[quote]LIFTICVSMAXIMVS wrote:
johnnytang24 wrote:
You still haven’t answered my question. Which line(s) are incorrect?
Everything. Your theory is wrong.
Your script thinks there are three options because you told it so. There are ONLY 2. Change that.
Sheesh! What is wrong with programmers?[/quote]
The script has only two options: Switch and Don’t Switch. Where do you see three options?
[quote]Sick Rick wrote:
Math is my mortal enemy…
You have 3 options, 1 is revealed, leaving you with 2 other options. The new chances would be 50/50.
The 1/3-2/3 chance reflects back to all three doors being shut. One is open, but it doesn’t affect the amount of options you have?
Sounds like bullshit to me. I’d stick with what my gut’d tell me at the time.[/quote]
And you know it is really silly because everyone is ignoring the fact that the odds are never 1/3.
It just seems like they are because the initial conditions trick you.
This problem kills me. Somehow it got brought up at a family gathering and nobody would believe me. I actually had to sit there and make two of my cousins play it for money to make everybody understand.
It is funny how counterintuitive it is.
You can do a proof by completion by just taking three cards out of a desk and placing them in front of you. Say you take out a King (the car) and a Seven and a Two.
There are now 6 ways this can play out, since there are only 6 ways you can order 3 cards. Assume you always initially choose the leftmost card.
Scenario One:
K-7-2. Host reveals the 7 or the 2. You switch, you lose.
Scenario Two:
K-2-7. Host reveals the 7 or 2. You switch, you lose.
Scenario Three:
2-K-7. Host reveals the 7. You switch, you win.
Scenario Four:
7-K-2. Host reveals the 2. You switch, you win.
Scenario Five:
7-2-K. Host reveals the 2. You switch, you win.
Scenario Six:
2-7-K. Host reveals the 7. You switch, you win.
That’s it. That’s the only way it can play it out because the host ALWAYS shows you one of the goats and he never shows you yours.
Another way is to think of it like this. There are 6 ways that the cards (prizes) can be ordered. The only way you can lose is if you initially pick the right prize. This only happens 2 out of 6 times. In the event where you choose the goat, which happens 4 out of 6 times, the host eliminates the other goat, which means you will always win 4 out of 6 times.
The two choices are NOT independent, they are inherently linked. The very fact that you have less than a 50/50 chance of initially winning is what causes you have to have a greater than 50/50 chance of winning after the switch. As the 100 (or 1000 or 1million or whatever) doors problem shows you, the worse your initial odds of winning are, the better your odds of winning after the switch are, simply because YOU CAN ONLY LOSE WHEN YOU INITIALLY PICK THE CAR.
That is really the way to think about the problem. It’s just common sense really. Since there is only one “good” prize, the only time you “get burned” by the switch is when you initially pick the right door. Think about it like this:
-Switching is good if you initially pick the goat, because you can never switch into a goat. The host eliminates the other goat, so if you initially picked the goat and you switch, you are guaranteed to get the car.
-Switching is bad if you initially pick the prize, because then the host is only eliminating one of the goats and you are guaranteed to get the goat.
So it comes down to two options. If you switch, you always win if you initially picked the goat and you always lose if you initially picked the car. If you don’t switch, you always win if you initially picked the car and you always lose if you initially picked the goat.
So, as counterintuitive as it may seem, THE SECOND CHOICE DOES NOT MATTER AT ALL. It is totally inconsequential. All that matters is your initial choice of door and what strategy you decide to play by. If you are a switcher, you will win every time you initially pick a goat (2/3) and lose every time you initially pick the car (1/3).
A final way to look at this, that ONLY uses common sense and really no math. You know that if you use a “staying” strategy, you will always win 1 out of 3 times, because there is only 1 car in the three doors. So you already admit the staying strategy will only win 1 out of 3 times and nothing the host can do will change that. Well then, you’d be an idiot to stay if he shows you a goat behind the other door, because you KNOW there is only one goat behind the two doors that are remaining. So switching has to be better than staying, because staying can be no better than 1/3 and it is obvious that switching is better. What is not obvious is HOW MUCH better switching is, but it is obvious that switching is at least a little bit better than staying.
Alright, I spent way too much time ranting about that.
Play the game at home. Get three cups and put a ball under one. You play the contestant and have someone else be the host. Please report back with your findings.
Anybody else notice that Regular Gonzalez started this thread but he is yet to contribute?
I have a feeling he is getting quite a laugh out of this.
Yeah, I can actually remember at least one other thread like this being on T-Nation, possibly two. There’s always people who just flat-out refuse to believe it.
