or whatever robots do). Otherwise, he will flip forever and ever until he flips a 'heads."

1. What are the odds that the Robot will win this game, and how can you prove it?

(EXTRA CREDIT!!)

2. Lets say the first 5 flips come up TAILS. What are the odds that the sixth flip will be a HEADS?

Good luck!

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[hide]

He has a 50% of the coin landing heads no matter what. no matter how many times he flips it he still has the same change of getting heads in the next flip.

50% of it being heads on the first flip

50% of it being tails.

there is no middle. Thats it.

extra credit one 50%[/hide]

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[hide]

The odds are 50%, since each flip is 50%.

Each time he flips the coin - it is 50%, so therefore all in all it is 50% since the probability never changes.

Extra credit:

50% - it is always 50%[/hide]

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Man, you posted a few seconds before me ;D

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thats 46 seconds exactly ;)

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[hide] this is a mathematical limis.

If eternity is a given fact, the odds to win are 100% [/hide]

[hide] 50% chance it will flip heads. [/hide]

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[hide] assuming that heads/tails is the only variable then 1/2 = 0.5 = 50% but this is not true as its infinite so 100%

its always 50%... i think [/hide]

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hello,

let's see if emprworm know's what he's talking about,

What are the odds for heads to flip at least once out of four times in a row?

Good day

Eric E.

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I think it is 93.75%

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Totally correct :)

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"What are the odds for heads to flip at least once out of four times in a row?"

15/16

Also correct,

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yes you are. on both counts. you were the first person to get the odds right and explain the proof (though I would prefer just a tad more explanation on how you can prove it.) HOwever, the extra credit was solved before you by someone else. But i'll give you both credit later tonight when i have more time to make the mods.

nicely done. please if you have time to spare, show me how went about proving the answer to part 1. If you dont have time, no biggie, ill give you credit anyway and post the proofs with the answer later.

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My common sense explanation:

An infinite number of throws yields the impossibility to never throw heads,

that is all to it.

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Extra credit was solved by Ex - 46 seconds before me, lol. I

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LOL I got here too late actually, it was Ex ;)

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Oh yeah, that's right, sry, confused - both of you guys have those similar planet avatars ;D

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Damn it, Emprworm, your riddles get solved too fast! :) I never get a chance to post the answer myself, unless I'm online when you post it... :(

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sorry, Edric. However, be not dismayed my friend. My next riddle will be unbelievably difficult. not mathematically, but logically. A pure philosophical pice of genious. (and no, I did not invent it :) )

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Jackpot for the One who solves this riddle,

Items at hand are scales and nine metal balls all equally looking but one is heavier,

only weighing twice is required to determine which one is heavier

how ?

and yes I invented this riddle myself :)

Good day

Eric E.

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[hide] Ok i got it. put 3 balls on each scale and 3 on the table. This will immediately tell you which group of 3 balls contains the heavy ball. Now, With 3 balls, you can figure out which is the heaviest with one weighing. Put one on each side of the scale and one on the table. If the scales balance, the ball on the table is heaviest.[/hide]

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okay,

Again, You are correct, well done.

Bring up your next riddle :)

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SOLUTION TO INFINITE FLIPS

Well this riddle went quick. So quick, in fact, the answer was posted literally minutes after the riddle was. Not good. I hope my next riddle doesn't go as fast. Well, for this riddle, I am giving credit for Part I to Eric E and ExAtreides for getting part II.

Answer is unhidden because it was too easy. Spoiler below

The probability of flipping a coin heads can be diagramed below.

WIth each flip, we look at all possible combinations

AFTER ONE FLIP:

HT - 50% chance of heads.

If we flip the coin twice, here are our new possible outcomes:

HH

HT

TH

TT

This time we have four possibilities. THe probability of flipping a coin twice in a row without heads is 1 out of 4.

So, after 1 turn it is 1/2, after two turns it is 1/4. So, overall, the probability of flipping a coin HEADS after just two turns is the sum of these probabilities: 1/2 + 1/4 = 3/4 or 75%.

