Welcome back to Teaser Tuesday! I think we've all gotten warmed up (hopefully SOME found the dice riddle challenging!), and now it's time to start making the regular teasers more difficult. No more softballs!
Note: If you could all please input your answers using the previously noted html code it will make your comment a button similar to the "Hint" buttons seen below, allowing others to answer without having a spoiler. Also, in your answer, do not include any ' or " or it will not work properly.
Question
On the game show Let's Make A Deal, Monty Hall shows you three doors. Behind one of the doors is a new car, behind another hides a goat, and behind the last door is your MD who will immediately fire you if his door is selected. You initially choose door #1, after which Monty shows you what's behind door #2 and it's the goat. He gives you the chance to stay with your original pick or select door #3. What do you do? Show your work for your answer to count.
Bonus Question
Suppose there is only one barber shop in your remote, and never traveled to, town, and it employs two barbers. One of the barbers has a nice, neatly trimmed head of hair. The other's hair is a complete mess. Which of the two barbers should you go to for a neatly trimmed haircut and why?
Good Luck!
Remember the first one from school and a movie. Second one is a total guess.
@yeahright - First, let me say, great work on the html! Works perfectly!
Both are correct. Good work!
I'm not very good at probability (never took it), but am a little confused. If door #2 is "eliminated", aren't the odds of you choosing the door with the car 1/2, no matter what your initial choice was?
I just see it as, since door #2 was shown to have the goat (the "negligible" choice), the only two doors left to choose (car, good) and (MD, bad) make your initial choice essentially the same?
Not sure if this is coherent or mathematically correct, but initially 33%(good)/33%(break even)/33%(bad) ---> door #2 = goat. Now, 50%(good)/50%(bad)? Your first choice could just as easily be the car as your second choice, regardless of now knowing what door 2 held?
Someone please explain this to me in simple terms, I'm talking in circles.
See if I can explain this in simple terms...
You have door A, B, C
Each door has 1/3 chance it is a car. Each door has 2/3 chance it is a goat because there are 2 goats behind 2 doors.
You select door A. The host reveals door C is a goat. You then know the car is behind either door A or door B because the host would not reveal the car.
Since there are STILL 3 doors, each door STILL has a 1/3 chance of being the goat.
Here is where it can get confusing: Since you picked A originally which was 1/3 chance, and door B chance + door C chance = 2/3 chance of being a car.
You want to switch from A to B because you know door C which is 1/3 of the 2/3 (B+C)
Remember it is a 2/3 chance it is behind B or C, but since you know C, that means B is 2/3 the car.
Fock, tried to SB but MS. I'll try to recompense
That's the intuitive way of thinking about it, and it's absolutely wrong. The key to solving the question is to remember that Monty Hall knows where the car is and the he's not picking a door at random.
Wikipedia has a pretty good explanation of the solution.
I didn't think before answering and got both wrong. I'm going to build a snowman.
Peace.
You would switch doors because you now have a 2/3 chance of choosing the one with the car given you know that the one of the doors is a goat.
You would get a haircut by the barber with the shitty haircut because the 2 barbers cut each others hair.
I say that you go to the barber with the good hair because a good barber would be able to do his own hair.
@mikesswimn ; I understand better now; however doesn't the fact that they are all 3 different options change the dynamics?
In the initial problem, I noticed that there are 2 doors with goats and one with a car.
Treat them as the same, I was merely trying to "banker up" the teaser :).
All answers to the goat problem are wrong. First of all, you need to specify whether keeping a job or a free car is more valuable to contestant. Second of all, 3 different objects behind 3 different doors changes the dynamics of the original problem of 2 goats and 1 prize.
You are wrong, the dynamics do not change. One door is a prize, the other two are non-prizes. It can be any variation of 'things'.
And keeping a job is not the best outcome, getting a free car and keeping your job is the best outcome. That's the point of the problem. I'm sure some people would prefer a goat.
What MBP is saying is that the statistics aren't the only factor in the decison making. Calculating risk is one thing, deciding if the risk is worth it is another. If there's a 2014 Lamborghini behind the door, then I play the game. If there's a beat up Focus with 200K on it, the potential payoff wouldn't be worth it to me. For someone making a million dollars a year at a job they could not replicate, no car is worth the risk. For someone making $15K a year at a job they could easily do elsewhere, any car is worth the risk.
You're right from a risk calculation standpoint and MBP is right from a weighted risk standpoint, so you two should argue and have a flame war. I'm bored. Entertain me.
Nope, you are wrong. One of the fundamental differences between the original problem and this one is that the two "non-prizes" are not fungible. When the host shows you a goat behind one of the doors in the original problem, he does not give you any definitive information about what's behind your door, but in this case he does. By showing you a goat in this example, the host is telling you that you do not have a goat, and so you can actually condition on this information.
And it is definitely relevant to know the relative value of the car vs. keeping your job. For example, let's suppose the value of getting fired is -2, and the value of the car is 1, then the prize can be considered to not get fired. In which case you are not trying to maximize the chances of finding a particular door. Instead you are trying to maximize the odds of avoiding a particular door (which is a fundamentally different game). Moreover, it calls into question how the host would proceed if the door you originally picked had a goat behind it. What would he show you then? This would be dependent on what is actually considered the prize of the game.
So let's go case by case here. Let's suppose the following:
Door A (Goat) Door B (Car) Door C (MD)
Suppose you pick Door B, then the host will undoubtedly show you Door A Suppose you pick Door C, then the host will undoubtedly show you Door A But what if you choose Door A, will the host show you Door B or Door C? This depends on what is considered the true prize. Moreover, suppose the car is considered the true prize and he shows you the MD, would you get fired anyway?
The bottom line is that this is an ill-posed problem that does not have a real solution because too many parameters are missing.
Too easy. Make them challenging at least.
This. Both question would be in any "brainteasers for Interviews" book.
Depends whether the goat went to a target or semi-target
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