Lecture Scene
- Thread starter kennybenny
- Start date
well i think i'm begining to understand now lol. that's a good example.
making me think about the situation where i18 for 10vsA where dealer peeks is tc=4 ...... if the dealer doesn't peek maybe you'd think twice about doubling at tc=4 sort of thing even where you don't know the exact cards left in the pack.
making me think about the situation where i18 for 10vsA where dealer peeks is tc=4 ...... if the dealer doesn't peek maybe you'd think twice about doubling at tc=4 sort of thing even where you don't know the exact cards left in the pack.
sagefr0g said:
well i think i'm begining to understand now lol. that's a good example.
making me think about the situation where i18 for 10vsA where dealer peeks is tc=4 ...... if the dealer doesn't peek maybe you'd think twice about doubling at tc=4 sort of thing even where you don't know the exact cards left in the pack.
making me think about the situation where i18 for 10vsA where dealer peeks is tc=4 ...... if the dealer doesn't peek maybe you'd think twice about doubling at tc=4 sort of thing even where you don't know the exact cards left in the pack.
RJT.
Actually, that's the precise reason why it's NOT 50/50. If the host were to open a RANDOM door, which happened to contain a goat, then out of the remaining two it would be 50/50. To better understand this, consider a more extreme example. I tell you I'm thinking of a number between 1 and 1,000,000. I ask you to take a guess at it. You say 451,999. Then I tell you my number is either 451,999 or 235,633. I give you the oppurtunity to switch. Do you? Of course. What I did is intentionally remove all the other numbers (doors with goats) and say "these are wrong." Does this help make it more clear?
The best way to get your head round this is to consider the 3 possible scenarios
Door 1 = Car
Door 2 = Goat
Door 3 = Goat
Scenario 1
You choose door 1
If you stick you get the car
If you change you get a goat
Scenario 2
You choose door 2
If you stick you get a goat
If you change you get the car
Scenario 3
You choose door 3
If you stick you get a goat
If you change you get a car.
If you stick with your original choice, you'll only get the car one out of 3 times.
If you change your original choice after one door has been removed you'll get the car in 2 out of the possible 3 scenarios.
RJT.
Door 1 = Car
Door 2 = Goat
Door 3 = Goat
Scenario 1
You choose door 1
If you stick you get the car
If you change you get a goat
Scenario 2
You choose door 2
If you stick you get a goat
If you change you get the car
Scenario 3
You choose door 3
If you stick you get a goat
If you change you get a car.
If you stick with your original choice, you'll only get the car one out of 3 times.
If you change your original choice after one door has been removed you'll get the car in 2 out of the possible 3 scenarios.
RJT.
When I figured it out for myself in laymen terms this is how I did so and I think it will make it easier for others to see who aren't heavy in the math.
Take 3 cups and put a key under cup 1. We will do 2 sets of 3 trials, the first set we will stick with our original choice after given the opportunity to switch. The 2nd trial we will switch to the other cup.
So we'll start with the key under cup 1, X=cups, 0=key
X X X
O
So by not changing we got it right and scored 1/1
Now the key has moved to cup 2
X X X
0
Now we got it wrong because we did not change so our score is 1/2
Last the key is in cup 3
X X X
0
We stuck with cup 1 and got it wrong again so our score is 1/3 or 33%
Now we will try changing our cup after one cup is revealed wrong:
Again we start with the key under cup 1 and our original choice is #1 but this time we change it to #2 or #3, the opposite of which of the two is revealed as the "goat" cup. Goat cup=*
X X X
O *
In this case we got it wrong because we changed from cup 1 to cup 3 after #2 was revealed to us. Our score is 0/1
2nd scenario:
X X X
0 *
Again our original choice was #1 but then when we saw #3 was wrong we switched to #2 and got it right so our score is 1/2.
3rd scenario
X X X
* 0
Here we pick #1, was shown #2 was nothing so we switched to #3 and got it right, so our final score is 2/3 or 67% compared to 33% in the first trial.
Take 3 cups and put a key under cup 1. We will do 2 sets of 3 trials, the first set we will stick with our original choice after given the opportunity to switch. The 2nd trial we will switch to the other cup.
So we'll start with the key under cup 1, X=cups, 0=key
X X X
O
So by not changing we got it right and scored 1/1
Now the key has moved to cup 2
X X X
0
Now we got it wrong because we did not change so our score is 1/2
Last the key is in cup 3
X X X
0
We stuck with cup 1 and got it wrong again so our score is 1/3 or 33%
Now we will try changing our cup after one cup is revealed wrong:
Again we start with the key under cup 1 and our original choice is #1 but this time we change it to #2 or #3, the opposite of which of the two is revealed as the "goat" cup. Goat cup=*
X X X
O *
In this case we got it wrong because we changed from cup 1 to cup 3 after #2 was revealed to us. Our score is 0/1
2nd scenario:
X X X
0 *
Again our original choice was #1 but then when we saw #3 was wrong we switched to #2 and got it right so our score is 1/2.
3rd scenario
X X X
* 0
Here we pick #1, was shown #2 was nothing so we switched to #3 and got it right, so our final score is 2/3 or 67% compared to 33% in the first trial.
Try looking at this with common street logic. First off, there are definitely 3 doors to choose from, and you have to pick 1 totally in the blind -- so you do. Since the blind probabilities of each door hiding the car are 33%, there is positively a one-third chance that you've picked the door with the car. If you always stay with that door come hell-or-high-water, you'll win the car one-third of the time -- period.
