Win a Car

bohsman

Your on a gameshow and youre told behind one of three doors theres a car, the other two are empty.

You pick door no1, before opening it the host opens door number 2 showing you its empty and then asks you do you stick with door no1 or switch to no3

What do you do and why?

Mark J

i`d push, odds are there and taking it down would substantially increase my stack.

push or fold, i`d say.

smurph

bohsman wrote:

Your on a gameshow and youre told behind one of three doors theres a car, the other two are empty.

You pick door no1, before opening it the host opens door number 2 showing you its empty and then asks you do you stick with door no1 or switch to no3

What do you do and why?

Stick with No. 1 because if you change, and the Car is behind door 1 that would just drive me mad. Its like playing cards, stick with your first thought, unless your thinking of beer or vodka, instead of the game.

PPP-Pit Boss

Quick thread closing.. (eek) I would stick as I constantly berate myself whenever I do not use my gut instinct. It makes notdifference whether you change or not the odds are equal for both. However the self flagellation that would have to occur if you didnt just stick to your guns ....
odd question BUT NONETHELESS VERY POKER RELATED :0)

Ste05

Always change.

Mark J

Mods!!!!!!!!!

Jaden

I'd switch. Obvious, if you think about it.

bohsman

PPP-Pit Boss wrote:

Quick thread closing.. (eek) I would stick as I constantly berate myself whenever I do not use my gut instinct. It makes notdifference whether you change or not the odds are equal for both. However the self flagellation that would have to occur if you didnt just stick to your guns ....
odd question BUT NONETHELESS VERY POKER RELATED :0)

Probability is not poker related then?

Ste and Jaden are right btw

frodi

It's now become a 50/50 shot. With no other info to go on I'd stick with my first choice. I fancied it when it was 2/1 now it is evens, apart from eliminating one choice nothing else has changed.

Mark J

probability is - not gameshows.

bohsman

ok so there are three poker hands face down, one is aces the other two are pocket twos, you get to pick one the hand is then played out and if you win you get a car....

changes the odds slightly of course but the answer is still to switch

Mark J

bohsman wrote:

ok so there are three poker hands face down, one is aces the other two are pocket twos, you get to pick one the hand is then played out and if you win you get a car....

changes the odds slightly of course but the answer is still to switch

that`s more like it!!!!

on topic!


[size=-10](the mods are going mental lately, better off deleting/editing your first post.)[/size]

Mark J

bohsman wrote:

but the answer is still to switch

why? it`s 50/50. plus, i couldn`t live with myself if i changed and was wrong.
whereas, if i stuck with my original, prob. be p*ssed but not as much as changing my choice and then missing.

Lafortezza

What kind of car is it?

bohsman

lafortezza wrote:

What kind of car is it?

A red one if it makes any difference to your answer

Mark J

i`ve read somewhere to use a 4 coloured one.

so as not to make mistakes.

padraig_f

Switch. If you always stick on the first one you pick, you have a 33% chance of being right. If you always switch, you have a 50% chance of being right.

I thought of a similar one for 'Who Wants to be a Millionaire?'. You're on £64,000. The next question you have no idea of, and you have one lifeline left, a 50-50. You use the lifeline and still have no idea, do you guess the answer or leave and take the £64k?

frodi

padraig_f wrote:

Switch. If you always stick on the first one you pick, you have a 33% chance of being right. If you always switch, you have a 50% chance of being right.

I thought of a similar one for 'Who Wants to be a Millionaire?'. You're on £64,000. The next question you have no idea of, and you have one lifeline left, a 50-50. You use the lifeline and still have no idea, do you guess the answer or leave and take the £64k?

The elimination of one wrong answer means that the chance of your original answer being right has improved from 33% to 50%. You HAD a 33% chance of being right, you now HAVE a 50% chance of being right. The same holds for door 3, it was 33% right it is now 50% right. Just by ruling out door 2 the chances have improved switching does not alter the new probablity.

ocallagh

lol oscar!

The solution to this problem will totally wreck everyones heads!!!! good luck explaining it on boards... David just showed me a fairly cool way of explaining it which involves a pack of cards

jtsuited

please explain why switching is correct. I'm completely stumped!!!

mewso

I agree with frodi here but I am willing to be put right on it. At the start you pick 1 from 3 and have a 33% of picking the correct door. When you pick and are shown the empty door you are then asked do you want to switch. This is no different than being presented with just the two doors from the start and being asked to choose one. i.e. this is a second distinctive choice with a 50% chance of getting it right. So switching or not makes no difference imo.

NickyOD

It's a hoax. There is no advantage to switching.

Daithio

The correct answer is to always switch but not for the reasons mentioned. The key is that, after you have picked your door, the host is always going to choose one of the remaining doors with a goat behind it. Basically by switching every time you are allowing yourself to open two doors rather than just one, and so doubling your chances of getting the right one from 33pc to 66 pc.

