Let's Make a Deal Event: Odds of Switching for a Win

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what explanation stands behind "let's make a deal"'s evenement? As i remember there are 3 doors : one with a car behind,and two with goats. After you choose 1 of three doors - they show you one door where there is a goat ,than give you a chance to switch the door and - as my friend claims - you should always switch a door because of the odds but he only read that in some article and can't explain.
How is it more probable that the "switch" is correct choice since it seems 50% chance that you have the right door already?
The only explanation I can see is that you already had one chance and you guessed "right" not taking the one witch the goat they show you , but the switch alone looks still as a fresh 50/50 to me... :(
Matt
 
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You just have to see that problem the right way. 1/3 of the time, you have already made the correct choice. But in 2/3 of the cases--since he has shown a goat, switching is to the correct door.

If it really bothers you, you can work it out with pieces of papers. Given three choices A,B,C, just place the correct choice at A 1/3 of the time, similarly at B and C. Then making the same choice on all occasions, say A, pick up the papers from the blank side, and switch. The secret is that the door opened shows a goat.
 
robert Ihnot said:
You just have to see that problem the right way. 1/3 of the time, you have already made the correct choice. But in 2/3 of the cases--since he has shown a goat, switching is to the correct door.

If it really bothers you, you can work it out with pieces of papers. Given three choices A,B,C, just place the correct choice at A 1/3 of the time, similarly at B and C. Then making the same choice on all occasions, say A, pick up the papers from the blank side, and switch. The secret is that the door opened shows a goat.

Does it mean that I need to see it as whole one event in order the switch on the end to be correct choice?Like in poker with a flush draw after the flop- 50/50(but I have to see the river) instead of 25% on turn +25% on river?
 
matt010nj, I don't understand your question but I have an analogy that might help your intuition...

Suppose that, instead of three doors, you have a million! Only one door out of the million doors has the car. Pick a door, any door! What are the chances that you've won a car?

After you've picked your door, let me help you out by opening 999,998 doors so that there are only two left closed: the door you chose and some other door. None of the 999,998 open doors revealed the car so you know that it must be behind one of the two doors left closed.

Do you want to change your guess?
 
mathman said:
http://en.wikipedia.org/wiki/Monty_Hall_problem

This is a well known "paradox". Look at the above.

ok I got it now,thanks
:) I Think in reality its similar to paradox about how hare never catches a turtle to whom he gave a head start :)
 
matt010nj said:
ok I got it now,thanks
:) I Think in reality its similar to paradox about how hare never catches a turtle to whom he gave a head start :)

It has no connection to the hare catching turtle paradox.
 
mathman said:
It has no connection to the hare catching turtle paradox.

why ? it is mathematicly/teoreticly correct but in real world just doesn't work just like turtle/hare(unless we treat it as whole event and ask about "switching" in advance).Am I wrong here?
My thinking is : at the point of choice - switch or not - we have two doors to choose from.
 
The fact that the host can't open your door and can't open the winning door biases the event
 
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