So, how did you do? Please explain if you were correct or not. Be sure to include mathematics to support you answer.
22 comments:
Anonymous
said...
JE So, my theory was correct. U start out with 1/3 chance of getting the prize and 2/3 of missing the prize. So after one door is opened, u could stay with the door with only 1/3 chance or you could change with 2/3s the chance with getting the prize.
I was wrong with my prediction. I was pretty close but it was also kind of off. I said that if i stayed i had a 1/3 chance of getting it but i actually got about 40% of getting it. also, i said if i changed the doors that i was to get about 50% right but i actually got about 70% right.
I was right, my prediction was that switching doors would give you a 66% chance to win and staying would give you a 33% chance to win. Of the 30 games I played for each strategy I got 60% chance to win when you switch and 30% chance to win when you stay.
My prediction was right because I stayed 101 times and I switched 101 times. Out of the 101 times I stayed I won 24 times which is 23.76% of the time.Out of the 101 times I switched I won 66 times which is 65.35% of the time. So my conclusion is that you have a greater probability of winning by switching your door than staying with the door that you originally picked.
I played the game a total of 200 times, switching doors 100 times and keeping my door the other 100 times. When i switched doors, I won 69 times meaning that there is a 69% chance you'll win if you switch doors. i won 39 times when i kept my door meaning there is a 39% chance you'll win if you keep your door. My conclusion is that it is better to switch your door because it is more likely you'll win.
I played the game and it was fun. I played the game 60 times. During the game I switched my door 30 times and kept my door 30 times. In doing that I won 13 games when I kept my door and 15 games when I switched my door. In the results of that, my experimental probability of winning was 43.33% when I kept my first door and 50% chance when I switched my door. Playing this game tells you that in switching your door you will have a better chance of winning then if you keep your first door.
When I played this game, I won more times when I switched the doors than when I stayed with the door I chose before. I played a total of 150 games, 75 games when I switched and 75 games when I stayed. I won 49 when I switched doors, which means I have a 69% chance of winning when I switch doors. I only won 25 games when I stayed with the same door which means I have a 25% chance of winning when I stayed with the same door.
My results say that I won more games when i switched. I played 200 games in all. 100 for switching and 100 for staying. I won 35 games when i stayed and 60 when i switched. That's a 35% chance of winning if i stay with the door and 60% chance of winning if i switch doors.
I played 80 games. In staying, I got 29 winning games, so 36.25% chance of winning. In switching, I got 48 winning games, a 60% chance of winning. This is because… When you open the door that is guaranteed wrong, you will have a fifty, fifty percent chance of getting the cruise and money. This means, however that there is a fifty percent chance your door is wrong, but it couldn’t be opened because you chose it. When you choose to switch, you can get rid of the chance of your door being wrong for the most part. That’s why there is a higher chance. You have to use experimental probability.
That game was fun! My results were completely unexpected. I played the game 80 times, the first 40 I stayed and the second 40 I switched. As I already explained you have a 50/50 chance of getting the prize after u click the first door, so the probability of getting the prize is 50%, not 33% like you may think. But my experimental probability of winning said otherwise. When I stayed I had an experimental probability for winning of 35%. This is less than 50% like the probability shows. When I switched I had an experimental probability to win of 57.5%. This also isn’t 50% like the probability. This is because the experimental probability is not the same as a theoretical probability. On a 50% chance it is completely random for the results. This is why my experimental probabilities are not 50%.
I believe I did pretty well. I won about 62% of the time. I won more times when I switched the door. I was sort of right but I used the experimental probability the computer gave me and I based my decision to switch the door on that. If the percent was higher to switch the door, I switched the door. If the percent was higher to stay with the same door, I stayed the same.
J.K. 2/13/09 I was not right. At first, I said I had a 1/3 chance of getting the door right if I stayed, and I was close enough to 33 percent to say I was right, even though I got 40% right. I was wrong, however, about the probability if I switched doors. I said 50% but it was closer to 2/3. I got 70% of the doors right when I switch. I was wrong about that part. If I stayed with the door I chose after a door was opened, I would keep that 33% chance. But if I changed the door, I would have a 66 percent chance because I already know that one of the doors is not the door. And the other door that might be the door I want only has a 33% chance of being right. Door 1= 0% because that door was opened already and was incorrect. Door 2= 33% that means Door 3= 66% because 0+33+66 is about 100% which is the total percentage.
