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re: Let's discuss the Monty Hall problem (probabilities, odds)
Posted on 8/12/17 at 1:11 pm to TigrrrDad
Posted on 8/12/17 at 1:11 pm to TigrrrDad
quote:
That's precisely what you are doing, because you won't show where my scenario is wrong
Dude. You've been shown
If you're not understanding now, I'm not going to be able to help you
And of course. There's the fact that this problem has already been tested hundreds, if not thousands of times.
But. You keep on keeping on
1
Posted on 8/12/17 at 1:12 pm to Spock's Eyebrow
quote:
If the host is an equal player, i.e. he's picking at random
That's not how Let's Make A Deal worked.
The thread title says "the Monty Hall problem" which is a well-known probability problem, not some random meme with no context. To pretend that the host would be choosing doors at random is disingenuous and you know it.
This post was edited on 8/12/17 at 1:13 pm
Posted on 8/12/17 at 1:12 pm to Jim Rockford
Nvm, already addressed
This post was edited on 8/12/17 at 1:14 pm
Posted on 8/12/17 at 1:13 pm to Sneaky__Sally
I'm considering every factor LISTED IN THE OP.
Posted on 8/12/17 at 1:14 pm to ShortyRob
quote:
Dude. You've been shown If you're not understanding now, I'm not going to be able to help you And of course. There's the fact that this problem has already been tested hundreds, if not thousands of times. But. You keep on keeping on
Again, you're talking about a different scenario than the OP. I listed every possible outcome.
Posted on 8/12/17 at 1:14 pm to TigrrrDad
quote:Because even though there are two doors, the odds that the car is behind door #1 are 33%, and the odds that the car is behind door #2 is 66%.
I choose door # 1. #3 is opened and reveals a goat.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
Show me how this is not 50/50.
You have a 50/50 choice, but not a 50/50 result. That's where you are going wrong.
Posted on 8/12/17 at 1:16 pm to PrimeTime Money
Someone simply add the other possible outcomes to my list.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
NOW ADD THE OTHER POSSIBLE OUTCOMES TO THIS LIST.
For it to be 66%, you have to be able to add two more winning outcomes to this list.
Do it.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
NOW ADD THE OTHER POSSIBLE OUTCOMES TO THIS LIST.
For it to be 66%, you have to be able to add two more winning outcomes to this list.
Do it.
This post was edited on 8/12/17 at 1:18 pm
Posted on 8/12/17 at 1:18 pm to TigrrrDad
quote:You still don't get it.
Someone simply add the other possible outcomes to my list.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
NOW ADD THE OTHER POSSIBLE OUTCOMES TO THIS LIST.
The car will only be behind door #1 33% of the time.
The car will be behind door #2 67% of the time.
It's not 50/50 that the car is behind door #1 and it's not 50/50 that the car is behind door #2.
This post was edited on 8/12/17 at 1:19 pm
Posted on 8/12/17 at 1:20 pm to PrimeTime Money
Show me. Add the two other outcomes to the list.
Posted on 8/12/17 at 1:20 pm to TigrrrDad
I'm not arguing the scenario that involves the car behind door #3 - I'm arguing the OP scenario.
Posted on 8/12/17 at 1:21 pm to PrimeTime Money
No use arguing with him. He's embarrassed that he's too stupid to understand probability so now he's pretending that Monty Hall would open a door at random instead of purposefully choosing one with a goat.
Posted on 8/12/17 at 1:22 pm to TigrrrDad
The person who pointed out that it's easier to visualize with more doors is dead on.
If there are 5 doors, you chances of being right when you pick are one in 5. The host is never going to reveal the car, so, in the games where you didn't pick right, he's going to reveal the three OTHER misses. IE, you picked one of the 4 bad doors.
When 3 doors are involved, you are doing to be right 1 out of 3 times when you pick.
Obviously, that means when you pick right and switch, you lose. So. You'll lose 1 out of 3 times.
But, in the cases where you pick wrong, Monty is then going to reveal the OTHER miss.
Hence, in those cases, when you switch you win.
Hence, if you play 10,000 times and switch every time, you'll win 2/3 of the time.
Frankly, the people talking odds after the first pick are missing the point.
The point is, move away from your odds of being right.
Your odds of being WRONG when you first picked are 2 in 3.
The host is ALWAYS going to reveal a miss which means that in those two instances, switching always wins.
