See If You Can Answer This Question Anot
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Originally posted by sherling22:is the calculations below correct?
(1) P(1st guess is right, and i choose not to switch)
= 1/3
if i choose to switch:
(2a) P(1st guess is wrong, but i switched to the correct door)
= P(1st guess is wrong) x P(i switched to the correct door)
= 2/3 x 1/2
= 1/3
(2b) P(1st guess is right, and i still switched to the correct door)
= P(1st guess is right) x P(i switched to the correct door)
= 1/3 x 1/2
= 1/6
(2c) P(i guess the right door)
= (2a) + (2b)
= 1/3 + 1/6
= 1/2 ----> which is greater than in (1)
so no matter whether my 1st guess is right or wrong, the probability of getting the right door if i choose to switch is higher.
i was overwhelmed coz i haven learn this yet 
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wait...you're saying the goat can eat, or the goat can be eaten?Originally posted by SingaporeMacross:goat can eat, car cannot. -
Ehhh dont have so complicated la..Originally posted by lpx88: i was overwhelmed coz i haven learn this yet
If you do not switch, its very simple, your chances of winning the car is 1/3.
If you do switch, the chances of choosing a goat in the 1st place is 2/3. Then, the show host will eliminate one of the goats, which makes your possibility of winning the car 2/3. -
no, even if the host eliminate one of the goats, there is still only 1 car, hence cannot have a "2" in 2/3.Originally posted by tIfosI:Ehhh dont have so complicated la..
If you do not switch, its very simple, your chances of winning the car is 1/3.
If you do switch, the chances of choosing a goat in the 1st place is 2/3. Then, the show host will eliminate one of the goats, which makes your possibility of winning the car 2/3.
the fact that when the host interfere in the game, he changed the game from a single event to a double event. although the additional event seems trivial to us, it is tedious to solve it using probability. i think DailyFreeGames.com gave a less erudite explanation, which he captured the gist of the problem
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there is 2 side to this coin, the chance of switching and getting the wrong door won't that now be 1 - 1/2 = 1/2Originally posted by sherling22:is the calculations below correct?
(1) P(1st guess is right, and i choose not to switch)
= 1/3
if i choose to switch:
(2a) P(1st guess is wrong, but i switched to the correct door)
= P(1st guess is wrong) x P(i switched to the correct door)
= 2/3 x 1/2
= 1/3
(2b) P(1st guess is right, and i still switched to the correct door)
= P(1st guess is right) x P(i switched to the correct door)
= 1/3 x 1/2
= 1/6
(2c) P(i guess the right door)
= (2a) + (2b)
= 1/3 + 1/6
= 1/2 ----> which is greater than in (1)
so no matter whether my 1st guess is right or wrong, the probability of getting the right door if i choose to switch is higher.
alternatively, the chance of not swtiching and still get the right answer right, is 1 - 1/2 = 1/2.
the whole idea is this, when one of the wrong door is opened, the whole equation/assumptions/factors had changed and hence a recalc is required to start from base. -
yep, you are rite that the probability of getting the wrong door after switching = 1/2Originally posted by vince69:there is 2 side to this coin, the chance of switching and getting the wrong door won't that now be 1 - 1/2 = 1/2
alternatively, the chance of not swtiching and still get the right answer right, is 1 - 1/2 = 1/2.
the whole idea is this, when one of the wrong door is opened, the whole equation/assumptions/factors had changed and hence a recalc is required to start from base.
but the probability of getting the wrong door if i choose not to switch = 2/3 (which is larger than 1/2)
so the other side of the coin also implies that if i switch door, i get a lower probability of getting a wrong door again -
By this time I should mention the following method/ example usually used to help convince the doubtful ones on the benefit of switching:
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Imagine there are 10 doors. Behind one of the doors there is a car, the other 9, goats.
You choose a door, say door number 1.
The host then opens 8 other doors revealing goats.
Is it to your advantage now to switch to the other, remaining door?
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Well this example help to drive the point very intuitively and it's easier to understand this way. -
Get the 2 goats, let them mate. After a few years, sell all your goats and trade in for a new car. Then drive the car around to look for more goats and sell them. So switch or not, does not matter because the most important thing at the end of the day go back home with the 2 goats or the car. If you can get all three, good. If can bring back the host too and sell him also.
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u sure the 2 goats different gender?Originally posted by dinky1409:Get the 2 goats, let them mate. After a few years, sell all your goats and trade in for a new car. Then drive the car around to look for more goats and sell them. So switch or not, does not matter because the most important thing at the end of the day go back home with the 2 goats or the car. If you can get all three, good. If can bring back the host too and sell him also. -
No matter which of the 3 doors you choose.. the host would still eliminate one with the goat behind..
