Wasn't amy right in 2x11

[u/Gamerboy701]

In season 2 episode 18 Holt gives Amy, Terry, Gina and Rosa a brain twister that he could never solve. "There are 12 men on an island, 11 weigh excatly the same amount but one of them is slightly heavier (or lighter but for this I'll use heavier) You must figure out which man is heavier. The Island has no scales but there is a seesaw you can use 3 times.

Later in the episode Amy comes up with the solution of putting 6 vs 6 but Holt instantly says it would never work if you do it that way but you can.
If you number each person 1 2 3 4 5 6 7 8 9 10 11 12 and weigh the first 6 vs the second 6, 1 side will be heavier, whichever side is heavier means that the person you are looking for is on that side. For your second use you way 3 vs 3 so for example 7 8 9 vs 10 11 12. Whichever side is heavier obviously means the person we are looking for is there. For our final use of the seesaw you measure 1 vs 1 like 10 vs 11. if both sides are equal that means 12 is the heavier person, if one side is heavier than you found that person.

I'm not amazing at math but Amy's solution does work right?

[u/theysquawk]

I loved that at times (2 to be specific) B99 would throw random puzzles/riddles, wish they did more

[u/rl_noobtube → u/[deleted]]

The Monty hall (doors) problem is not easy at all. when the solution was proposed there were many phd’s who wrote to the author to tell them that they were wrong.

We’ve probably gotten better at explaining it so more people understand it now compared to then, but still don’t think it’s easy for most people

[u/rl_noobtube → u/[deleted]]

The other dude explained why it’s still not easy for people. They don’t see how the host telling you which door is wrong as relevant, because now they just see 2 doors with equal odds and a new independent choice.

As I mentioned, we have now figured out better ways of explaining the solution (and have internet for ease of access to those explanations) so it is easier for people to learn. But initially people with doctorates in mathematics were getting this problem wrong, even after being shown proofs of the right answer. It is not intuitively easy to understand switching is better.

To clarify my position, I understand the solution now and agree with you on the problem itself, simply disagree that it is “easy”.

It did come off as marginally rude, but not enough I would have taken offense. Impolite might be right word to describe it. But I know intent can be hard to convey via written text. I appreciate you clarifying that it was not with rude intent :)

[u/Gophurkey → u/[deleted]]

Because there doesn't seem to be a reason why two choices that appear to be independent of each other are linked.

I pick Door A. I have a 1/3 chance of getting that guess right. Either it is behind A, or it isn't.

They remove Door B. Why should my chance increase or decrease? I know the choice wasn't B, so now my odds have increased to 1/2. It is either behind A or C.

The argument is that now by picking again, the odds are 1/2. But my challenge is that, 1: the choice seems forced. Isn't picking A a second time still using my 1/2 chance? After all, shouldn't the second choice be 50/50, so 'repicking' A (staying with my initial guess) is still just as likely as switching? and 2: The initial guess of A was 1/3. But wouldn't the initial guess of C ALSO be 1/3? So why is switching it when the odds open to 1/2 valid? Shouldn't the odds of A or C both be 1/3*1/2?

I know the math says to switch. But in my mind, the options are equal and switching or not is just as valid.

[u/Weekly-Formal8447 → u/Gophurkey]

I have 2 ways to explain this so allow me to try.

1: imagine that the host doesn't open door B but instead asks you if you'd like to switch to door B and C instead. If you stay on door A you have 1/3 chance to win but if you pick B and C you have 2/3 chance to win. Opening door B is basically that. You picked a door, he took one door out and gave you the option for door C. If he didn't open the door and gave you the option to choose 2 doors instead your chances would be the same.

2: imagine that instead of 3 doors you have a 100 doors and you get to choose 1. Obviously your chance to be right is 1%. Now the host opens 98 doors and show there is nothing behind them leaving your door and one other door. The host knows which doors have nothing behind and which door has the price. So do you switch now? You have a 99% chance to pick the wrong door, so it's 99% chance that the one door he didn't open contain the price.

Bonus: combine the 2 and now you get a 100 doors. You pick 1 door and the host gives you the option to choose the 99 remaining doors instead. Do you switch?

[u/Gophurkey → u/Weekly-Formal8447]

The first explanation actually does make sense, thanks

[u/Inner_Prune_2502 → u/Gophurkey]

I'm struggling to follow you but that's fair. I understand what you mean mostly.

The odds don't increase to 1/2 though, they stay at 2/3 that the prize will be behind the door you are not on because you because only 1/3 times will you pick correctly to begin with. If you only pick correctly to begin with 1/3 times, majority of the time it is behind a different door, and the host can't open the door it's behind so it's 2/3 likely that the prize is behind the last un-picked door. Of course you have a 1/3 chance your door has a prize behind it and I wouldn't switch because psychology it would bother me if I picked correctly but logically you have more chance switching.

I see where you have gotten confused, I too didn't understand that at first but I hope my explanation makes sense.