I did well due to the probability increasing after they showed you a door with a pig behind it. At this point there is a half and half chance of getting the grand prize. You can do this experiment thousands of times and get the prize less then half of the time but that does not mean the probability has changed at all. At the beginning of each game you still have a probability of 50 50.
In a way, I am right. The way I went with was a more of a gamble. The real way to win the game was to choose the middle door and switch the door. Out of the 10 trials of switching it was an 80% chance of winning the prize. When I stuck with the same door, out of the 50 trials made, I was only lucky 26% of the time. The prediction of winning this game is too switch the door, starting out you have a 33% of winning, you pick one it gives you the wrong door and to choose from two doors, now you have a 50% chance of winning. The computer gave me the experimental probability and based my decision on the percentage to switch the door or stay the same!
I was incorrect but I was correct about the estimation of the percents. A game stayed is very close to 33% while games switched is very close to 60%.
Games stayed: 44 Games stayed and won: 16 Experimental to win: 36.36%
Games switched: 17 Games switched and one: 10 Experimental probability to win: 58.82%
The reason why the experimental probability changed is because many different factors. First off it an estimation problem and I’m taking an educated guess on the experimental probability for games stayed and games switched. Also bias can have an effect on the probability because one would really want there answers to be correct. Number of times I did the experiment can have an affect, there a huge difference if I did it thee times or a thousand times. That’s why this is called an estimation problem.
C.H. I was right in some places and wrong in others. Initially you have a 33.3% chance of clicking on the door with a cruise behind it while you have a 66.6% chance of getting a pig. When you click your door the door that opens will always be a pig. Now that there are two doors left with a pig behind one and a cruise behind another you have a 50% chance of getting a cruise and a 50% chance of a pig. every time you start the game over the percentages go back to as they were before you picked a door. Even though that this is the theoretical probability it seemed that when you were given the chance to stay or switch it seemed that there was closer to a 60 or 70% chance of getting the pig instead of the cruise.
If you choose door 1 it is a 33% chance of winning. When a wrong door opens up, the unopened door has a 50% chance of being right. The first one you chose still has a 33% chance of winning. So the other door has a higher chance of winning because it is 50% rather than 33% of winning.
When you choose you first door, it has a 33% chance of being correct. Then when they open the other door, the third door now has a 50% chance, while the door you chose still has a 33% chance of being correct. This means, the other door has a higher chance of winning, because it is 50% than 33%.
Games stayed: # of games played: 65 # of games won: 13 Percentage: 20%
Games switched: # of games played: 65 # of games won: 42 Percentage: 64.62%
My prediction was right! As you can see above that switching had the higher percentage, but this is not always the case for every trial. Let’s talk about coins. You have a 50-50 chance of getting heads, but once you flip it 100 times, you can get 43 heads or 65 heads instead of 50. So you could play this door game 65 times again and get different data, you could switch and win 50 times and when you stay you win 20, I think it differs every time! The theoretical data is you have 33.3 chance at first and then you have a 66.6 chance of getting the right door if you were to switch!
I was right, my prediction was that switching doors would give you a 66% chance to win and staying would give you a 33% chance to win. The reason for this is that when you first pick the door you have a 33% chance of winning, by always keeping that door you keep your 33%. If you switch door you technically have a 66% chance of getting the winning door, you may think because you are deciding between the original door you picked and the new door its 50%, but its not. It is 66% because technically the only way you can lose is if the original door you picked is the winning door, here is an example.
Door 1 Door 2 (door you picked) Door 3
This is a 33% chance of the correct door.
Door 1 Door 2 (door you picked) Door 3 wrong door (opened)
Your door is still a 33% chance of being the correct door; just because you now know one of the wrong doors it doesn’t make your guess any better than it originally was. So if you think about it like this, the wrong door is obviously the wrong door, the door you picked is still the original 33% chance so that gives the other door a 66% chance.
Of the 30 games I played for each strategy I got 60% chance to win when you switch and 30% chance to win when you stay.
22 comments:
JE
So, my theory was correct. U start out with 1/3 chance of getting the prize and 2/3 of missing the prize. So after one door is opened, u could stay with the door with only 1/3 chance or you could change with 2/3s the chance with getting the prize.
