+270 From now on I will try to post question everyday except weekend.
This is question for today. Hope you enjoy it.
A contestant on a quiz show is asked to choose one of three doors. behind one of the doors is the star prize of a sports car, but behind each of the other two doors there is a toy car.
The contestant chooses one of the three doors.
The host then opens one of the remaining doors and reveals a toy car. The host then asks the contestant if they want to stick with their first choice or switch to the other unopened door.
State would you would recommend the contestant to do in order to have the great probability of winning the sports car.
+23252 I fully agree with the 1/3, 2/3 probabilities.
At the beginning, the player has a 1/3 chance of getting the correct door; the host has a 2/3 chance of having the correct door.
The host gives the player the option of trading sides. If the player trades, then the player gets the host's 2/3 chance of winning. (Showing an incorrect door doesn't change the chances; the host must always start with (at least) one losing door. All this does is fake people into believing that the chances are 50-50.)
+3502 you cant cause no matter how many curtains you switch too. Your probability of winning is still a split 50/50 because there are only two doors left
+7188 I'd say....peek then choose the winner! That probably isn't an option of that though, is there?
+270 Guys, to give you guys hint,
one of two doors is more likely to get sports car.
+23252 Switching from the one that you originally chose to the one not shown by the host gives you the best overall probability of winning.
+22 You got this question from Google, I serched up "Math Problems" and that came up. Make your own questions?
+270 Egeclaw, this question is very well-known.
And I got this from textbook which is Edexcel AS and A Level Modular Mathematics Statistics 1.
What I'm doing here is just wanna share good questions and have fun.
I should have mentioned it, sorry for that.
+22 Not beeing mean, I just thought you were taking questions from Google. I mean most people here have probably searched them and know the answers.
+270 Yeah. The point is, we are trying be the result when someone search math questions.
+270 So let's assume that Door A is Sports car. Door B and C are toy.
At first time, if you chose door A, and if you switch afterward, you will 100% get toy.
At first time, if you chose door B, and if you switch afterward, you will 100% get sports car.
The same with C, at first time, if you chose door C, and if you switch afterward, you will 100% get sports car.
Therefore if you switch afterward, the probability of getting sports car is $$\frac{2}{3}$$.
If you don't understand, I can explain further.
+7188 Okay. So 1/3 is your chance at the beginning. If wrong, 1/2 is your chance afterward. You add them, and get 2/3
+3502 i still dont get it the question asks if you are left with two doors 50/50 and your trying to get a sports car (So obviously your not going to pick curtain 3) Then you would have two left to pick from curtain A and B 50/50?
+270 I don't understand you, and 1/3 + 1/2 does not equal 2/3.
To explain further, you can use your common sense to get the gut.
If there are 100 billion doors, 99 billion doors are toy cars, and 1 door is sports car.
If you randomly choose one, it's really unlikely to get sports car.
After you choose, host opened all the door with toy cars except one door and the door you chose.
Which one is likely to be sports car?
Use your common sense.
+3502 woah woah woah now what are you trying to say by use you common sense? I dont know where you come fom but that aint used lightly around here
+7188 Whoops! I still don't get it! I made a mistake....
I'm not worrying anymore
+23252 I fully agree with the 1/3, 2/3 probabilities.
At the beginning, the player has a 1/3 chance of getting the correct door; the host has a 2/3 chance of having the correct door.
The host gives the player the option of trading sides. If the player trades, then the player gets the host's 2/3 chance of winning. (Showing an incorrect door doesn't change the chances; the host must always start with (at least) one losing door. All this does is fake people into believing that the chances are 50-50.)
