See If You Can Answer This Question Anot

BadzMaro SG

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.

JennTS SG

I say flip a 3 headed coin.

hegu SG

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 Smile

hegu SG

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

^tamago^ SG

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.

fishyang SG

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

demonage SG

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

mamamamama SG

All of us just wanna give our few cents worth..

Skibi SG

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.

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