Namaiki Asahi-chan wo Wakarasetai - Ch. 2 - Senpai, You Idiot

[MangaDex]

 

[Zenerz2]

An Otaku who is also good at study. Close enough, Welcome Back Satoshi

 

[stonks4251]

Idiots the both of them lol
Thanks for the translation

 

[Mangamoz]

Holy shit HE'S the real mesugaki!

 

[MrGrinns]

If one die lands on top of the other die, the bottom die remains in play, regardless of if the face is obscured. Thus the lowest possible number remains two

 

[Angelonius]

Idiots the both of them lol
Thanks for the translation

They need to get married.

 

[Azimfirst]

Its 16% I guess?

 

[aveney41]

She's precious haha

 

[TrapTrash1678]

I fucking lost it when she started rolling the dice.

 

[MasterPannya]

I swear to God this is a shoujo manga.

 

[YosepRA]

Ehh.... 1/6? There are 6 sides on each die, power it by 2 because we have 2 dies, there goes 36 combinations.

Favorable outcomes are (1, 1), (2, 2), ..., (6, 6), therefore we have 6 favorable outcomes.

6 / 36 = 1 / 6.

Maybe. 😭

 

[Versa]

Ehh.... 1/6? There are 6 sides on each die, power it by 2 because we have 2 dies, there goes 36 combinations.

Favorable outcomes are (1, 1), (2, 2), ..., (6, 6), therefore we have 6 favorable outcomes.

6 / 36 = 1 / 6.

Maybe. 😭

I believe your correct, though perhaps overcomplicated it. The first die doesn't actually matter to 'doubles', any of its faces can be part any resulting double. The only factor is that the second die must land on the same side as the first so a 1/6 right there.

 

[wallsg]

I believe your correct, though perhaps overcomplicated it. The first die doesn't actually matter to 'doubles', any of its faces can be part any resulting double. The only factor is that the second die must land on the same side as the first so a 1/6 right there.

Some people don't get statistics and probability though. Many don't understand the difference between "what are the odds you'll flip 50 heads in a row" and "after flipping 49 heads in a row what are the odds of the next flip being a head".

Myself, I can't get my head around the Monty Hall problem. Just because you're shown that one of the unselected doors is the booby prize why does it make the other unselected door have a 2/3 probability of winning and the selected door only a 1/3 probability of winning? My mind will not budge off of 50/50. I can see the results empirically at Better Explained, but still... The best explanation for me: "To lose by switching you would have had to have picked the good door the first time, which only has a 1/3 chance."

 

[jc9]

Some people don't get statistics and probability though. Many don't understand the difference between "what are the odds you'll flip 50 heads in a row" and "after flipping 49 heads in a row what are the odds of the next flip being a head".

Myself, I can't get my head around the Monty Hall problem. Just because you're shown that one of the unselected doors is the booby prize why does it make the other unselected door have a 2/3 probability of winning and the selected door only a 1/3 probability of winning? My mind will not budge off of 50/50. I can see the results empirically at Better Explained, but still... The best explanation for me: "To lose by switching you would have had to have picked the good door the first time, which only has a 1/3 chance."

this is tricky because eliminating the door makes people believe that suddenly they are in a completely new situation and they have to choose one of 2 doors, when in reality it's allowing them to choose - simultaneously - ALL of them, except the one they selected at first.

to unpack it:

• you pick one door essentialy at random. each individual door has 1/n chance of success (1/3 in original game). other, unselected doors combined have (n-1)/n chance of winning, 2/3 in original game.
• eliminating all/all but one duds from unselected doors and allowing you to switch is the same as changing the rules to "you can switch to select all of these n-1 doors, and if there's a prize in one of them you win" - from math pov it's exactly the same thing.
• your choice comes to staying with your initial single door with 1/n chance, or switching to n-1 doors, each of which had initially 1/n chance also - but you get n-1 of them instead of only 1. so with 3 doors, you have your initial single door with 1/3 chance, or you can switch it for 2 doors, each with 1/3 chance, for a total of 2/3 probability of winning

with 100 doors it's even easier to understand (and math stays the same) - you have a choice between a single door with 1/100 chance, or 99 doors with 99/100 chance in total.

hell lets use lotto with same rules as another example (because math will be the same again) - you select numbers, and have 1 in millions/billions chance of winning the grand prize. switching door in this case is esentially allowing you to switch your coupon to a grand prize for all numbers EXCEPT the one you initially selected.

