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Riddle/Logic problem.

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Originally posted by: royaldank
This was posted at another site I read this morning. Thought it was interesting.
------------
You are sitting blindfolded at a table with some large number n of coins (n > 10) on it.
Each coin has a heads side and a tails side.
You are told that exactly 10 of the coins are currently heads up.
You are wearing gloves that will let you pick up, flip or move a coin, but not feel any differences in texture between sides.

Please make two groups of coins that contain exactly the same number of Heads-up coins.

You cannot see the coins, nor distiguish heads versus tails through touch.
Click to expand...

quote:

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Originally posted by: nineball9
Pull out 10 coins to another group and flip them over.
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Good job. You are correct.
Click to expand...

The problem is lacking details to support this answer.

If I have 50 coins and 10 are heads up... How do I know what pile to put what and then what do I flip? I'm gloved and blindfolded. I don't know where the original 10 heads up coins are located. If I have 50 coins and 40 are tails up (not specified) then the odds are unlikely I can luck into moving the correct coins.

What am I missing?
 
Originally posted by: TheGameIs21
The problem is lacking details to support this answer.

If I have 50 coins and 10 are heads up... How do I know what pile to put what and then what do I flip? I'm gloved and blindfolded. I don't know where the original 10 heads up coins are located. If I have 50 coins and 40 are tails up (not specified) then the odds are unlikely I can luck into moving the correct coins.

What am I missing?
Click to expand...


Only 10 are heads up.

Given 50 coins, select any 10. Say you select 7 tails and 3 heads. That leaves 7 heads left in the main pot. When you flip all 10 coins you pulled aside, it will have 7 heads and match the main pot.
 
Originally posted by: royaldank
Originally posted by: TheGameIs21
The problem is lacking details to support this answer.

If I have 50 coins and 10 are heads up... How do I know what pile to put what and then what do I flip? I'm gloved and blindfolded. I don't know where the original 10 heads up coins are located. If I have 50 coins and 40 are tails up (not specified) then the odds are unlikely I can luck into moving the correct coins.

What am I missing?
Click to expand...


Only 10 are heads up.

Given 50 coins, select any 10. Say you select 7 tails and 3 heads. That leaves 7 heads left in the main pot. When you flip all 10 coins you pulled aside, it will have 7 heads and match the main pot.
Click to expand...

DOH... Thanks.
 
Originally posted by: SagaLore
Originally posted by: royaldank
Originally posted by: ArmenK
Edit: So you DO have to count them. If there are only 10 the solution is to split all the coins into two piles.
Click to expand...

My fault, I guess it needs to have n>10 in the problem. I copied and pasted from another site.

For those that might not understand, take n = 1000. Pull out 10 (1 is a head). Flip these 10 coins over (9 are now heads). At this point, you have 9 heads in the initial pot, and 9 heads in the group you flipped.
Click to expand...

I don't see how that completes this:

Please make two groups of coins that contain exactly the same number of Heads-up coins.

That statement either means you just need 10 coins, or it means that you need at least 10 coins with the heads up. If the second is true, then having a group with 9 heads doesn't solve this.
Click to expand...

u obviously do not understand the riddle/problem/question.
 
Originally posted by: rayray2
/head explodes
Click to expand...

maybe this will help...

H=head
T=tails
n=30 (20 T's, 10 H's)

first:
TTTTTTTTTTTTTTTTTTTTHHHHHHHHHH (20 T's, 10 H's)

second:
pull at random any 10 coins (for this example, i will say i pulled 7 T's and 3 H's. This will go in group 2):

group 1:TTTTTTTTTTTTTHHHHHHH (13 T's, 7 H's)
group 2:TTTTTTTHHH (7 T's, 3 H's)

Third:
flip the coins from group 2 over

group 1:TTTTTTTTTTTTTHHHHHHH (stays the same as in step 2; 13 T's, 7 H's)
group 2:HHHHHHHTTT (7 T's when flipped becomes 7 H's, 3 H's when flipped becomes 3 T's)

therefore; group 1 and group 2 both have 7 H's which solves the riddle.
 
Originally posted by: chuckywang
I have another riddle sort of like this one, but it's kinda dirty...I'll post it if you all want me to.
Click to expand...

?

Same riddle as OP only you have 2 types of women some with shaved some with bush?
 
Originally posted by: element
Originally posted by: chuckywang
I have another riddle sort of like this one, but it's kinda dirty...I'll post it if you all want me to.
Click to expand...

?

Same riddle as OP only you have 2 types of women some with shaved some with bush?
Click to expand...

No, it's not like that....it's a different riddle fundamentally, but it still requires the same sort of thinking.
 
Originally posted by: chuckywang
Originally posted by: element
Originally posted by: chuckywang
I have another riddle sort of like this one, but it's kinda dirty...I'll post it if you all want me to.
Click to expand...

?

Same riddle as OP only you have 2 types of women some with shaved some with bush?
Click to expand...

No, it's not like that....it's a different riddle fundamentally, but it still requires the same sort of thinking.
Click to expand...

