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# Whatchya Gonna Do (2 goats and a car)

196 replies to this topic

### #21 Martini

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Posted 05 February 2008 - 09:12 PM

Look at the final scenario not the initial ones. You have two doors. One has a goat behind it; one has a car. It's 50/50. The rest is just smoke and mirrors.

Did you try to follow my scenario? Did you try the computer simulation I linked to? There's no smoke and mirrors. Try to follow this:

Behind door #1 there is a goat named Betsy. Behind door #2 there is a goat named Fred. Behind door #3 there is a car.

Three possible scenarios if you choose to switch doors when asked:

1. You pick door #1 (Betsy). The host will show you Fred behind door #2 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

2. You pick door #2 (Fred). The host will show you Betsy behind door #1 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

3. You pick door #3 (the car). The host will show you a goat behind one of the remaining doors and then asks if you'd like to switch your door for the other one (which also has a goat behind it). You choose to switch. You win a goat.

Out of the three possible scenarios in which you choose to switch doors, you will win the car in two of them.
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### #22 Maximus

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Posted 06 February 2008 - 12:22 AM

Although the simulation does seem to help with your side of the arguement (everyone but a few)

I am still struggling to understand it surely at first there is a one in three chance of the car being behind the door you pick, but when the host reveals the goat the chances then chance from one in three to one in TWO as there are now only 2 possible places the car could be?????
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### #23 Jkyle1980

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Posted 06 February 2008 - 02:19 AM

Door numbers do not matter. There are four possible outcomes:

You switch and get a car.
You switch and get a goat.
You do not switch and get a car.
You do not switch and get a goat.

50/50. It's even.
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### #24 Martini

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Posted 06 February 2008 - 03:10 AM

Door numbers do not matter. There are four possible outcomes:

You switch and get a car.
You switch and get a goat.
You do not switch and get a car.
You do not switch and get a goat.

50/50. It's even.

That's not how you go about figuring out probability. Using your rationale:

There are two possible outcomes when buying a lottery ticket:

50/50. It's even.

Did you or did you not try the simulation I linked to? I went over the three possible scenarios for you and outlined how two out of three of those scenarios the car is won. You're not even bothering trying to refute it.
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### #25 Martini

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Posted 06 February 2008 - 03:12 AM

Although the simulation does seem to help with your side of the arguement (everyone but a few)

I am still struggling to understand it surely at first there is a one in three chance of the car being behind the door you pick, but when the host reveals the goat the chances then chance from one in three to one in TWO as there are now only 2 possible places the car could be?????

Yes, after one door is eliminated, there are two possible doors which the car could be behind. However, the host does not eliminate a door randomly as he never eliminates one with the car behind it.

Think about it this way: When you pick one of the doors, you have a 1 in 3 chance that you picked the door with the car behind it. If the host gave you a choice to switch your door for both of the others, wouldn't you do it so you can have a 2 in 3 chance of winning? Well, that's essentially what the host is doing, only he's revealing what's behind one of the losing doors first.
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### #26 Jkyle1980

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Posted 06 February 2008 - 03:45 AM

Yes, after one door is eliminated, there are two possible doors which the car could be behind. However, the host does not eliminate a door randomly as he never eliminates one with the car behind it.

Think about it this way: When you pick one of the doors, you have a 1 in 3 chance that you picked the door with the car behind it. If the host gave you a choice to switch your door for both of the others, wouldn't you do it so you can have a 2 in 3 chance of winning? Well, that's essentially what the host is doing, only he's revealing what's behind one of the losing doors first.

But he's not telling you to trade your original door for the other two. He's giving you the revealed door free no matter what based on your hypothetical. If it there no matter what, then it should no longer be figured into the probability.

(This is by far my favorite post, btw.)
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### #27 Martini

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Posted 06 February 2008 - 04:38 AM

But he's not telling you to trade your original door for the other two. He's giving you the revealed door free no matter what based on your hypothetical. If it there no matter what, then it should no longer be figured into the probability.

(This is by far my favorite post, btw.)

No, he's not allowing you to trade your original door for the other two. That's why I said that it's essentially what he's doing. I gave Maximus and you a way of looking at a similar scenario that leaves you with the same probability of winning the car in hopes that you would see why switching doors is the wise choice.

We agree that at first there is a 1 in 3 chance of picking the door with the car behind it. We also agree that if the host lets you switch your door for the other two, your probability increases to 2 in 3, correct? Then it should be evident to you that switching is the best choice and garners you the same increased probability of winning the car whether or not he allows you to keep what's behind both doors or first reveals a goat that you can't keep.

You said this earlier:

Look at the final scenario not the initial ones. You have two doors. One has a goat behind it; one has a car. It's 50/50. The rest is just smoke and mirrors.

