[quote="ExtraCredit"]Tamari didn’t say 1 must board every day.
Tamari said in one of her answer that from day 2 and on another person will realize that they are blue, and go on the plane....when she named the people ABC and D and E
Tamari said in one of her answer that from day 2 and on another person will realize that they are blue, and go on the plane....when she named the people ABC and D and E
I think that was just going to a different scenario. She was trying to explain what if there were 3 or 4 or 5. I. The original post it doesn’t say that one should board daily.
Anyway, I wasted too much energy on this already.
This is the same answer as till now. I’ll repeat the same question. Why no action at all on day 20 or 30?
I was upset when I couldn't solve this riddle to begin with. Now I take pride in understanding it. And the explanation I read was not nearly as clear as the one I gave you.
Hidden:
Seeing 39 people, they are not expecting ANY action until day 39. It's only day 39 that will be the test, whether there are taka 39 people, and if no action then, must be 40.
I know you keep on saying not to go with small numbers, but the logic always boils down to the last person: Knowing that at least one person is blue:
If there is 1 person, he sees no one else, and therefore knows that he's blue, so he leaves day 1.
If there are 2 people, each sees 1. Were there only 1 person, that person should be leaving on day 1. So if nothing happens on day 1, they both know they're blue and leave on day 2.
SO IF THERE ARE 2 PEOPLE, THEY BOTH LEAVE ON DAY 2.
If there are 3 people, everyone sees 2, they all know nothing will happen on day 1. The question is only if something will happen on day 2, because WERE THERE ONLY TWO PEOPLE, THEY WOULD BOTH LEAVE DAY 2. If nothing happens on day 2, then all three chap that they must also be blue, which will result in all 3 leaving on day 3.
SO IF THERE ARE 3 PEOPLE, THEY ALL LEAVE DAY 3, RIGHT?
Now, if there are 4 people, they each see 3. IF THERE ARE 3 PEOPLE, THEY ALL LEAVE DAY 3, RIGHT? If nothing happens on day 3, all 4 chap they're blue and will leave on day 4.
SO IF THERE ARE 4 PEOPLE, THEY ALL LEAVE DAY 4.
If there are 5 people, they each see 4. WERE THERE 4 PEOPLE, THEY ALL LEAVE DAY 4. If nothing happens on day 4, they each know they're blue, so they all leave day 5.
Extra Credit, do you really need me to go till 40? It's all the same pattern.
According to the number of blue people each individual sees, that's the only day number they look out to see if anything happens.
I was upset when I couldn't solve this riddle to begin with. Now I take pride in understanding it. And the explanation I read was not nearly as clear as the one I gave you.
Hidden:
Seeing 39 people, they are not expecting ANY action until day 39. It's only day 39 that will be the test, whether there are taka 39 people, and if no action then, must be 40.
I know you keep on saying not to go with small numbers, but the logic always boils down to the last person: Knowing that at least one person is blue:
If there is 1 person, he sees no one else, and therefore knows that he's blue, so he leaves day 1.
If there are 2 people, each sees 1. Were there only 1 person, that person should be leaving on day 1. So if nothing happens on day 1, they both know they're blue and leave on day 2.
SO IF THERE ARE 2 PEOPLE, THEY BOTH LEAVE ON DAY 2.
If there are 3 people, everyone sees 2, they all know nothing will happen on day 1. The question is only if something will happen on day 2, because WERE THERE ONLY TWO PEOPLE, THEY WOULD BOTH LEAVE DAY 2. If nothing happens on day 2, then all three chap that they must also be blue, which will result in all 3 leaving on day 3.
SO IF THERE ARE 3 PEOPLE, THEY ALL LEAVE DAY 3, RIGHT?
Now, if there are 4 people, they each see 3. IF THERE ARE 3 PEOPLE, THEY ALL LEAVE DAY 3, RIGHT? If nothing happens on day 3, all 4 chap they're blue and will leave on day 4.
SO IF THERE ARE 4 PEOPLE, THEY ALL LEAVE DAY 4.
If there are 5 people, they each see 4. WERE THERE 4 PEOPLE, THEY ALL LEAVE DAY 4. If nothing happens on day 4, they each know they're blue, so they all leave day 5.
Extra Credit, do you really need me to go till 40? It's all the same pattern.
According to the number of blue people each individual sees, that's the only day number they look out to see if anything happens.
I feel bad for you. Must be boring to write the same megilla so many times. Of course you don’t need to explain this till 40 because I understood this right away.But bottom line, the first day they all see there are at least 39 so they decide now let’s take a nap till day 39?? Can’t they skip the nap step? That is my question all this time.
Ps I’d love to read the explanation where you read it originally. It might register better.
Last edited by ExtraCredit on Sun, Oct 18 2020, 1:49 pm; edited 2 times in total
You have this new routine of going shopping every Sunday. You decide to head towards the local bus station at a random time every Sunday and board whichever bus comes first.
There is the G20 bus that will get you to Plaza mall or the D20 bus that will get you to American Mall. The plan seems perfect because you know that both buses run every 20 minutes.