Is it the 'how can two guys each bang two girls with only two condoms among them two"
 
n coins

10H
(n-10)T

randomly pull out 10 coins

Group 1:
XH
YT

Group 2:
(10-X)H
(n-10-Y)T

flip Group 1:
XT
YH

proove that there are an equal number of H
YH = (10-X)H

Remember that
XH + YT = 10

since we flipped Group 1 it would be
XT+ YH = 10

Rearranging, we get:
YH = 10 - XT

😉
 
Originally posted by: TuxDave
Originally posted by: chuckywang
Originally posted by: element
Originally posted by: chuckywang
I have another riddle sort of like this one, but it's kinda dirty...I'll post it if you all want me to.
Click to expand...

?

Same riddle as OP only you have 2 types of women some with shaved some with bush?
Click to expand...

No, it's not like that....it's a different riddle fundamentally, but it still requires the same sort of thinking.
Click to expand...

Is it the 'how can two guys each bang two girls with only two condoms among them two"
Click to expand...

Yes it is.
 
I've got a good one.

Everyone knows the old "9 marbles, 8 of same weight, one slightly heavier, use a balance scale and 2 weighs to determine which marble is the heavy one" riddle right? Well here's a harder version:

You have a balance scale and 12 golf balls, 11 of which are the same weight, one of which has a deviant weight, and are allowed to use the scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether it?s heavier or lighter than the rest.
 
Originally posted by: silverpig
I've got a good one.

Everyone knows the old "9 marbles, 8 of same weight, one slightly heavier, use a balance scale and 2 weighs to determine which marble is the heavy one" riddle right? Well here's a harder version:

You have a balance scale and 12 golf balls, 11 of which are the same weight, one of which has a deviant weight, and are allowed to use the scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether it?s heavier or lighter than the rest.
Click to expand...

I came across this puzzle in a science magazine years ago. It is diabolically difficult! I couldn't solve it then, and after reading the solution, I sat down with paper and pencil to try to figure out how it worked. While I understood the logic of the solution, I could not create a truth table or mathematical solution. (Sometime later, I read some math Martin Gardner did on these types of puzzles. While I understood the math, it was difficult to follow, nor was it anything I could have derived on my own.)

After rereading the puzzle today, and even though I remembered the gist of how to solve it, I still could not do it with paper and pencil! Alas, I resorted to google for the solution. (Hint: if you want the solution, the original puzzle - as I remember it anyway - used counterfeit coins instead of golf balls.)

 
Originally posted by: royaldank
This was posted at another site I read this morning. Thought it was interesting.
------------
You are sitting blindfolded at a table with some large number n of coins (n > 10) on it.
Each coin has a heads side and a tails side.
You are told that exactly 10 of the coins are currently heads up.
You are wearing gloves that will let you pick up, flip or move a coin, but not feel any differences in texture between sides.

Please make two groups of coins that contain exactly the same number of Heads-up coins.

You cannot see the coins, nor distiguish heads versus tails through touch.
Click to expand...

Hmm, heads and tails huh.

If it feels like a blow, its the head otherwise its tails. 😉
 
Originally posted by: silverpig
I've got a good one.

Everyone knows the old "9 marbles, 8 of same weight, one slightly heavier, use a balance scale and 2 weighs to determine which marble is the heavy one" riddle right? Well here's a harder version:

You have a balance scale and 12 golf balls, 11 of which are the same weight, one of which has a deviant weight, and are allowed to use the scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether it?s heavier or lighter than the rest.
Click to expand...

Here's my theory. Please shoot holes in it.

Weigh #1 - Make 3 groups of 4 and balance 2 of them.

Scenario 1 - Same weight
- You know the unbalanced 4 have the item.
- Weigh #2 Take 3 of the unbalanced 4 and weigh them against any 3 from the 8 that were balanced
- Scenario 1-1 - The 3 shows heavier, first off you know the ball is heavier. If lighter, you know it's lighter.
- Take any two and weigh them against each other.
- Scenario 1-1-1 - Same weight The unweighed of the 3 is the ball and you already know whether it's light or heavier
- Scenario 1-1-2 - Different weight. Based on the results from 1-1 you know which ball it is.
- Scenario 1-2 - You weighted 3 balls and they are all the same. You know the unweighed 4th ball is the deviant ball. Weigh it against one of the original 8 to find out if it's heavier or lighter.

Scenario 2 - Different Weight. For this branch, L = potential light ball, H = potential heavy ball, S = known safe ball
- You know the unbalanced 4 are safe
- Take 2 of the "lighter" balls and 3 of the "heavier" balls. Weigh them like this:
LH LH -Scale- SS HS
- As a result of this configuration there are 3 possible scenarios:
1. Same weight. This means one of the remaining balls is it (you have L L H left). See #3 below
2. Left side is heavier. One of the two H's is deviant. Measure one of the Hs and you have your answer
3. Right side is heavier. You know it's either the H on the right side or one of the two L's on the left side. You have L L H. Take the L and the H and measure against S S. If the LH is heavier, the H is the culprit. If LH is lighter, the L is. If they are the same, the unmeasured L is it.
 
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