probability question. If someone gets it right they will astound You! 10p to first correct.?

Question

You're on a game show and there are 3 doors. You're asked to pick a door knowing that one has a prize and 2 have donkeys.
You pick a door and the game show attendant opens a different door to reveal a donky. He then gives you the chance to change to the other door.
Should you? what are the chances of winning on your chosen door, and the other door.

Answer 1

CLASSIC, CLASSIC QUESTION! google "Monty Hall"

YES, ALWAYS change!

chances you picked the right one to start with is only 1/3.
but now, you have eleminated one, so now you have 1/2 chance!

SO: The odds would look like this:

picked the RIGHT one first and want to keep is only (1/3)

BUT WRONG one you picked to start with is (2/3)

So odds are you have the wrong one! SO CHANGE!

=)

Answer 2

Yes, change to the other door.
Your first chosen door has a probability of 1/3 (unchanged).
The door you switch to has a winning probability of 2/3.
(And the one that was shown to you has a probability of 0.)

It's amazing how many different (and wrong) answers were given (although the right one was given frequently).

Here's an example to make it easier to understand why switching is the best strategy:

Suppose there are 100 doors and you choose one (number 13, say). Then the host opens all the doors except numbers 13 and 67, and all have goats behind them.
He didn't open number 13 because you chose it. But why did he not open 67?
Well, there's one chance in 100 that you chose the winning door, so he just selected door number 67 at random.
But there's a 99% chance that you did NOT select the winning door, and the host has left door number 67 closed in order to keep the "big deal of the day" hidden.
Are you going to make the switch and get the big deal (99% of the time)? Or are you going to say it doesn't matter, that it's 50-50.

(Incidentally, in the 3-door situation, if the door that is opened is chosen at random, and THERE IS ACTUALLY A CHANCE that it will be the big deal, BUT IT TURNS OUT NOT TO BE THE BIG DEAL, then your chances have just increased from 1/3 to 1/2, and it doesn't make a difference whether you switch or not.)

Answer 3

I'm answer 90 something, but this is a weird question to understand and here's the straight forward intuitive answer:

The first door you picked, rather than thinking about the odds that you were right, think about that fact that there were 2/3 chance you picked a donkey. So after the other donkey is exposed, if you switch your choice, 2/3 of the time you are switching ~away~ from a donkey you chose (since we all agree the odds of selecting a donkey are 2/3), and only 1/3 of the time will you be switching away from the prize. Make sense?

The question is misleading to our logic, which I personally find it fascinating that by and large, us individuals on the planet seem to share this instinctual logic, wrong though it may be. The hardest argument to argue against, for me, is the fact that once only two doors remain, in relationship to just those two doors, the odds of getting the prize would be 50/50, but how do you reconcile the fact that originally you were 2/3 likely to be wrong and then magically just by opening some ~other~ door your odds of having chosen correctly in the first place suddenly just become 50/50? Does that really make any sense, but again, standard human logic (including mine) just stare at the two doors with tunnel vision, unable to imagine how any of the other factors may have affected the point you're at. Interesting...

--charlie

Answer 4

It isn’t 50/50 people. It's called conditional probability.

It’s 2/3 if you switch.

Think of it this way. You only had a 1/3 chance of picking the right door when there were 3 doors. What does it matter if the host opens one? It doesn’t. He can always open one with a donkey. It doesn’t change the fact you only have a 1/3 chance of your original door being right. If you don’t believe me run a Monte Carlo experiment and find out.

If you want a slightly tougher problem make it four doors with two stages. Do you switch or stay after he opens the first door? And then do you switch or stay after he opens the second door? What strategy maximizes your odds?

Answer 5

The chances if you keep your door are 1/3, 2/3 if you change.

The reason is, before you see the donkey, you have 1/3 of a chance. The attendant will NEVER open to show the prize. It is not a random opening of a door. So all the attendant is showing you is that at least one of their two doors has a donkey. So that is no new information.

Another way, call the doors 1, 2 & 3. Suppose you pick 1.

Suppose the prize is behind door #1. Then you would win if you keep your door.

Suppose the prize is behind door #2. The attendant will show door #3. You would win if you switch.

Suppose the prize is behind door #3. The attendant will show door #2. You would win if you switch.

