Let's Make a Deal!
Forums › General Discussion › Let's Make a Deal!-
On the old TV game show, a contestant is presented with 3 doors. Behind one door is a prize.
Step 1: player picks random door.
Step 2: host opens the non-winning door that the user did NOT pick and asks: "Keep your door or switch to the other?" Two doors remain closed, one has a prize.
Step 3: player stays on their original door, (usually), it is opened and they either win or they don't.
Is it better to switch, or stick, or doesn't matter?
I have a program on the web to test this experiment after discussion, just like Mythbusters. 😄
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Better to switch.
;) -
Well either way it's a 50-50 chance, I can't imagine one being better than the other
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This will be fun!
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Only Monty Hall knows the answer to this.
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Depends, am I allready holding a wad of cash in my grubby paws and what kind of costume am I wearing?
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This problem freaks me out because the answer is so obvious, and wrong. It's not called a paradox for nothing. It fooled millions of tv viewers.
Clearly, 2 doors closed, one prize, 50-50 chance! Better to stick with your first intuition, right?
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33.3% chances per door.
Lose one, the rest are 50/50right?
Wrong.
Staying as one 33.3 is removed from play, the door you previously chose is still 33.3%, while the other door jumps to 66.6%.
It's better to switch. -
🔰Bཞuęℵǿཞ🔰 wrote:
That's just silly. 2 doors, one prize, 50-50.33.3% chances per door.
Lose one, the rest are 50/50right?
Wrong.
Staying as one 33.3 is removed from play, the door you previously chose is still 33.3%, while the other door jumps to 66.6%.
It's better to switch. -
Once the 1st "wrong door" is removed the odds change for the door you originally chose. It doesn't stay at 33.3%. Both remaining door odds change to 50%. it's the same with betting. Odds are never fixed.
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Brown🎵Note😲 wrote:
And that's why it was a game show!🔰Bཞuęℵǿཞ🔰 wrote:
That's just silly. 2 doors, one prize, 50-50.33.3% chances per door.
Lose one, the rest are 50/50right?
Wrong.
Staying as one 33.3 is removed from play, the door you previously chose is still 33.3%, while the other door jumps to 66.6%.
It's better to switch. -
Add XXL wrote:
Yeah.Once the 1st "wrong door" is removed the odds change for the door you originally chose. It doesn't stay at 33.3%. Both remaining door odds change to 50%. it's the same with betting. Odds are never fixed.
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But it's not 2 doors. It's a 3 for equation, with one variable moved to .1 for you. Looks like this.. variables for percent chance.
A(x)+B(y)+C(z)=100%A(33.3)+B(33.3)+C(33.3)=100(99.9)
Let's say A is opend.
A(0)+B(33.3)+C(66.6)=100(99.9)
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You guys are still forgetting. The person usually guessing what door to pick already has a nice prize and that factors in a lot.
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🔰Bཞuęℵǿཞ🔰 wrote:
Stop clouding the issue with crazy math.But it's not 2 doors. It's a 3 for equation, with one variable moved to .1 for you. Looks like this.. variables for percent chance.
A(x)+B(y)+C(z)=100%A(33.3)+B(33.3)+C(33.3)=100(99.9)
Let's say A is opend.
A(0)+B(33.3)+C(66.6)=100(99.9)
One door will always be thrown out, so it's always a choice between 2 doors. How can your thinking about a door change the odds?
Nope. Flip a coin.
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Amish Hitman wrote:
Oh, yeah, that's true. What's in the box might be better than what's behind door number 1.You guys are still forgetting. The person usually guessing what door to pick already has a nice prize and that factors in a lot.
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You guys are so cute!
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Brown🎵Note😲 wrote:
Dam you I know what I'm saying, how does it not make sense to you?🔰Bཞuęℵǿཞ🔰 wrote:
Stop clouding the issue with crazy math.But it's not 2 doors. It's a 3 for equation, with one variable moved to .1 for you. Looks like this.. variables for percent chance.
A(x)+B(y)+C(z)=100%A(33.3)+B(33.3)+C(33.3)=100(99.9)
Let's say A is opend.
A(0)+B(33.3)+C(66.6)=100(99.9)
One door will always be thrown out, so it's always a choice between 2 doors. How can your thinking about a door change the odds?