But after a few months you realize that you have visited American mall only once in 5 weeks. Why is this happening if you are indeed heading to the station at random times?
Hidden:
The D20 bus arrives 4 minutes later than the G20. So only if you arrive in those 4 minutes between these two buses, you'll get to American Mall, whereas you have 16 minutes between G20 and B20, which gives you much greater chance of getting to Plaza.
I feel bad for you. Must be boring to write the same megilla so many times. Of course you don’t need to explain this till 40 because I understood this right away.But bottom line, the first day they all see there are at least 39 so they decide now let’s take a nap till day 39?? Can’t they skip the nap step? That is my question all this time.
Ps I’d love to read the explanation where you read it originally. It might register better.
Answer to bolded: YES. Only day 39 will either confirm or nullify their assumption.
The D20 bus arrives 4 minutes later than the G20. So only if you arrive in those 4 minutes between these two buses, you'll get to American Mall, whereas you have 16 minutes between G20 and B20, which gives you much greater chance of getting to Plaza.
I feel bad for you. Must be boring to write the same megilla so many times. Of course you don’t need to explain this till 40 because I understood this right away.But bottom line, the first day they all see there are at least 39 so they decide now let’s take a nap till day 39?? Can’t they skip the nap step? That is my question all this time.
Ps I’d love to read the explanation where you read it originally. It might register better.
I get it. It’s similar reasoning to the riddle of 5 people wearing diff colored hat. The first person cannot guess color, he first has to wait for other 4 to say they don’t know theirs.
With every day that passes and no one leaves the island, it’s a silent confirmation that there is one more blue-eyed person that the previous day.
I get it. It’s similar reasoning to the riddle of 5 people wearing diff colored hat. The first person cannot guess color, he first has to wait for other 4 to say they don’t know theirs.
With every day that passes and no one leaves the island, it’s a silent confirmation that there is one more blue-eyed person that the previous day.
It's different in that everyone sees everyone else. And therefore they all know it's either 39 or 40. Were there 39, they all have to leave day 39 according to the logic previously explained. All days leading up to 39 would have only been considered had there been fewer people with blue eyes. Day 38 would be the test were there 39 people. Day 37 would be the test were there 38 people etc.
So they know no one will leave till day 39.
You have this new routine of going shopping every Sunday. You decide to head towards the local bus station at a random time every Sunday and board whichever bus comes first.
There is the G20 bus that will get you to Plaza mall or the D20 bus that will get you to American Mall. The plan seems perfect because you know that both buses run every 20 minutes.
But after a few months you realize that you have visited American mall only once in 5 weeks. Why is this happening if you are indeed heading to the station at random times?
Hidden:
If the statistical random distribution with big numbers is 4:1, this would imply that D20 leaves 4 minutes after G20... so the waiting time for G20 is 16 minutes, but the waiting time for D20 is only 4 minutes, so the chances to catch G20 vs. D20 are 4:1
I feel bad for you. Must be boring to write the same megilla so many times. Of course you don’t need to explain this till 40 because I understood this right away.But bottom line, the first day they all see there are at least 39 so they decide now let’s take a nap till day 39?? Can’t they skip the nap step? That is my question all this time.
Ps I’d love to read the explanation where you read it originally. It might register better.
No, I don't think they can skip the nap step...
Because none of the 40 knows it's him...
So they have a collective epiphany on day 39 when no-one boarded the plane...
And since they know there are at least 39, they do not expect any epiphany before day 39...
If the statistical random distribution with big numbers is 4:1, this would imply that D20 leaves 4 minutes after G20... so the waiting time for G20 is 16 minutes, but the waiting time for D20 is only 4 minutes, so the chances to catch G20 vs. D20 are 4:1
The difference between my answer and yours is the difference between a student and her professor.....
How can you get to the number 21 using 1, 5, 6 and 7? You may use each digit only once and standard arithmetic operations (addition, subtraction, multiplication and division) as many as you want. You may also use parentheses. Any other symbols are not allowed I.e. decimals.
You CANNOT combine two numbers. For example, 15 + 6 = 21...you can't combine 1 and 5 to make 15.
I'm glad to see someone else got it!
Extracredit, wake up!
Sorry, got busy baking and eating cookies. Let me check out the link. Before I do, I want to ask chaniemommy why it takes 39 days of napping to get the epiphany. But hold the answer till I check out the link. Might not need explanations anymore.
Sorry, got busy baking and eating cookies. Let me check out the link. Before I do, I want to ask chaniemommy why it takes 39 days of napping to get the epiphany. But hold the answer till I check out the link. Might not need explanations anymore.
Why, if there 2 persons with blue eyes, does not everybody jump on the plane the second day?
The only thing I gained from the link is that I’m asking exactly what he’s asking. His questions 2&3 are Word for word my questions too. I need an answer to those. the explanation he gives wasn’t different than your first one. In fact I gave that answer in the first place when you wanted us to assume there are only 2 blue eyed. I need answers to #2&3 but I’m exhausted from that one already. On to your 21 one.