 

[wallsg]

meant to be a reply

 

[wallsg]

this is tricky because eliminating the door makes people believe that suddenly they are in a completely new situation and they have to choose one of 2 doors, when in reality it's allowing them to choose - simultaneously - ALL of them, except the one they selected at first.

to unpack it:

• you pick one door essentialy at random. each individual door has 1/n chance of success (1/3 in original game). other, unselected doors combined have (n-1)/n chance of winning, 2/3 in original game.
• eliminating all/all but one duds from unselected doors and allowing you to switch is the same as changing the rules to "you can switch to select all of these n-1 doors, and if there's a prize in one of them you win" - from math pov it's exactly the same thing.
• your choice comes to staying with your initial single door with 1/n chance, or switching to n-1 doors, each of which had initially 1/n chance also - but you get n-1 of them instead of only 1. so with 3 doors, you have your initial single door with 1/3 chance, or you can switch it for 2 doors, each with 1/3 chance, for a total of 2/3 probability of winning

with 100 doors it's even easier to understand (and math stays the same) - you have a choice between a single door with 1/100 chance, or 99 doors with 99/100 chance in total.

hell lets use lotto with same rules as another example (because math will be the same again) - you select numbers, and have 1 in millions/billions chance of winning the grand prize. switching door in this case is esentially allowing you to switch your coupon to a grand prize for all numbers EXCEPT the one you initially selected.

I still think "To lose by switching you would have had to have picked the good door the first time, which only has a 1/3 chance (of happening)." is more intuitive. (I will grant you that it is not a rigorous explanation.) KNOWING something is right and FEELING something is right can sometimes be two different things. It is so "obvious" that when all of those bad n-2 doors were removed that you only have two doors remaining so each has an equal chance of being the winner. When you can fix in your head that the chance you picked the right door in the first place was 1/n and switching would win in all other (n-1)/n cases then it's a little more intuitive.

 

[BattleKangaroo]

Probability of getting doubles:
Number of favourable outcomes / Total number of possible outcomes = 6/36 = 1/6 = 16.66%.

 

[nguuuquaaa]

Some people don't get statistics and probability though. Many don't understand the difference between "what are the odds you'll flip 50 heads in a row" and "after flipping 49 heads in a row what are the odds of the next flip being a head".

Myself, I can't get my head around the Monty Hall problem. Just because you're shown that one of the unselected doors is the booby prize why does it make the other unselected door have a 2/3 probability of winning and the selected door only a 1/3 probability of winning? My mind will not budge off of 50/50. I can see the results empirically at Better Explained, but still... The best explanation for me: "To lose by switching you would have had to have picked the good door the first time, which only has a 1/3 chance."

Let's use a bigger number to visualize it easier. Say, 100 doors.

Split the doors into 2 groups: group A with the door you choose, and group B with the other 99 doors.
Obviously group B has a 99/100 chance of just having the winning door in it, doesn't matter where.
The thing with this problem is that people think that the host did something that didn't matter (open 98 losing doors), but in truth is that the host "filtered" 98 losing doors from group B, leaving 1 single door as the representative of group B.
Switching door meaning going from group A to group B. So 99/100 chance of winning if switching.

Tl;dr switching is like selecting 99 doors at the same time.

 

[wakawa00]

There's an error in page 11, on the third panel both speech bubbles have the same duplicate text

 

[JacksDemonHole]

So I'll take it that this is a gambling manga after all??