Fun math problem...
- Thread starter huntedannoyed
- Start date
Recommended Videos
I've heard this before and the answer is that you have a better chance of winning by swapping doors, but I still don't know why.mintsauce said:
No, it's 2/3.huntedannoyed said:If the guy opens one door, there are now only two doors to choose from; 50-50, up from 1/3.Lukeje post=18.74799.847111 said:No... it goes from 1/3 to 2/3 if you swap (you are doubling your chance of winning, as you are now betting on TWO doors, not just one).huntedannoyed post=18.74799.847043 said:This was popularized in the movie "21." You should change your answer if given the choice because now your odds are 50% instead of 33.33% Funny enough, I tried this with three playing cards and it is totally accurate. Which leads to this thought... If you ever get down to two cases on "Deal or no Deal," should you switch your case if you get down to two... Yes, you should.Steve Dark post=18.74799.847032 said:Easy. Fill the 3, put it in the 5. Fill the 3 again and top up the 5 from it. You now have 1 in the 3. Empty the 5 and put the 1 in the 5. Then fill up the 3 again. Add the 3 to the 1 in the 5 and you get 4 in the 5.
Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?
You have a 1/3 chance of being right the first time, and a 2/3 chance of being wrong.
If you were right the first time and swap, you lose.
If you were wrong the first time and swap, you win.
Hence the chance of winning if you swap is the chance you were wrong the first time, which is 2/3.
111221SkinnySlim said:
the next line is describing the previous one so the line above has 3 1s 2 2s and 1 1, stripping out the letters
312211 so the sequence continues as such:
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
etc etc
And yeah the Monty hall thing is crazy, but it works in practice almost flawlessly, always change the box if you get to the last two in Deal or no Deal.
I've been thinking about that gameshow puzzle and I don't think there is an advantage in switching and before you shout out 'READ THE WIKI ARTICLE' I have already.
It's true that the chance of your door winning is 1/3 initially. It isn't true in my opinion that switching doors after one is revealed to be a losing door increases the chance of you winning. I have two trains of thought that lead me to this idea that swtiching is a useless idea.
1) Before one of the losing doors was opened, the chance of winning was 1/3 and the chance of another door being correct was 2/3. Once the third door is shown to be incorrect, the probability of being correct CHANGES in my opinion. The chance that your door is the winner is 1/2, and the chance of the other door being correct is also 1/2. So, the 2 remaining doors have an equal chance of winning the prize at the point of being asked to confirm your selection. The original probabilities no longer apply, because you are now in a whole new line of probability where you are only choosing between 2 doors, the third door is irrelevant. Therefore there is no reasoning behind the fact that the other door is more likely to be the winner.
2) If your door has a 1/3 chance of winning, so does the 2nd and 3rd doors. The probability of one of the other 2 doors winning rather than a particular door is 2/3. Once the 3rd door is removed as a choice, the theory states that the original probabilities still hold true, which means that the chance of the other door being correct (singular because 1 door has been removed) is 2/3. However, this would apply to both doors so the probability that door 1 (your door) winning is 2/3 (from the perspective of door 2) and the probability of door 2 winning is 2/3 also (from your perspective). This leads to a total probability of something winning equalling 4/3 which is impossible. So I say that this theory is rubbish because it leads to an impossible situation ( Reducio Ad Absrdum in effect I think).
Well, thats my reasoning. If anyone can see any reasons why my 2 arguments are incorrect and there actually is an advantage so swapping please feel free to show how I'm wrong and that you're right. I just can't see it, though if I ever do get on a gameshow I will switch. I just want to know why these two points don't apply.
It's true that the chance of your door winning is 1/3 initially. It isn't true in my opinion that switching doors after one is revealed to be a losing door increases the chance of you winning. I have two trains of thought that lead me to this idea that swtiching is a useless idea.
1) Before one of the losing doors was opened, the chance of winning was 1/3 and the chance of another door being correct was 2/3. Once the third door is shown to be incorrect, the probability of being correct CHANGES in my opinion. The chance that your door is the winner is 1/2, and the chance of the other door being correct is also 1/2. So, the 2 remaining doors have an equal chance of winning the prize at the point of being asked to confirm your selection. The original probabilities no longer apply, because you are now in a whole new line of probability where you are only choosing between 2 doors, the third door is irrelevant. Therefore there is no reasoning behind the fact that the other door is more likely to be the winner.
2) If your door has a 1/3 chance of winning, so does the 2nd and 3rd doors. The probability of one of the other 2 doors winning rather than a particular door is 2/3. Once the 3rd door is removed as a choice, the theory states that the original probabilities still hold true, which means that the chance of the other door being correct (singular because 1 door has been removed) is 2/3. However, this would apply to both doors so the probability that door 1 (your door) winning is 2/3 (from the perspective of door 2) and the probability of door 2 winning is 2/3 also (from your perspective). This leads to a total probability of something winning equalling 4/3 which is impossible. So I say that this theory is rubbish because it leads to an impossible situation ( Reducio Ad Absrdum in effect I think).
Well, thats my reasoning. If anyone can see any reasons why my 2 arguments are incorrect and there actually is an advantage so swapping please feel free to show how I'm wrong and that you're right. I just can't see it, though if I ever do get on a gameshow I will switch. I just want to know why these two points don't apply.
This.Scarecrow38 said:
When you drop from 3 possible doors to 2 possible doors with one of them holding the prize the probability is 50% thus it doesn't matter if you swap or don't.
?dit: I admit I should have read the proper Hall-challenge. With the host knowing the correct door, I turn to the swaping-side.
OK, here's that catch that should be listed here about the Monty Hall problem so everyone who doesn't feel like going to the wiki should see:
Monty Hall KNOWS which door the prize is behind.
This is important because if you haven't chosen the correct door yet, he only can open ONE door that does not have the prize behind it.
To put it in better perspective, imagine there are 100 doors, you pick one and Monty shows you 98 empty doors. Now, your chance is 1 in 100 that you have picked the winning door. Then after you have been shown 98 empty doors that he was aware were empty, intuitively (at least to me) it is almost guaranteed the prize is behind that other door. The small size of 1 in 3 makes it seem implausible this works, but a large number puts it in perspective.
Mathematically, if you pick one door it is a 1/100 chance it is behind that door and 99/100 that it is behind ONE of the other 99 doors. Since you've just been shown that 98 of those doors are empty, it is 99/100 that it is in that remaining door.
Monty Hall KNOWS which door the prize is behind.
This is important because if you haven't chosen the correct door yet, he only can open ONE door that does not have the prize behind it.
To put it in better perspective, imagine there are 100 doors, you pick one and Monty shows you 98 empty doors. Now, your chance is 1 in 100 that you have picked the winning door. Then after you have been shown 98 empty doors that he was aware were empty, intuitively (at least to me) it is almost guaranteed the prize is behind that other door. The small size of 1 in 3 makes it seem implausible this works, but a large number puts it in perspective.
Mathematically, if you pick one door it is a 1/100 chance it is behind that door and 99/100 that it is behind ONE of the other 99 doors. Since you've just been shown that 98 of those doors are empty, it is 99/100 that it is in that remaining door.