… but every time I think I’m out, they pull me back in!
Nobody answered the more interesting Lady, or the Tiger question
[quote]jtrinsey wrote:
Alright, I spent way too much time ranting about that.[/quote]
Join the club.
Thanks jtrinsey. You saved me the trouble and made a very clean presentation. To rephrase, it does not matter when you make the choice about your strategy, to switch or not. The host’s revealing of the goat in no way gives you a more informed way of deciding. Once your choice of strategy is made, winning or losing depends only on your choice of door. 2/3s of those choices result in winning the prize.
[quote]jtrinsey wrote:
YOU CAN ONLY LOSE WHEN YOU INITIALLY PICK THE CAR.
[/quote]
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.
[quote]LIFTICVSMAXIMVS wrote:
jtrinsey wrote:
YOU CAN ONLY LOSE WHEN YOU INITIALLY PICK THE CAR.
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.[/quote]
What he was saying is that if you play the switch strategy, then you only lose if you initially pick the car. (whatever happened to the millions dollars?)
[quote]jtrinsey wrote:
This problem kills me. Somehow it got brought up at a family gathering and nobody would believe me. I actually had to sit there and make two of my cousins play it for money to make everybody understand.
It is funny how counterintuitive it is.
You can do a proof by completion by just taking three cards out of a desk and placing them in front of you. Say you take out a King (the car) and a Seven and a Two.
There are now 6 ways this can play out, since there are only 6 ways you can order 3 cards. Assume you always initially choose the leftmost card.
Scenario One:
K-7-2. Host reveals the 7 or the 2. You switch, you lose.
Scenario Two:
K-2-7. Host reveals the 7 or 2. You switch, you lose.
Scenario Three:
2-K-7. Host reveals the 7. You switch, you win.
Scenario Four:
7-K-2. Host reveals the 2. You switch, you win.
Scenario Five:
7-2-K. Host reveals the 2. You switch, you win.
Scenario Six:
2-7-K. Host reveals the 7. You switch, you win.
[/quote]
I have to say, it wasn’t until this post that I finally saw it (granted I never actually sat down and worked out the possible choices - I’m in the middle of grading stupid first year students, so forgive me if I was just trying to get the short version from reading as opposed to performing calculations). I was with LIFTICVSMAXIMUS in initially thinking the the probability was only ever actually 1/2 until I actually sat down, looked at this presentation, and worked it out on my own. I continued this solution to the (3!*3) = 18 possible solutions just to be complete.
Although you’re tired of ranting about this, I just wamted to thank you for actually taking the time to present something step-by-step. Just saying the probability is 1/3 and 2/3 even when there appears to be a choice between only two items is very counter-intuitive as the wiki stated. When something is this counter-intuitive, it is extremely counter-productive to simply state that the probability is still 1/3 and 2/3 and the other person is wrong. I never affords the person who is incorrect the ability to see their mistake. Both parties just get frustrated and neither gets their point across.
Thanks again!
This is just funny. I thought the same way as lifty yesterday until I realised I was wrong with the 100 doors problem. Let me try to clarify this Lifty.
There are 100 doors and you pick the first one. The game show guy opens all the other doors except for one door all the way down the hall. Do you really thing that the door still closed all by itself is not the correct door? The one you chose, the first one in the hall, has 1/100 chance of being right. The door still closed and surrounded by open doors has 99/100 chance of having the car.
Were not trying to fool you. Once you see the point, you are going to feel dumb. It’s okay though, everybody thinks it’s 1/2 to start with.
[quote]LIFTICVSMAXIMVS wrote:
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.[/quote]
You don’t get to pick after the goat is eliminated; you just get to switch or not.
Your explanation above does not model the situation correctly.
[quote]LIFTICVSMAXIMVS wrote:
jtrinsey wrote:
YOU CAN ONLY LOSE WHEN YOU INITIALLY PICK THE CAR.
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.[/quote]
Actually this example will work. Let’s assume, as you have, that there is a 50/50 chance of switching.
you lose if: Pick the car and switch
1/3 chance of picking the car * 1/2 chance of switching = 1/6
you lose if: Pick the goat
2/3 chance of picking a goat * 1/2 chance of not switching = 2/6
you win if: Pick the car
1/3 chance of picking the car * 1/2 chance of not switching = 1/6
you win if pick the goat and switch
2/3 chance of picking a goat * 1/2 chance of switching = 2/6
As you can see, you switch 3/6 times. You win 2 of those times. You don’t switch 3/6 also, and then you only win once.
So, when switching you win 2/3 times.