So each possibility has 1/3 of a chance. So you have a 2/3 chance to win if you switch.

Answer 6

If you picked a door, and the game show attendant opened a different door, (as stated) why didn't he open the door that you picked in the first place? Must mean that the one you picked has the prize! So after he gives you the chance to change to the other door, pick the same one you picked in the first place.

Answer 7

It's really easy.

Now you're left with 2 unkown doors. One has a donkey and the other a prize. So it's a 1/2 chance of winning the prize.

Those guys who say you should choose a different door are wrong. They are mistaking probability with conditional probability.

Before you had a 1/3 chance of being right and a 2/3 chance of being wrong. Now you have a 1/2 chance either way. You could be right or wrong whether you change it or not.

**************EDIT********************8

OK I understand now. It is not a purely random problem. Your initial choice influences the host's behavior. For example, if your initial choice was a horse, then the host is forced to reveal the other horse. He will never reveal the car.

The link below shows the probability tree.
1) If you choose horse A, then the host reveals horse B, and switching will win the car.
2) If you initally chose horse B, then the host reveals horse A, and switching o nce again reveals the car
3) If you choose the car, then the host reveals 1 of the horses (each with 1/2 prob), and switching loses.

So 2/3 times, switching results in a win.
Great question. You got me at first.

Answer 8

The chances of winning on the chosen door was originally 1/6, because there were 3 doors orginally (1/3 chance of picking the right one) and in the nex tround there are two doors (1/2 chance). 1/3*1/2=1/6. However, in the second round, the odds of the other door winning would be 1/2. So 1/6 probability if you stick with the original door all the way through, and a 1/2 probability of the other remaining door being the winner.

Answer 9

Admittedly, this question could have been better worded.

But those who are familiar with this kind of problem will tell you, that the correct strategy is to pick the other door.

The probability that it's behind the other door is 2/3

Consider this: instead of 3 doors, there are 100 doors. You pick one, and then the lovely assistant opens 98 other doors one by one, all of which contain donkeys. How then are the chances, that the one door she neglected to open(besides yours) contains the prize?

The question here, is not how likely that you chose the correct door on your first try........ The question is: how likely is it that you *didn't* choose it?

It's twice as likely that that the prize is behind one of the two *other* doors, then the one you originally picked......

~Soylent Yellow

Answer 10

Some people on here are saying that changing doors will give you a better probability of winning. How so?

Think about it:

Say there were two people to each pick a door. Person A chose door 1. Person B chose door 2. The host opens door 3 to expose one of the donkeys.

Now what some people are saying here is that if you switch doors, it will increase your probability of winning.

OK, so lets make them switch. Person A and Person B trade doors. Person A now has door 2, and person B now has door 1.

It is impossible that the probability for BOTH of them went up, because the outcome is that only one will win. But despite this, some people say that switching doors increases your probability.
But with this example you can see that the outcome is that both person A and person B have an equal 50/50 chance of winning. Thr probability did NOT increase for either of them just because they switched doors. Its GOT to stay the same: the law of averages makes that clear.

So switching doors does NOT increase your chances.

The fact that one door was opened, did though. Both person A and person B's chances of winning increased from 33.3% up to 50% when the 3rd door was open revealing the donkey.
The probability stayed at 50% when they traded. You can see it this way, also:

Person A: you, the contestant
Person B: Fate

Whose to say which one "probability" is going to favor? You switch doors, and Fate automatically switches doors, also.

You both switched, and some here state that switching increases the chances of winning. But whose chances, yours or Fate's? If probability goes up for one, it has to go down for the other. If the only factor in determining who's probability goes up is switching doors, the fact that both of you switching doors makes the probability stay even. Thus, chances remain at 50/50.

See, probability stays at 50/50 whether you switch doors or not.

Answer 11

the probability of choosing any of these doors
is
(1/3)+(1/3)+(1/3) = 1
probability of choosing the one with prize = 1/3
since the guy showed me one of the donkey rooms.. there
are 2 left and one of them has a donkey in it while the other one for sure has the prize.
so at this point it is 50/50 ..
so yes I would take my chance and go with the door i picked before.. I would do so coz I know that there is a 50% chance of it having the prize..