But wait! The game show host is about to give you more information.
After you've made your pick, 2 times out of 3 the host will have a car and a goat left over.
One time out of 3, he'll be left with two goats.
But no matter which time is which, he'll always deliberately expose a goat!
So then, 2 times out of 3, he must be exposing the only goat available, because the other door will have the car behind it. Since you realize this, the probabilities of the doors have just gone from 33/33/33 to 33/0/66.
You're holding a 33% shot, just as you were in the beginning, the exposed goat door is now 0%, and the unchosen door has become a 66% shot because 2 times out of 3 the host will have only 1 goat to expose.
It's important to understand that if the host opened 1 of the 2 unchosen doors blindly and it just happened to be a goat, then you'd be left with a 50/50 proposition. But he in fact chooses which door to open, making sure it's a goat, and that gives you additional information.
But wait! The game show host is about to give you more information.
After you've made your pick, 2 times out of 3 the host will have a car and a goat left over.
One time out of 3, he'll be left with two goats.
But no matter which time is which, he'll always deliberately expose a goat!
So then, 2 times out of 3, he must be exposing the only goat available, because the other door will have the car behind it. Since you realize this, the probabilities of the doors have just gone from 33/33/33 to 33/0/66.
You're holding a 33% shot, just as you were in the beginning, the exposed goat door is now 0%, and the unchosen door has become a 66% shot because 2 times out of 3 the host will have only 1 goat to expose.
It's important to understand that if the host opened 1 of the 2 unchosen doors blindly and it just happened to be a goat, then you'd be left with a 50/50 proposition. But he in fact chooses which door to open, making sure it's a goat, and that gives you additional information.
In my opinion, this is the easiest way to understand it. The person who chose a door has more information than a random person who walks in later. It's absolutely true that if a random person were to walk into a room with two unopened doors, that random person would have a 50%-50% chance of picking the right one. But the person there since the beginning has a better chance because they're getting conditional probability, not straight probability.
zengrifter said:
Quote:
Originally Posted by E-town-guy
The fact is, if you choose door #1 then they eliminate door #3 how can there be a benefit to switching to #2. Both #1 and #2 have the same odds 50/50. There is just as much reason to switch as there is not to.
Thats what I said! zg
Originally Posted by E-town-guy
The fact is, if you choose door #1 then they eliminate door #3 how can there be a benefit to switching to #2. Both #1 and #2 have the same odds 50/50. There is just as much reason to switch as there is not to.
Thats what I said! zg
you can't just look at the situation with just the 2 doors left.
you have to look at the situation BOTH before the goat is exposed and after the goat is expose. that way, there is ONLY THREE scenarios that can happen.
1) choose right door (door #1), one of the two goat is exposed and if you decide to change your decision, you will loose (door #2 and #3 are goats)
2) choose wrong door (door #2), the other goat (door #3) is exposed and if you decide to change your decision you win (door #1)
3) choose wrong door (door #3), the other goat is exposed (door #2) and if you decide to change your decision you win (door #1)
ONE of these 3 scenarios will occur, and there is no other combination of events which can happen.
of the 3 scenarios, two of which you are destined to win if you change your decision. thus it is 2/3 of chance of winning if you change your decision.
you have to look at the situation BOTH before the goat is exposed and after the goat is expose. that way, there is ONLY THREE scenarios that can happen.
1) choose right door (door #1), one of the two goat is exposed and if you decide to change your decision, you will loose (door #2 and #3 are goats)
2) choose wrong door (door #2), the other goat (door #3) is exposed and if you decide to change your decision you win (door #1)
3) choose wrong door (door #3), the other goat is exposed (door #2) and if you decide to change your decision you win (door #1)
ONE of these 3 scenarios will occur, and there is no other combination of events which can happen.
of the 3 scenarios, two of which you are destined to win if you change your decision. thus it is 2/3 of chance of winning if you change your decision.
Here's a really easy way to "understand" it.
You agree that your odds of picking a goat are 2/3 when you first guess, right? Of course you do.
The gameshow host will reveal a goat no matter what, right? Of course.
Based on this alone, IF YOU PICK A GOAT INITIALLY, THE GAMESHOW HOST IS FORCED TO ELIMINATE THE ONLY OTHER "WRONG" ANSWER. THUS, IF YOU ALWAYS SWITCH YOUR ANSWER (and happened to pick a goat), YOU ALWAYS WIN.
You are banking on choosing a goat with your first choice. If you do, and switch, you always win.
Not too tough to grasp.
You agree that your odds of picking a goat are 2/3 when you first guess, right? Of course you do.
The gameshow host will reveal a goat no matter what, right? Of course.
Based on this alone, IF YOU PICK A GOAT INITIALLY, THE GAMESHOW HOST IS FORCED TO ELIMINATE THE ONLY OTHER "WRONG" ANSWER. THUS, IF YOU ALWAYS SWITCH YOUR ANSWER (and happened to pick a goat), YOU ALWAYS WIN.
You are banking on choosing a goat with your first choice. If you do, and switch, you always win.
Not too tough to grasp.
At the tables its nothing more than maybe 7th grade math, it is the back end that has the real math behind it all. QFIT can attest to this I am sure, it is a whole other world.. Honestly I dont care about the math, what I do care about is how to play the hand in front of me and how I should play the hand correctly.