Which means that there is actually only 1 out of 2 chances of choosing the right door..
Let's look at it in another scenario..
The host would first reveal one of the doors with a goat behind it..
After which the host invites you to pick a door out of the remaining 2 doors..
There you go! 50% chances of getting the car
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Use real-life situations, not mathematics!
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Door no door still the same
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Ok my turn......
dun switch. 1st door is bigger than the other 2 doors coz 1st door can conceal a car while the other 2 doors can only conceal a goat each.
dat is why no1 is chosen in the first place .
Just my non-sense
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Sorry, if i still dont get it. For this scenario, even if the host opens the other 8 doors, leaving with now 2 doors, then the "current" base probability becomes 1/2 and 1/2 anyway.Originally posted by Li Ka Shing:By this time I should mention the following method/ example usually used to help convince the doubtful ones on the benefit of switching:
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Imagine there are 10 doors. Behind one of the doors there is a car, the other 9, goats.
You choose a door, say door number 1.
The host then opens 8 other doors revealing goats.
Is it to your advantage now to switch to the other, remaining door?
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Well this example help to drive the point very intuitively and it's easier to understand this way.
So switching wont cause any difference? -
OK think of it this way, when u choose 1 door out of 10, your chance of hitting a car is what.......only 1 in 10, correct anot.......Now when the host opens 8 other doors revealing goats, that means now it's left with only 2 doors, your chosen door and the remaining unopened door. One of them will have a car. Now how confident are you that your originally chosen door, which has a 1 in 10 chance, actually has a car behind? Well the chance is only 1 in 10, not so good right? You're not so confident, right? But now after the host elimitates 8 other doors, that means now the car has to be behind one of the two doors. In other words, he has helped eliminate other possibilities and improve the chance that the car is actually at the other remaining door. By switching you are taking advantage of this and improving your chance.
OK think of this another example.
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Suppose now there are 100 doors. Behind 1 door there is a car, the other 99, goats. Now you choose a door, say door number 1. After that the host opens 98 other doors revealing goats. He then ask you whether you want to change your mind to switch to the other remaining door. Is it to your advantage to switch?
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It's getting more and more clearer this way, use bigger and bigger numbers. -
By giving up 1 door .. it increases ur chance to 50/50 . So whether its advantagous or not , it comes down to whether u want to listen or trust on your own judgement or not.
50/50 of course its advan. Even if u increase doors , there is still a 1% probablity that the host is wrong. 1% is still it takes for something bad to happen. Something not 100% is not guranteed. It just means increasing ur odds in favour to it but still an unknown factor.
Like physics.. all can be calculated but there is always the 0.1% of human error n ur thing doesnt work. So it comes down to whether u think that u are going for majority and higher probability or the still 1% of error. -
I say flip a 3 headed coin.
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i think we really need a try out here:
i'll think of a number 1 to 3 now, you guys will guess what number i'm thinking, i'll reveal a number that is not the correct answer after you've made your choice -
i've a better idea, let's put it another way...based on the 10 doors problem
i think of a number 1 to 10, you are supposed to guess that number
10 numbers are really hard to choose one from, say you choose 1.
now i say: 2,3,4,5,6,8,9,10 are NOT the correct answers
can you see something now? a very obvious hint that the 'gap' is probably the correct answer.
now all is left is 1 and 7. would you like to change your ans to 7? or insist on 1?
by common sense, i'm sure the choice to change or not is more or less decided.
can you still say the odds is still 50:50?
[in fact the odds are 1:9]
the reason why we dun see this in the 3 doors problem is becos the number of doors is too few (it's the minimum for this problem to work). ppl usually cannot detect the small difference of probability 2/3 and 1/2 through their intuition or common sense -
Read this excerpt:
"There are three possible scenarios, each with equal probability (1/3):
[*]The player picks goat number 1. The game host picks the other goat. Switching will win the car.
[*]The player picks goat number 2. The game host picks the other goat. Switching will win the car.
[*]The player picks the car. The game host picks either of the two goats. Switching will lose.
In the first two scenarios, the player wins by switching. The third scenario is the only one where the player wins by staying. Since two out of three scenarios win by switching and each scenario is equally likely, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three."
And also this:
"It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch."
More here. -
lol. this problem and solution can be found in THE CURIOUS INCIDENT OF THE DOG IN THE NIGHT TIME by Mark Haddon. This guy with autism(i think) has gifts in math and he already wrote the ans there
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guys u all discuss this also no use sia it has already been proven by mathematics and by human sia wth still down here argue
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All of us just wanna give our few cents worth..
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I think the solution is quite difficult to understand if you read it and try to visualise in your head. Just draw a car and two goat (use X and O for easier time) and follow the solution, you should get the answer in no time.