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Started By
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re: Let's Find Out How Smart You Baws Are
Posted on 3/24/24 at 1:43 am to drexyl
Posted on 3/24/24 at 1:43 am to drexyl
quote:
Take the switch every time
Not sure about this, just seems like a reverse of the gambler's fallacy. I mean statistically speaking tails hitting 20 times in a row is almost impossible, but after 19 in row it's still 50/50 to hit tails the next time.
In the shell game just because you were more likely to lose at first, it should still be 50/50 for where the correct one is. Each act is independent.
This post was edited on 3/24/24 at 2:08 am
5
Posted on 3/24/24 at 3:06 am to AUstar
Well staying or switching is also a 50/50 probability of picking the correct one.
Posted on 3/24/24 at 3:10 am to Guess
quote:
In the shell game just because you were more likely to lose at first, it should still be 50/50 for where the correct one is. Each act is independent.
Here is a Simulation where you can run it thousands of times very quickly and it will show you the results. You need to run it a few thousand times for it to steady out.
(Set simulation to 1000 and instant results.)
This post was edited on 3/24/24 at 3:11 am
Posted on 3/24/24 at 7:35 am to Guess
quote:
Not sure about this, just seems like a reverse of the gambler's fallacy. I mean statistically speaking tails hitting 20 times in a row is almost impossible, but after 19 in row it's still 50/50 to hit tails the next time.
The key difference is that in the gambler's fallacy scenario, there is no predicated choice; it isn't being manipulated by someone with certain knowledge of the outcome. In the shell scenario, someone is intentionally "flipping tails" each time, until the last remaining shell. The host's chance of flipping over the shell with money is 0% every time, for every shell up until the last one. Your chances of having the shell with money is still 1/N (N=number of shells). Nothing has changed that.
Basically, the more shells the host flips over, the better your chances are of winning by switching your pick to one of the remaining shells.
This post was edited on 3/24/24 at 7:44 am
Posted on 3/24/24 at 9:06 am to Guess
quote:
Each act is independent.
Incorrect
Posted on 3/24/24 at 9:09 am to Nutriaitch
quote:
there is also a 2/3 chance the one on the table is wrong. EXACTLY like the one in your hand
That makes no fricking sense
Posted on 3/24/24 at 10:08 am to FutureMikeVIII
quote:
That makes no fricking sense
See the simulation the guy posted above. Go run it 100 times and you will see that keeping your original choice loses about 67% of the time.
For instance, I just ran it 100 times and ended up losing 63% of the time if I kept my original selection. The more times you run it, the closer it will get to the 33/67 probability. This is a basic fact of statistics - the more data you have, the more the trends show up in the data.
Seeing a simulation is what convinced Princeton mathematician Paul Erdos. He was also skeptical until he saw a guy run it on a computer.
This post was edited on 3/24/24 at 10:12 am
Posted on 3/24/24 at 10:15 am to Guess
quote:
Not sure about this,
Then ignore the statistics.
quote:
In the shell game just because you were more likely to lose at first, it should still be 50/50 for where the correct one is. Each act is independent.
Wrong.
Posted on 3/24/24 at 10:27 am to WestSideTiger
quote:
Well staying or switching is also a 50/50 probability of picking the correct one.
Nope.
If you stick with your original choice your odds always remain 1/3 because you are not taking advantage of the new information. Just imagine 100 shells. Do you think your odds ever improve past the initial 1% just because losers keep getting revealed? No, because you didn't have that information when you chose that shell. But if you wait for more information, your odds can improve drastically by making a new choice (imagine waiting until 98 losers are revealed).
Posted on 3/24/24 at 10:44 am to Guess
quote:
Not sure about this, just seems like a reverse of the gambler's fallacy. I mean statistically speaking tails hitting 20 times in a row is almost impossible, but after 19 in row it's still 50/50 to hit tails the next time.
So many people are referencing the gambler’s fallacy. That’s not what’s happening here at all.
If you play the game a million times, you will still win around 667,000 times. If you play ten million times, you will win around 6,667,000 times.
The probability of winning is 2/3 if you switch. Always.
Posted on 3/24/24 at 10:52 am to Nutriaitch
quote:
and if you choose the right shell at the beginning, you will always switch to the losing shell.
