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tothepoint




 
 
 
 

Post  Mon, Oct 26 2020, 9:36 am
Answer

Hidden: 

Here are the possible combinations and their sums:

1+1+72=74
1+2+36=39
1+3+24=28
1+4+18=23
1+6+12=19
1+8+9=18
2+2+18=22
2+3+12=17
2+4+9=15
2+6+6=14
3+3+8=14
3+4+6=13

Since 266 & 338 are the only combination with same sum, the fact that there is an oldest daughter boils it down to final answer of 338.
(I thought about twins too.. I guess they are usually referred to as ‘my twins’)
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tothepoint




 
 
 
 

Post  Mon, Oct 26 2020, 9:45 am
And another one!

100 people are waiting to board a plane. The first person’s ticket says Seat 1; the second person in line has a ticket that says Seat 2, and so on until the 100th person, whose ticket says Seat 100.

The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible; if their seat is taken, they will randomly choose one of the remaining seats (after the first person, the second person takes a seat; after the second person, the third person takes a seat, and so on). What is the probability the 100th person sits in Seat 100?
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ChanieMommy




 
 
 
 

Post  Mon, Oct 26 2020, 11:20 am
tothepoint wrote:
And another one!

100 people are waiting to board a plane. The first person’s ticket says Seat 1; the second person in line has a ticket that says Seat 2, and so on until the 100th person, whose ticket says Seat 100.

The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible; if their seat is taken, they will randomly choose one of the remaining seats (after the first person, the second person takes a seat; after the second person, the third person takes a seat, and so on). What is the probability the 100th person sits in Seat 100?


Hmmm... good question
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ExtraCredit




 
 
 
 

Post  Mon, Oct 26 2020, 2:08 pm
tothepoint wrote:
And another one!

100 people are waiting to board a plane. The first person’s ticket says Seat 1; the second person in line has a ticket that says Seat 2, and so on until the 100th person, whose ticket says Seat 100.

The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible; if their seat is taken, they will randomly choose one of the remaining seats (after the first person, the second person takes a seat; after the second person, the third person takes a seat, and so on). What is the probability the 100th person sits in Seat 100?

Hidden: 

50%
Udder yu, udder nisht Hiding
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tothepoint




 
 
 
 

Post  Tue, Oct 27 2020, 12:08 pm
ExtraCredit wrote:
Hidden: 

50%
Udder yu, udder nisht Hiding


Correct! Do you understand the reasoning though?
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ExtraCredit




 
 
 
 

Post  Tue, Oct 27 2020, 2:43 pm
tothepoint wrote:
Correct! Do you understand the reasoning though?

Hidden: 

My reasoning is not much more than I already wrote. There’s just a 50% chance that the first or subsequent out of place passengers will take his seat. I meant it quite literally that it’s udder yu udder nisht, but Im sure you have a smarter explanation. Let’s hear.
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tothepoint




 
 
 
 

Post  Tue, Oct 27 2020, 5:37 pm
Answer to airplane
Hidden: 

Ok. So let us say that I’m person #1 to sit down and you’re the last. Also if you sit in your own seat then you “win”, otherwise you “lose”.

Let’s say that there are only two seats, yours and mine. If I sit in my own seat, you win. If I sit in your seat, you lose. So you have a 50% chance of winning.

Now let’s go back to 100 seats. The previous paragraph still holds true: you have a 50% chance of winning if we only consider your seat and mine. Now if I sit anywhere else, I’m just postponing the decision. Let’s say I sit in the seat of the person who’s 13th in line. Persons 2 through 12 will sit in their own seats, then when person 13 comes in he can either sit in my original seat (and you win) or yours (and you lose). Or of course he could sit anywhere else and postpone the decision again.

If this keeps going, then eventually there are only two seats left and person 99 is forced to choose either your seat or mine, again with 50% chance. There are only two seats that matter throughout the game: yours and mine. Any sitting in other seats is just postponing the decision of which of the two interesting seats gets sat in first. Note also that you’ll only ever end up in your seat or mine, no one else’s
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ExtraCredit




 
 
 
 

Post  Wed, Oct 28 2020, 5:31 am
tothepoint wrote:
Answer to airplane
Hidden: 

Ok. So let us say that I’m person #1 to sit down and you’re the last. Also if you sit in your own seat then you “win”, otherwise you “lose”.

Let’s say that there are only two seats, yours and mine. If I sit in my own seat, you win. If I sit in your seat, you lose. So you have a 50% chance of winning.

Now let’s go back to 100 seats. The previous paragraph still holds true: you have a 50% chance of winning if we only consider your seat and mine. Now if I sit anywhere else, I’m just postponing the decision. Let’s say I sit in the seat of the person who’s 13th in line. Persons 2 through 12 will sit in their own seats, then when person 13 comes in he can either sit in my original seat (and you win) or yours (and you lose). Or of course he could sit anywhere else and postpone the decision again.

If this keeps going, then eventually there are only two seats left and person 99 is forced to choose either your seat or mine, again with 50% chance. There are only two seats that matter throughout the game: yours and mine. Any sitting in other seats is just postponing the decision of which of the two interesting seats gets sat in first. Note also that you’ll only ever end up in your seat or mine, no one else’s

I basically said this but with way less words! Laugh
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ExtraCredit




 
 
 
 

Post  Wed, Nov 18 2020, 6:29 pm
To the point, nothing new?
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Live+Love




 
 
 
 

Post  Wed, Nov 18 2020, 8:11 pm
EC since you asked...

I have 26 donuts, 3 of my friends ask me for one, another 2 strangers ask for 2 how many do I have left?

Hidden: 

That's right! 26
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ExtraCredit




 
 
 
 

Post  Wed, Nov 18 2020, 8:16 pm
Hidden: 

19?
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Live+Love




 
 
 
 

Post  Wed, Nov 18 2020, 8:18 pm
ExtraCredit wrote:
Hidden: 

19?


Close guess! though im a lot more generous than you think! 😜😝
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ExtraCredit




 
 
 
 

Post  Wed, Nov 18 2020, 8:21 pm
mocha wrote:
Close guess! though im a lot more generous than you think! 😜😝

Did you already post the answer? I didn’t click on it yet cuz want to figure it out.
How about
Hidden: 

18? She decided to give 2 for each friend and only 1 for a stranger?
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ExtraCredit




 
 
 
 

Post  Wed, Nov 18 2020, 8:23 pm
Or
Hidden: 

16. If she gives strangers 2 she certainly gives friends 2?
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Live+Love




 
 
 
 

Post  Wed, Nov 18 2020, 8:44 pm
and I thought you're better at this... Can't Believe It Can't Believe It




JK I was trying to make you laugh and instead I made you work so hard forgive me, please...
it is actually my first checking out this page
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ExtraCredit




 
 
 
 

Post  Wed, Nov 18 2020, 8:46 pm
mocha wrote:
and I thought you're better at this... Can't Believe It Can't Believe It




JK I was trying to make you laugh and instead I made you work so hard forgive me, please...
it is actually my first checking out this page


So now I clicked into your answer!
What a stingy friend! LOL
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doodlesmom




 
 
 
 

Post  Sun, Nov 22 2020, 7:28 pm
Bumpity bump....I can use some brain stimulation....sorry I’m not offering one
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ExtraCredit




 
 
 
 

Post  Sun, Nov 22 2020, 7:33 pm
doodlesmom wrote:
Bumpity bump....I can use some brain stimulation....sorry I’m not offering one

We need chaniemommy back! Crying
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ExtraCredit




 
 
 
 

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




 
 
 
 

Post  Sun, Nov 22 2020, 10:32 pm
January 1, January 2, January 3
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