A brain teasing probability puzzle

Araysar

Loser Araysar

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Siddar_sl said:
No there were three doors two goats and one car. You had a 33% chance of picking the car and Host has a 66% chance of having the car at the end of round one. The slight of hand is that host has tricked you into thinking its a 50/50 chance in round two. Its not the 66% chance that host has the car from round one is still in play in round two even if your wish to ignore it.
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You only had a 33% chance if the game stopped after 1st round and then the host revealed whether you were right or wrong.

That never happens. You're automatically shuttled off to 2nd round, your initial choice never being acknowledged as right or wrong and one door is eliminated for you and you're asked to pick one of two doors now.

It is the illusion of choice.
 
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felldoh

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I love it. Someone is (apparantly) genuinely arguing that if you have three boxes, two of which contain goats, you have a 50% chance of selecting the one containing the car.
 
Araysar

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felldoh_sl said:
I love it. Someone is (apparantly) genuinely arguing that if you have three boxes, two of which contain goats, you have a 50% chance of selecting the one containing the car.
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You only have 2 boxes, hence 50/50. I can see why you are confused.

This is a perfect example of mathematical misdirection. The Monty Hall problem is very similar to this problem in regards to the sleight of hand.


Three men require lodging, and at the chosen hotel, management asserts the cost for one room's rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?
 
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felldoh

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So if you choose one of the 3 boxes, don't ever switch it, you are suggesting that 50% of the time you will choose the car?
 
Araysar

Loser Araysar

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felldoh_sl said:
So if you choose one of the 3 boxes, don't ever switch it, you are suggesting that 50% of the time you will choose the car?
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I'm suggesting that the 2nd round still offers you a choice. If you choose not to change your door, you're still making an implicit choice. That choice is 50/50.

There are 2 times in this problem when you get to make a choice, but only the second time counts. The first one is pure illusion because it delivers you to exact same set of options in your second choice.

One car, one goat, one choice. 50/50
 
Araysar

Loser Araysar

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Here's lets do this.

Let's say Monty Hall has 1,000 doors.

And he makes me select one. I choose a door and then MH discards the other 998 doors.

Were my odds 1 in a 1000? Or are my odds 1 in 2?

If the final outcome is always keeping my original door, or selecting the other door that Monty offers me after he eliminated the other 998 doors, how can my odds be 1 in a 1000 when the final choice (the only one that matters) is always picking one door out of 2?


You guys are hilarious. Idiocy by consensus.
 
Tuco

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Araysar_sl said:
Three men require lodging, and at the chosen hotel, management asserts the cost for one room's rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?
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for fun:

Start:
ManA: $10
ManB: $10
ManC: $10
Hotel: $0
Bellhop: $0

Pay for Lodging:
ManA: $0
ManB: $0
ManC: $0
Hotel: $30
Bellhop: $0

Hotel Refund Lodging:
ManA: $1
ManB: $1
ManC: $1
Hotel: $25
Bellhop: $2

End delta:
ManA: -$9
ManB: -$9
ManC: -$9
Hotel: +$25
Bellhop: +$2
=
0

No problem.
 
Araysar

Loser Araysar

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Exactly. No problem as long as you dont fall for the sleight of hand.

Now try this one, Tuco.

Let's say Monty Hall has 1,000 doors.

And he makes me select one. I choose a door and then MH always discards the other 998 doors and offers me to select the remaining door or keep my original door.

Are my odds 1 in a 1000? Or are my odds 1 in 2?
 
T

The Master

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I think the fact that Ara compared the problem to the distraction in a magic trick pretty much lets us know what level he puts mathematical proofs on....
 
Araysar

Loser Araysar

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The Master_sl said:
I think the fact that Ara compared the problem to the distraction in a magic trick pretty much lets us know what level he puts mathematical proofs on....
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I bet you think that quarters really do exist behind people's ears. So adorably gullible.
 
Tuco

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Araysar_sl said:
Exactly. No problem as long as you dont fall for the sleight of hand.

Now try this one, Tuco.

Let's say Monty Hall has 1,000 doors.

And he makes me select one. I choose a door and then MH always discards the other 998 doors and offers me to select the remaining door or keep my original door.

Are my odds 1 in a 1000? Or are my odds 1 in 2?
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Let's be exhaustive here:
1. Your odds of selecting the correct door in the first round are 1/1000.
2. Your odds of winning in the first round are 0.
3. Your odds of winning if you keep your choice in the second round are 1/1000
4. Your odds of winning if you choose are 1 - 1/1000 or 999/1000.
 
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felldoh

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Araysar_sl said:
I'm suggesting that the 2nd round still offers you a choice. If you choose not to change your door, you're still making an implicit choice. That choice is 50/50.

There are 2 times in this problem when you get to make a choice, but only the second time counts. The first one is pure illusion because it delivers you to exact same set of options in your second choice.

One car, one goat, one choice. 50/50
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The second round offers you a choice, but the odds never changed. There is still a (1/x) chance that your initial selection was the car, and a (x-1)/(x) chance that the car is the host's group of boxes. If you deny this to be true, then you believe that you have a 50% chance of initially selecting the car from a pool of that car + an infiniate number of goats.
 
iannis

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Define fat.

Because i'm just sayin, some guys see a little bit extra and are conditioned to be all like "eww fat". Her tummy doesn't have to be a washboard here.

As long as I don't have to push any overhang to get at the goods. That's kinda gross.
 
Araysar

Loser Araysar

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Tuco_sl said:
Let's be exhaustive here:
1. Your odds of selecting the correct door in the first round are 1/1000.
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Correct

Tuco_sl said:
2. Your odds of winning in the first round are 0.
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But Tuco, why wouldnt your odds be 1 in 1000? Why is it 0?

Tuco_sl said:
3. Your odds of winning if you keep your choice in the second round are 1/1000
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How can that be when you are presented with a binary option at the end? Is there a time when Monty hall doesnt remove 998 doors?


Tuco_sl said:
4. Your odds of winning if you choose are 1 - 1/1000 or 999/1000.
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1 in 2 because 998 doors are always removed for you. Regardless of any choices you make, your choice is always 1 car and 1 goat at the end, whether its 3 doors, 300 doors, 3,000 doors, 3,000,000 doors.
 
Siddar

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Araysar_sl said:
Here's lets do this.

Let's say Monty Hall has 1,000 doors.

And he makes me select one. I choose a door and then MH discards the other 998 doors.

Were my odds 1 in a 1000? Or are my odds 1 in 2?

If the final outcome is always keeping my original door, or selecting the other door that Monty offers me after he eliminated the other 998 doors, how can my odds be 1 in a 1000 when the final choice (the only one that matters) is always picking one door out of 2?


You guys are hilarious. Idiocy by consensus.
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The odds of you having the right door is .1% the odds of Monty having the right door are 99.9%.

The key is that Monty always throws away all wrong choices except for the possibility of being stuck with one wrong choice at end becase you chose the one correct answer before Monty. The odds of Monty having the right choice are the reverse of you making the correct on. The more you lower your odds of picking the right answer the higher you make Monty's.
 
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