For each turn we take, we continue to add the sum of the probable sets. For 3 turns it would be: 1/2 + 1/4 + 1/8

4 turns: 1/2 + 1/4 + 1/8 + 1/16

After just 10 turns, the odds of flipping a heads at least once is 99.9%

But there is STILL A CHANCE that heads wont come up!!

So if we flip forever, is there not at least a slim chance that heads will not come up? The answer is no. The odds that you will flip heads is 100%. IN calculus, this is known as a limit.

Consider this basic problem:

What is 1 - .999999999999999999999999999999999999999... (infinite .9's) ?

What happens when you subtract .99999...(forever) from 1?

take a look:

1.000000000000000000000000000000000000000000000000 (forever)

- .999999999999999999999999999999999999999999999999 (forever)

______________________________________________________

= .000000000000000000000000000000000000000000000000 (forever)

The answer is 0!! You will never ever write down a "1" in your answer. You will always, forever, be writing down ".00000..." which means your answer is 100% purely zero. Therefore 1 - .9999 (forever) equals ZERO. There is no such thing as an 'infinitely small' value. such a value is equal to zero in all mathematics. Flipping a coin forever, the odds of getting heads is 100%.

Now for the extra credit:

We already saw that the possible combinations for two flips are:

HH

HT

TH

TT

For three flips it is:

HHH

HHT

HTT

HTH

THH

THT

TTH

TTT

So the odds of getting TT is 1/4. If our first two flips = TT, then what are our odds of getting another T? Some might think that 'well, look! There is only 1 out 8 chances of getting a TTT, so the odds of another T must be very small!"

Completely wrong. Since we ALREADY got a TT, the odds of another T are simply the same as always: 50%. They will always be 50%.

The answer to the extra credit is 50%.

Good luck on the next one....

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EMPRWORM'S RIDDLE #11

Please if you know this riddle, DONT POST and ruin it. Thanks guys. This is one of my personal favorites.

Spikes and Spheres

You are in a most unfortunate circumstance. You at the front of a line of 3 people in a barbaric arena and part of a most twisted game for the emperor's enjoyment. The emperor has gathered together the three wisest men in the kingdom, all known for their superb wit and extreme genious. You are one of them (of course). Player B is standing 1 meter behind you. And a third person, Player C is standing 2 meters behind you. You are all wearing very special robes. On the back of each of your robe is a picture of either a green spheroid or a silver ball of spikes.

Player C is in the back of the line. He can see both your robe and Player B's robe. Player B can see the picture on your robe only, and you cant see anyone's robe! NO one can see the picture on their own robe (since it is on the back).

The game is this: The first person to correctly identify whether or not there is an image of a green sphere or silver ball of spikes on the back of their robe will be rewarded with 1 million gold coins. If you answer WRONGLY you will be immediately executed on the spot. If you do not answer at all, no harm will come to you. You will go home in peace. So it would be unwise to just guess at random.

The Rules of the game are:

- There is AT LEAST 1 Green Sphere.

- There may be only one spike ball and two green spheres, there may be two spike balls and one green sphere, or there may be 3 green spheres. But you may be certain that there are not three spike balls.

The game will be played as follows: when the clock starts, Player B and C will have 1 minute to try and figure out what is on their robes. The first one to figure it out, simply yells out the answer. If 1 minute passes, and neither of them have declared an answer, you will get one shot to either make a guess or pass. But if you guess wrongly you will die.

Doesn't sound very fair that the two people behind you can see a robe and you cant, but oh well, life isn't always fair.

THE GAME BEGINS, AND THE CLOCK STARTS

1 minute passes. Neither Player B nor Player C made a guess (the two guys behind you both pass). You figure that they would rather pass and live then guess wrong and die. Now the emperor turns to you.

"Well, are you going to guess what is on the back of your robe now?"

The executioner raises his ax, ready to strike you down if you speak wrongly.

You know that those two men behind you are geniouses and happen to be very very greedy for money. The greediest people you have ever known, as well as the most intelligent.

D0 you PASS, and go home poor, but alive, or do you declare what is on the back of your robe?

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