Exactly, and there’s a 1/3 chance of picking the right shell at the beginning.
So, you have a 1/3 chance of losing if you switch, which implies a 2/3 chance of winning if you switch. That’s the point.
This post was edited on 3/26/24 at 4:06 pm
Posted on 3/24/24 at 10:53 am to FutureMikeVIII
quote:
there is also a 2/3 chance the one on the table is wrong. EXACTLY like the one in your hand
quote:
That makes no fricking sense
instead of everyone just saying I'm wrong, explain how.
A.
B.
C.
it's in one of those 3.
1/3 chance for each.
take away C.
now it's in either A or B
one of those 2.
the odds of which one doesn't change based on which is in your hand or which is on the table.
A and B have the exact same odds of being correct and the exact same odds of being wrong.
thing away C doesn't change odds, it's just to make you THINK they did.
A and B are equal.
you're thinking that you're most likely wrong from the beginning.
the dealer taking away a wrong answer makes you think that the other one must be right.
but in reality, B still has only a 33% chance of having been right.
just like A does.
my question that nobody has answered is why does B get all of C's odds when C is removed and A gets nothing?
that's the part I'm not understanding.
A & B were equal when C was still in the mix.
Why does removing C make them suddenly so drastically different?
the only difference between the two is that one is in my hand and the other is on the table.
A (in my hand) started the game with a 2/3 chance of being wrong. Remove C and ok, the one on the table must have 2/3 chance of being right now.
except: B (on the table) started the game with a 2/3 chance of being wrong. remove C. now the other one, A, must have 2/3 chance of being right.
why does the 2/3 logic only apply one way and not the other?
Posted on 3/24/24 at 11:03 am to AUstar
quote:
For instance, I just ran it 100 times and ended up losing 63% of the time if I kept my original selection. The more times you run it, the closer it will get to the 33/67 probability. This is a basic fact of statistics - the more data you have, the more the trends show up in the data.
Yeah that’s the law of large numbers . There’s no escaping it.
Posted on 3/24/24 at 11:14 am to Nutriaitch
quote:
thing away C doesn't change odds, it's just to make you THINK they did.
It does change the odds, though. The dealer isn’t randomly removing one, he always removes an empty one.
Posted on 3/24/24 at 11:25 am to TutHillTiger
quote:
What is the airspeed velocity of an unladen swallow?
African or European?
Posted on 3/24/24 at 12:39 pm to Nutriaitch
quote:
B still has only a 33% chance of having been right.
just like A does.
my question that nobody has answered is why does B get all of C's odds when C is removed and A gets nothing?
Because you said it yourself: B has a 33% chance of being right.
There are only two empty shells. Any time you pick an empty shell at the start, the dealer will always reveal the other empty shell.
Break it down into two scenarios. You either pick the money shell, or an empty shell at first. Nothing else is possible.
SCENARIO A: You pick an empty shell at the start
Assume you pick an empty shell at the beginning. If you switch from an empty shell, you automatically pick the money shell. Why? Because the dealer already revealed the other empty shell, so you can’t possibly switch to it.
This scenario has a 2/3 chance of happening since 2 of 3 shells are empty.
SCENARIO B: You pick the money shell
If you pick the money shell at the beginning, then when the time comes, you will automatically switch to an empty shell. This is because the dealer reveals one of the empty shells, so you will end up switching to the other.
This scenario has a 1/3 chance of happening since 1 of the 3 shells has the money.
Posted on 3/24/24 at 1:23 pm to AUstar
quote:I understand the Monty Hall problem. I was responding to the dude that said both of the remaining shells would have 2/3 probability of being right. Which is obviously illogical.
See the simulation the guy posted above. Go run it 100 times and you will see that keeping your original choice loses about 67% of the time.
Posted on 3/24/24 at 3:01 pm to AUstar
quote:
Only baws who scored 30+ on the ACT
thats cute.
Posted on 3/26/24 at 2:09 pm to jtweezy
obviously, i disagree. your first pick was a 33% probability, since there are three shells. take away a shell, and your pick now has a 50% chance of being right and a 50% chance of being wrong. the only thing that changes the probability is the number of shells.
Posted on 3/26/24 at 2:14 pm to AUstar
switch.
It's the door problem originally.
It's the door problem originally.
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