Nope. Flip a coin.
NO MATTER IF THE DOOR IS OPEND, it's still 3 doors it could have been behind, not only 2.
3 doors, 3 variables, even if the one is 0 it still devides the 100% by 3 -
🔰Bཞuęℵǿཞ🔰 wrote:
What he said33.3% chances per door.
Lose one, the rest are 50/50right?
Wrong.
Staying as one 33.3 is removed from play, the door you previously chose is still 33.3%, while the other door jumps to 66.6%.
It's better to switch. -
Bruenor, I don't understand your logic. You have one prize, three doors;1/3. You choose a door and do not "replace" it; 1/2. 50% chance then, because you have narrowed it down to 2 doors. But I've never heard if this gameshow so I'm just going on what I've seen in this thread.
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🔰Bཞuęℵǿཞ🔰 wrote:
2 doors, one prize, 50-50.Brown🎵Note😲 wrote:
Dam you I know what I'm saying, how does it not make sense to you?🔰Bཞuęℵǿཞ🔰 wrote:
Stop clouding the issue with crazy math.But it's not 2 doors. It's a 3 for equation, with one variable moved to .1 for you. Looks like this.. variables for percent chance.
A(x)+B(y)+C(z)=100%A(33.3)+B(33.3)+C(33.3)=100(99.9)
Let's say A is opend.
A(0)+B(33.3)+C(66.6)=100(99.9)
One door will always be thrown out, so it's always a choice between 2 doors. How can your thinking about a door change the odds?
Nope. Flip a coin.
NO MATTER IF THE DOOR IS OPEND, it's still 3 doors it could have been behind, not only 2.
3 doors, 3 variables, even if the one is 0 it still devides the 100% by 3 -
Look, even though I know it's wrong, it's hard to argue.
It always comes down to 2 doors, one prize, pick one.
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THE EQUATION DOSENT CHANGE.
It has to equal 100%.(in this case 99.9 rounded)The original equation is
33.3+33.3+33.3=99.9
Removing one from play, its variable factor does not go to both to make 50/50. It only transfers to 1 making it
0+33.3+66.6=99.9
This is a dam advanced variable replacement method, way past your basic algebra, which causes you to make it 50/50. It's not.
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🔰Bཞuęℵǿཞ🔰 wrote:
The point is, you can't explain it in a way that 99% of us will believe without proof.THE EQUATION DOSENT CHANGE.
It has to equal 100%.(in this case 99.9 rounded)The original equation is
33.3+33.3+33.3=99.9
Removing one from play, its variable factor does not go to both to make 50/50. It only transfers to 1 making it
0+33.3+66.6=99.9
This is a dam advanced variable replacement method, way past your basic algebra, which causes you to make it 50/50. It's not.
You can say that all day, but it is counter to common sense, even if it is wrong.
(A)(B+C)
33.....66You pick A.
B or C is removed. Say C.You now have a free choice:
(A)(B)
50..50 -
I'm sorry I sat down real quick and looked at the math. Bruenor is right
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Dammit no you don't.
Let's suppose that you guessed correctly. Then it makes no difference what the game show host does, the other door is always the wrong door. So in that case, by keeping your choice, the probability that you win is 1/3 x 1 = 1/3.
But let's suppose you guessed incorrectly. In that case, the remaining door is guaranteed to be the correct door. Thus, by keeping your choice, the probability of winning is 2/3 x 0 = 0.
Your total chances of winning by keeping your guess is: 1/3 + 0 = 1/3.
Again, let's suppose that you guessed correctly. By changing your guess the probability that you win is 1/3 x 0 = 0.But let's suppose you guessed incorrectly. Again, this means that the remaining door must be the correct one. Therefore by changing your choice, the probability of winning is 2/3 x 1 = 2/3.
Your total chances of winning by changing your guess is: 2/3 + 0 = 2/3.
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If you always change your guess, you'll win 2/3 times.
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Always switch, same as the one with all the red boxes.
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Switch
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The probability of winning by using the switching technique is 2/3 while the odds of winning by not switching is 1/3
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Shibby420 wrote:
No it's not, it's a 33:66 chanse, and it's better to switchWell either way it's a 50-50 chance, I can't imagine one being better than the other
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