What if there were 100 doors? One door has a car behind it, the other 99 have goats. (Remember, the host doesn't reveal what's behind doors randomly before giving you a chance to switch; he never reveals the car.) You pick a door. The host reveals 98 doors all with goats behind them. He now gives you a chance to switch your door for the one that he didn't open. Do you still think "One has a goat behind it; one has a car. It's 50/50"?
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### #28 bonanova

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Posted 06 February 2008 - 05:11 AM

I am still struggling to understand it surely at first there is a one in three chance of the car being behind the door you pick, but when the host reveals the goat the chances then chance from one in three to one in TWO as there are now only 2 possible places the car could be?????

You're on the right track - you should struggle with the idea of a 1/3 chance changing to a 1/2 chance.
Why should seeing a goat that you already knew was there affect your 1/3 chance?
It shouldn't. And it doesn't.

The odds are 100% that the car is behind one of the doors:

Door 1: The door you picked has a 1/3 chance. The other two doors, combined, have the remaining 2/3 chance.
Door 2: The opened door that revealed a goat has zero. Now you know where the 2/3 chance lies - the third door.
Door 3: The third door has the remaining odds: 1 minus 1/3. Roughly 2/3.

Should you make the switch?

Hint:
The number of choices open to you is not what is important here, because the success probabilities of the choices are not equal.
Lottery tickets have only two outcomes - you win or you lose. But not with equal probability.

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Vidi vici veni.

### #29 Jkyle1980

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Posted 06 February 2008 - 05:59 AM

No, he's not allowing you to trade your original door for the other two. That's why I said that it's essentially what he's doing. I gave Maximus and you a way of looking at a similar scenario that leaves you with the same probability of winning the car in hopes that you would see why switching doors is the wise choice.

We agree that at first there is a 1 in 3 chance of picking the door with the car behind it. We also agree that if the host lets you switch your door for the other two, your probability increases to 2 in 3, correct? Then it should be evident to you that switching is the best choice and garners you the same increased probability of winning the car whether or not he allows you to keep what's behind both doors or first reveals a goat that you can't keep.

You said this earlier:

What if there were 100 doors? One door has a car behind it, the other 99 have goats. (Remember, the host doesn't reveal what's behind doors randomly before giving you a chance to switch; he never reveals the car.) You pick a door. The host reveals 98 doors all with goats behind them. He now gives you a chance to switch your door for the one that he didn't open. Do you still think "One has a goat behind it; one has a car. It's 50/50"?

We do agree that a 2/3 chance of winning is better than a 1/3 chance of winning. However, we do not agree that switching to the other door increases your probability of winning to 2/3.

You're on the right track - you should struggle with the idea of a 1/3 chance changing to a 1/2 chance.
Why should seeing a goat that you already knew was there affect your 1/3 chance?
It shouldn't. And it doesn't.

The odds are 100% that the car is behind one of the doors:

Door 1: The door you picked has a 1/3 chance. The other two doors, combined, have the remaining 2/3 chance.
Door 2: The opened door that revealed a goat has zero. Now you know where the 2/3 chance lies - the third door.
Door 3: The third door has the remaining odds: 1 minus 1/3. Roughly 2/3.

Should you make the switch?

Hint:
The number of choices open to you is not what is important here, because the success probabilities of the choices are not equal.
Lottery tickets have only two outcomes - you win or you lose. But not with equal probability.

The three doors have a 1/3 chance of having a car before the first goat is revealed. After the goat is revealed, that 1/3 is split between the other two doors so that they each have 1/2 chance of winning.

"Pick a number between 1 and 3. By the way, it's not 1." gives you a 50/50 chance of choosing 2 or 3.

"Pick a number between 1 and 3." You pick 2. "By the way, it's not 1." Your decision here to stay with 2 or switch to 3 is the same as your original decision from the previous scenario. You are still just picking randomly between two objects, one winning and one losing. It is 50/50
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### #30 Gambler

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Posted 06 February 2008 - 06:55 AM

... the host, Monty Barker, showed her three doors ...

Monty who?

Ok, let me take a stab ...

If you win, you take the car. If you don't get the car, Monty wins.

You can have chose a door which has 1 in 3 chances to win. Monty then has, the

After you pick, just because Monty talks slick, opens your door, his door, Carol Merrill's door (Carol who?), or starts singing the blues, the chances of your door does not change. It's statistical; when you picked it, it had 1 in 3 and nothing can change that statistical fact.

Once that is accepted, you can quickly see that eliminating one of the remaining doors leaves the other remaining door with 2 in 3. This is because your door keeps 1 in 3 and the total always remains 3 in 3 so the remaining door must have 2 in 3.

I like
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