Like I said. If there were 5 doors, it would be clear as day to you. If you picked and then he revealed 3 doors........and you stood pat, you're a moron.
How about 10 doors?
100?
Do you get it now? The host is HELPING you when he reveals YOUR other misses.
If there are 5 doors, you chances of being right when you pick are one in 5. The host is never going to reveal the car, so, in the games where you didn't pick right, he's going to reveal the three OTHER misses. IE, you picked one of the 4 bad doors.
When 3 doors are involved, you are doing to be right 1 out of 3 times when you pick.
Obviously, that means when you pick right and switch, you lose. So. You'll lose 1 out of 3 times.
But, in the cases where you pick wrong, Monty is then going to reveal the OTHER miss.
Hence, in those cases, when you switch you win.
Hence, if you play 10,000 times and switch every time, you'll win 2/3 of the time.
Frankly, the people talking odds after the first pick are missing the point.
The point is, move away from your odds of being right.
Your odds of being WRONG when you first picked are 2 in 3.
The host is ALWAYS going to reveal a miss which means that in those two instances, switching always wins.
Like I said. If there were 5 doors, it would be clear as day to you. If you picked and then he revealed 3 doors........and you stood pat, you're a moron.
How about 10 doors?
100?
Do you get it now? The host is HELPING you when he reveals YOUR other misses.
Posted on 8/12/17 at 1:22 pm to dbeck
He's going to open a door with a goat every time, right? Every single time, correct?
Posted on 8/12/17 at 1:23 pm to dbeck
If it is 66%, there will be two more win outcomes. Show me.
Posted on 8/12/17 at 1:23 pm to TigrrrDad
OK
IF MONTY OPENS AT RANDOM
Your probability of being correct with your initial pick is 1/3, and the same 1/3 if you stick with it after Monty's pick
There is 1/3 chance of you losing by Monty revealing the car
And there is 1/3 chance of you winning by switching if he reveals a goat
SO
If you look at it this way, your odds can be "50/50" because your odds of winning are the same AT THAT POINT no matter which door you choose. BUT, if Monty chooses randomly, then your odds of making it that far are only 2/3. So you're looking at 50% of 2/3, which is 1/3 odds of winning.
IF MONTY OPENS AT RANDOM
Your probability of being correct with your initial pick is 1/3, and the same 1/3 if you stick with it after Monty's pick
There is 1/3 chance of you losing by Monty revealing the car
And there is 1/3 chance of you winning by switching if he reveals a goat
SO
If you look at it this way, your odds can be "50/50" because your odds of winning are the same AT THAT POINT no matter which door you choose. BUT, if Monty chooses randomly, then your odds of making it that far are only 2/3. So you're looking at 50% of 2/3, which is 1/3 odds of winning.
Posted on 8/12/17 at 1:23 pm to TigrrrDad
quote:Then you've chosen to feign ignorance of what the Monty Hall problem is in order to preserve your dignity.
I'm not arguing the scenario that involves the car behind door #3 - I'm arguing the OP scenario.
That's fine.
You're not the first down this road.
Posted on 8/12/17 at 1:24 pm to TigrrrDad
You are ignoring the fact that the car was placed at the beginning of the problem. The host revealing the incorrect door and then gives you the option to stay or choose. What he is doing from a probability perspective is saying "would you like to choose the door you picked or both of the doors you originally didn't pick - thus, switching gives you the probability of both doors in your favor - or 2/3 chance.
Posted on 8/12/17 at 1:24 pm to TigrrrDad
quote:
If it is 66%, there will be two more win outcomes. Show me.
Posted on 8/12/17 at 1:25 pm to Korkstand
quote:
Korkstand
Gets it.
Your odds were 1 in 3 that you picked right at the outset. This means that 2 out of 3 times you've missed If Monty is just a doofus and opens at random and reveals a goat, he's revealed your MISS in the situations where you missed.
Posted on 8/12/17 at 1:25 pm to ShortyRob
You don't get it. I'm not arguing the odds of the 3 doors game when switching - I'm arguing the specific scenario in the OP. The explanation of the "switch odds" includes the car being behind #3. That is NOT the scenario in the OP. You people are arguing a whole different scenario than the OP. I listed every possible outcome of the Op scenario. If it is different, you should be able to list 2/3 of the scenarios as wins.
This post was edited on 8/12/17 at 1:27 pm
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