I was wrong with my prediction. I was pretty close but it was also kind of off. I said that if i stayed i had a 1/3 chance of getting it but i actually got about 40% of getting it. also, i said if i changed the doors that i was to get about 50% right but i actually got about 70% right.
M.R.
I was right, my prediction was that switching doors would give you a 66% chance to win and staying would give you a 33% chance to win. Of the 30 games I played for each strategy I got 60% chance to win when you switch and 30% chance to win when you stay.
N.E.
My prediction was right because I stayed 101 times and I switched 101 times. Out of the 101 times I stayed I won 24 times which is 23.76% of the time.Out of the 101 times I switched I won 66 times which is 65.35% of the time. So my conclusion is that you have a greater probability of winning by switching your door than staying with the door that you originally picked.
AH
I played the game a total of 200 times, switching doors 100 times and keeping my door the other 100 times. When i switched doors, I won 69 times meaning that there is a 69% chance you'll win if you switch doors. i won 39 times when i kept my door meaning there is a 39% chance you'll win if you keep your door. My conclusion is that it is better to switch your door because it is more likely you'll win.
CR
I played the game and it was fun. I played the game 60 times. During the game I switched my door 30 times and kept my door 30 times. In doing that I won 13 games when I kept my door and 15 games when I switched my door. In the results of that, my experimental probability of winning was 43.33% when I kept my first door and 50% chance when I switched my door. Playing this game tells you that in switching your door you will have a better chance of winning then if you keep your first door.
J.M.
When I played this game, I won more times when I switched the doors than when I stayed with the door I chose before. I played a total of 150 games, 75 games when I switched and 75 games when I stayed. I won 49 when I switched doors, which means I have a 69% chance of winning when I switch doors. I only won 25 games when I stayed with the same door which means I have a 25% chance of winning when I stayed with the same door.
J.S.
My results say that I won more games when i switched. I played 200 games in all. 100 for switching and 100 for staying. I won 35 games when i stayed and 60 when i switched. That's a 35% chance of winning if i stay with the door and 60% chance of winning if i switch doors.
AM
I played 80 games.
In staying, I got 29 winning games, so 36.25% chance of winning.
In switching, I got 48 winning games, a 60% chance of winning.
This is because…
When you open the door that is guaranteed wrong, you will have a fifty, fifty percent chance of getting the cruise and money. This means, however that there is a fifty percent chance your door is wrong, but it couldn’t be opened because you chose it. When you choose to switch, you can get rid of the chance of your door being wrong for the most part. That’s why there is a higher chance. You have to use experimental probability.
I was incorrect but I was correct about the estimation of the percents. A game stayed is very close to 33% while games switched is very close to 60%.
Games stayed: 44
Games stayed and won: 16
Experimental to win: 36.36%
Games switched: 17
Games switched and one: 10
Experimental probability to win: 58.82%
That game was fun! My results were completely unexpected. I played the game 80 times, the first 40 I stayed and the second 40 I switched. As I already explained you have a 50/50 chance of getting the prize after u click the first door, so the probability of getting the prize is 50%, not 33% like you may think. But my experimental probability of winning said otherwise. When I stayed I had an experimental probability for winning of 35%. This is less than 50% like the probability shows. When I switched I had an experimental probability to win of 57.5%. This also isn’t 50% like the probability. This is because the experimental probability is not the same as a theoretical probability. On a 50% chance it is completely random for the results. This is why my experimental probabilities are not 50%.
R.H.
I believe I did pretty well. I won about 62% of the time. I won more times when I switched the door. I was sort of right but I used the experimental probability the computer gave me and I based my decision to switch the door on that. If the percent was higher to switch the door, I switched the door. If the percent was higher to stay with the same door, I stayed the same.
J.K. 2/13/09
I was not right. At first, I said I had a 1/3 chance of getting the door right if I stayed, and I was close enough to 33 percent to say I was right, even though I got 40% right. I was wrong, however, about the probability if I switched doors. I said 50% but it was closer to 2/3. I got 70% of the doors right when I switch. I was wrong about that part. If I stayed with the door I chose after a door was opened, I would keep that 33% chance. But if I changed the door, I would have a 66 percent chance because I already know that one of the doors is not the door. And the other door that might be the door I want only has a 33% chance of being right. Door 1= 0% because that door was opened already and was incorrect. Door 2= 33% that means Door 3= 66% because 0+33+66 is about 100% which is the total percentage.
