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ExtraCredit




 
 
 
 

Post  Sun, Nov 22 2020, 10:37 pm
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zaq wrote:
January 1, January 2, January 3

You’ve been promoted to the next grade!
My guess was ‘yesterday, today, tomorrow’
Hope I’m promoted too.


Last edited by ExtraCredit on Sun, Nov 22 2020, 10:45 pm; edited 1 time in total
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Mrs. XYZ




 
 
 
 

Post  Sun, Nov 22 2020, 10:41 pm
ExtraCredit wrote:
My DC asked me this so excuse the level: name 3 consecutive days without using Sunday Wednesday or Friday!


Hidden: 

yesterday, today, tomorrow?
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shanie5




 
 
 
 

Post  Sun, Nov 22 2020, 10:44 pm
ExtraCredit wrote:
My DC asked me this so excuse the level: name 3 consecutive days without using Sunday Wednesday or Friday!


Hidden: 

Hoshana Raba, shmini atzeres, simchas torah
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ExtraCredit




 
 
 
 

Post  Sun, Nov 22 2020, 10:44 pm
Mrs. XYZ wrote:
Hidden: 

yesterday, today, tomorrow?

Yes mam. Do you have something new for us here? You were once active here...
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ExtraCredit




 
 
 
 

Post  Sun, Nov 22 2020, 10:44 pm
shanie5 wrote:
Hidden: 

Hoshana Raba, shmini atzeres, simchas torah

Lol! Original!
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Mrs. XYZ




 
 
 
 

Post  Sun, Nov 22 2020, 10:53 pm
ExtraCredit wrote:
Yes mam. Do you have something new for us here? You were once active here...


(I was not following the whole thread so not sure if this exact one was mentioned)

There are 12 gold coins, all weigh the same except for 1 which weighs either more or less.

Using a balance scale 3 times, find out which one is diff. and if it weighs more or less.
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ExtraCredit




 
 
 
 

Post  Sun, Nov 22 2020, 11:12 pm
ExtraCredit wrote:
Ok I think I got this.

Hidden: 

#1 you weigh 4&4. I’ll go with the hardest scenario and say that the counterfeit one is here and we now don’t know whether it’s one of the 4 weighing more or 4 weighing less. (If the scale was even then it’s easy to finish off. I’ll only explain the hardest scenario)

#2 on side A- you weigh 3 of the previous light ones plus 2 heavy ones. On side B- you weigh 1 light one plus the 4 good ones from #1. (If it’s even you are left with only 2 heavy ones to weigh at the 3rd weighing and see which is heavier)
*In the event side A is lighter you do for #3 2 of the light ones and see whether one is lighter, or the third is the counterfeit.
*in the event side A is heavier, you stay with a shaila between the 2 heavy ones. You weigh those on round three. If one is heavier then that’s the counterfeit, if they’re even then the light from side B is the counterfeit.

Ok, I’m dizzy but I’m glad I got it!

Yes I did it here
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ExtraCredit




 
 
 
 

Post  Mon, Dec 14 2020, 2:28 pm
100 children sit in a circle. Each one has a little red ball. They have to pass the ball, either to the child at their right or to the child at their left (choosing randomly, both having the same chance of occuring).

Each child passes the ball exactly one time.
How many children are expected to remain without a ball?
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windchime




 
 
 
 

Post  Mon, Dec 14 2020, 4:28 pm
ExtraCredit wrote:
100 children sit in a circle. Each one has a little red ball. They have to pass the ball, either to the child at their right or to the child at their left (choosing randomly, both having the same chance of occuring).

Each child passes the ball exactly one time.
How many children are expected to remain without a ball?


25?
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ExtraCredit




 
 
 
 

Post  Mon, Dec 14 2020, 6:57 pm
windchime wrote:
25?

Hooray you got it!
Can you explain why you chose that number?
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windchime




 
 
 
 

Post  Tue, Dec 15 2020, 1:04 pm
ExtraCredit wrote:
Hooray you got it!
Can you explain why you chose that number?


Sure, it's probability. So the chance that 50 players will receive 2 balls is equal to the chance of each player receiving 1 ball. So the number in question (50) is averaged out. That comes to 25.
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ExtraCredit




 
 
 
 

Post  Tue, Dec 15 2020, 1:52 pm
windchime wrote:
Sure, it's probability. So the chance that 50 players will receive 2 balls is equal to the chance of each player receiving 1 ball. So the number in question (50) is averaged out. That comes to 25.

So interesting to see how you figured it out.

Here’s the explanation I got: (I didn’t figure it out but someone gave me this answer)

the four possibilities here are:
either I get from left only
or I get from right only
or I get from both
or I get from none.

That's why you get an expectation value of 25% for the case "none".

Honestly I’m not good with probability. I focused on what would be the most possible kids left without a ball? The answer to that is 50.
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windchime




 
 
 
 

Post  Tue, Dec 15 2020, 2:14 pm
ExtraCredit wrote:
So interesting to see how you figured it out.

Here’s the explanation I got: (I didn’t figure it out but someone gave me this answer)

the four possibilities here are:
either I get from left only
or I get from right only
or I get from both
or I get from none.

That's why you get an expectation value of 25% for the case "none".

Honestly I’m not good with probability. I focused on what would be the most possible kids left without a ball? The answer to that is 50.


Thanks for sharing!
That's interesting and probably the traditional way to solve the problem.
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ExtraCredit




 
 
 
 

Post  Tue, Dec 15 2020, 2:21 pm
windchime wrote:
Thanks for sharing!
That's interesting and probably the traditional way to solve the problem.

Taken from here

https://mindyourdecisions.com/.....-out/
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