Similar to the boxer analogy, just because there are four options does not mean each has a 25% chance of coming true.
[quote]pookie wrote:
LIFTICVSMAXIMVS wrote:
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.
You don’t get to pick after the goat is eliminated; you just get to switch or not.
Your explanation above does not model the situation correctly.
[/quote]
Let me clarify.
There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.
I haven’t seen this approach tried yet.
First, start with the situation where you have the one prize and two goats behind the trhee doors. You chose a door. Nothing else happens. How often do you win? 1/3 of the time.
Second, same as the first, except I show you a goat behind a door you didn’t pick. You are not allowed to change your choice. How often do you win? 1/3 of the time. It’s exaclty the same as the first scenario, except I give you some useless information. You can’t change your choice and I can always find a goat behind a door you didn’t pick.
Third, same as the second, except now you are allowed to change your choice. How often do you win? It depends on if you change your choice or not. You can ignore the information I have given you and stick with your original choice, but that’s the same as the second scenario. You will win 1/3 of the time. Or you can change your choice and win the remaining 2/3s of the time.
[quote]LIFTICVSMAXIMVS wrote:
pookie wrote:
LIFTICVSMAXIMVS wrote:
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.
You don’t get to pick after the goat is eliminated; you just get to switch or not.
Your explanation above does not model the situation correctly.
Let me clarify.
There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.[/quote]
Let US clarify.
What were the odds of each of your conditions, i.e. initially picking a car or initially picking a goat.
Again, and this has been your problem all along, you are forgetting the initial choice, and odds thereof, that led you to your 50% chance conclusion.
Tedro mapped it out for you.
[quote]LIFTICVSMAXIMVS wrote:
jtrinsey wrote:
YOU CAN ONLY LOSE WHEN YOU INITIALLY PICK THE CAR.
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.[/quote]
Yes, the odds of you picking the right door the second time are 50:50. If you randomly decide whether to switch or stay, you’ll switch half the time and stay half the time, right? So half the time you’ll have a 1/3 chance of winning, and half the time you’ll have a 2/3 chance of winning. 1/6 + 2/6 = 1/2. There it is, your solution reconciled with ours.
The two solutions going on here are not mutually exclusive, they’re just completely different probabilities. The probability of winning if you switch is 2/3. The probability of winning if you repick randomly is 1/2. There’s nothing to argue about.
[quote]ktennies wrote:
wfifer wrote:
elano wrote:
LIFTICVSMAXIMVS wrote:
pookie wrote:
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.
After door #3 was open by the host his chances became 1/2.
What does that change if he picks door #2?
If you have a choice of 4 multipe choice questions, you randomly pick an answer because you don’t know jack about that topic. The millionaire woman says " do you want to use a 50/50" so 2 of the other answers are crossed out. There is a 3/4 chance that the other answer besidse the one you picked is correct.
Right?
Damn, that’s an interesting application. It works assuming they always leave the answer you’re leaning toward, which seems to be the case. If it was random your odds would be back to 1/2.
Well the host isn’t going to cross out the correct answer. The point is to get rid of two wrong answers to make the question easier.
But as for the question: yes. By switching answers, you would have a 3/4 chance of a correct answer (because you have a 3/4 chance of being wrong on the first guess). The original answer still has a 1/4 chance of being correct.
However, this wouldn’t work on the show, because I believe you have to use the “50/50” before you choose an answer on “Who Wants to Be a Millionaire” :)[/quote]
I’m saying it works because contestants talk before they use the 50/50. I’ve never seen them eliminate the answer the contestant says he likes. Sometimes the contestant will be between two answers already and he’ll say so. Those answers are always the two left after the 50/50. From all the episodes I’ve seen, it doesn’t appear to be random at all.
If this assumption is correct, it’s in your best interest (assuming you know nothing of the topic) to tell them you’re leaning toward one of the answers. Then switching has a probability of success of 3/4. If you do know the topic and are leaning toward say, choice A, tell them you think it’s one of the other choices. In other words, try to pick a goat. This should actually increase your chance of winning even more.
[quote]LIFTICVSMAXIMVS wrote:
pookie wrote:
LIFTICVSMAXIMVS wrote:
No. One goat is eliminated.
there are exactly 2 ways out of 4 to win after the first goat is eliminated. That is 50:50.
You don’t get to pick after the goat is eliminated; you just get to switch or not.
Your explanation above does not model the situation correctly.
Let me clarify.
There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.[/quote]
Your explanation fails to take into account that there are 2 ways of doing (2) and (4) but only 1 way of doing (1) and (3).
Your explanation above does not model the situation correctly.