M.G
I did well due to the probability increasing after they showed you a door with a pig behind it. At this point there is a half and half chance of getting the grand prize. You can do this experiment thousands of times and get the prize less then half of the time but that does not mean the probability has changed at all. At the beginning of each game you still have a probability of 50 50.
B.B.
In a way, I am right. The way I went with was a more of a gamble. The real way to win the game was to choose the middle door and switch the door. Out of the 10 trials of switching it was an 80% chance of winning the prize. When I stuck with the same door, out of the 50 trials made, I was only lucky 26% of the time. The prediction of winning this game is too switch the door, starting out you have a 33% of winning, you pick one it gives you the wrong door and to choose from two doors, now you have a 50% chance of winning. The computer gave me the experimental probability and based my decision on the percentage to switch the door or stay the same!
C.C.
I was incorrect but I was correct about the estimation of the percents. A game stayed is very close to 33% while games switched is very close to 60%.
Games stayed: 44
Games stayed and won: 16
Experimental to win: 36.36%
Games switched: 17
Games switched and one: 10
Experimental probability to win: 58.82%
The reason why the experimental probability changed is because many different factors. First off it an estimation problem and I’m taking an educated guess on the experimental probability for games stayed and games switched. Also bias can have an effect on the probability because one would really want there answers to be correct. Number of times I did the experiment can have an affect, there a huge difference if I did it thee times or a thousand times. That’s why this is called an estimation problem.
C.H.
I was right in some places and wrong in others. Initially you have a 33.3% chance of clicking on the door with a cruise behind it while you have a 66.6% chance of getting a pig. When you click your door the door that opens will always be a pig. Now that there are two doors left with a pig behind one and a cruise behind another you have a 50% chance of getting a cruise and a 50% chance of a pig. every time you start the game over the percentages go back to as they were before you picked a door. Even though that this is the theoretical probability it seemed that when you were given the chance to stay or switch it seemed that there was closer to a 60 or 70% chance of getting the pig instead of the cruise.
J.S.
If you choose door 1 it is a 33% chance of winning. When a wrong door opens up, the unopened door has a 50% chance of being right. The first one you chose still has a 33% chance of winning. So the other door has a higher chance of winning because it is 50% rather than 33% of winning.
J.M.
When you choose you first door, it has a 33% chance of being correct. Then when they open the other door, the third door now has a 50% chance, while the door you chose still has a 33% chance of being correct. This means, the other door has a higher chance of winning, because it is 50% than 33%.
JAL
Games stayed:
# of games played: 65
# of games won: 13
Percentage: 20%
Games switched:
# of games played: 65
# of games won: 42
Percentage: 64.62%
My prediction was right! As you can see above that switching had the higher percentage, but this is not always the case for every trial.
Let’s talk about coins. You have a 50-50 chance of getting heads, but once you flip it 100 times, you can get 43 heads or 65 heads instead of 50.
So you could play this door game 65 times again and get different data, you could switch and win 50 times and when you stay you win 20, I think it differs every time! The theoretical data is you have 33.3 chance at first and then you have a 66.6 chance of getting the right door if you were to switch!
I was right, my prediction was that switching doors would give you a 66% chance to win and staying would give you a 33% chance to win. The reason for this is that when you first pick the door you have a 33% chance of winning, by always keeping that door you keep your 33%. If you switch door you technically have a 66% chance of getting the winning door, you may think because you are deciding between the original door you picked and the new door its 50%, but its not. It is 66% because technically the only way you can lose is if the original door you picked is the winning door, here is an example.
Door 1
Door 2 (door you picked)
Door 3
This is a 33% chance of the correct door.
Door 1
Door 2 (door you picked)
Door 3 wrong door (opened)
Your door is still a 33% chance of being the correct door; just because you now know one of the wrong doors it doesn’t make your guess any better than it originally was. So if you think about it like this, the wrong door is obviously the wrong door, the door you picked is still the original 33% chance so that gives the other door a 66% chance.
Of the 30 games I played for each strategy I got 60% chance to win when you switch and 30% chance to win when you stay.
AD
Games Stayed- 30
Games Won- 7
Winning Probability- 23%
Games Switched- 30
Games Won- 19
Winning Probability- 63%
The results show that switching doors is the better choice because the results were more than doubled in games won.
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