Crumbs

It's no hoax. Switching is always the right play.

It originates from the famous "Monty Hall" problem (I think). Google it.

impr0v

Always switch. There was 33% chance that the car was behind the door you picked, and a 66% chance that it was behind the other two. Since one of those doors is now open, there's a 66% chance it's behind the unopened one.

Earwig Eddie

my leaving cert maths teacher used to hurt my brain with this one. as far as i remember, improv has it correct above. though the arguments can go on for hours...

mewso

This is being phrased to make a point and I know where I read this (a limit poker book I bought recently, forget the name).

Someone explain how the door you haven't picked goes from 33% to 66% while the one you did pick stays at 33%. Surely the one you picked has also increased in probability to 50% and the one you didnt to 50% also.

*Edit - best explanation I could find - http://www.comedia.com/hot/monty-answer.html

Daithio

Musician, after you have picked your door, there is a 66 pc chance that one of the other two doors is the car. When the host eliminates one of these, the 33 pc chance that it had gets added to the remaining door that you didn't choose, not divided equally between the door you did choose and the last remaining one. This is because the host chose the door with the goat from the two remaining, not from the three like you did at the start. Take 3 cards, one of them a joker, pretending that this one is the car. Do out the experiment 20 times, always switching, and you'll see that you'll get the joker (or car) 66pc of the time.

sendic

http://math.ucsd.edu/~crypto/Monty/monty.html

mewso

Yeah thanks daithio I should have been more clear about the link above. It's an explanation as to why switching is correct and it's relatively clear so I have seen the light.

Duff Man Jr.

impr0v wrote:

Always switch. There was 33% chance that the car was behind the door you picked, and a 66% chance that it was behind the other two. Since one of those doors is now open, there's a 66% chance it's behind the unopened one.

Give that man a biscuit!

DeVore

I got this wrong when Oscar initially told me it because I misunderstood that the game show host will ALWAYS show you a booby prize (lets call it a goat).

The easiest way of understanding the reason why was worked out in a Barcelona pub.

Suppose you ALWAYS SWITCH.
If you pick the car initially and then switch you lose.
If you pick either goat initially and then switch you win. (you may need to think about that)

The chances of picking a goat initially are 2/3

Hence the "ALWAYS SWITCH" team has a 2/3 chance of winning a car.


Now lets look at the "ALWAYS STAY" team:
If they pick the car initially, they win.
ANYTHING ELSE THEY LOSE.

The chances of picking a car initially is 1/3, hence the "ALWAYS SWITCH" team wins twice as many cars as the "ALWAYS STAY" team.

DeV.

Dub13

with all those hot Spanish chicks around this is the kind of crap you lot were talking.... :eek:

DubTony

.... and people on this forum complained about bad beat posts?

:eek: :eek: :eek: :eek: :eek: :eek: :eek: :eek: :eek: :eek: :eek:

karlh

<jimbowen> OOOOOhhh hard luck pet, well you still get yer Bully tankard and B.F.H., but lets have a look at what ya could have won....... </jimbowen>

Fatboydim

Did this with the missus and a pack of cards. Ten times always switch, Ten times always stay. It's too small a sample to be really accurate but out of interest it worked out that always stick won 2/ 10 and always switch 7 / 10.

Which meant in reality that the king was picked out first time twice in the first instance and three times in the second.

So adopting a control - I conducted the test where the choice made by the host was random. In the control case always switching won 2 / 10 . Always sticking won 5 / 10.

So in conclusion the odds are only going to change if the host knows what's behind the doors.

How this helps you in poker however I have no idea.

Crumbs

DubTony wrote:

.... and people on this forum complained about bad beat posts?

Fatboydim wrote:

How this helps you in poker however I have no idea.

It's a very good exercise in conditional probability which is an important part of poker.

The application of this particular problem mightn't be relevant but the reasoning behind it is.

frodi

sendic wrote:

http://math.ucsd.edu/~crypto/Monty/monty.html

I change my position. This site explains it very well. The crucial bit of info is that the host KNOWS that there is a goat behind the opened door. I would have prefered a blue car however those goats look attractive. :eek:

Iago

I still don't get this...

If I choose the door with a car behind it...the host is going to open a door without a car in it so my odds of being right have changed from 3:1 to 2:1

If I choose a door without a car behind it...the host is going to open one of the two doors that doesn't have the car so my odds are still reduced from 3:1 to 2:1

How does switching change those odds? If I play this game 100 times and everytime the host opens a door that doesn't have a car in it, regardless of whether I chose rightly or wrongly then that shouldn't affect the overall outcome. My odds of being right, once that initial door has been revealed are still 50:50...

The 2nd choice is the key decision point, the argument that you are now making a 50:50 choice rather than 33:33:33 choice is the main contention behind this problem. However you have to take this decision in isolation, essentially you have a new problem, one door has a goat, one has a car pick the right one. It's a 50:50 toss up.

Unless the host will only open a door with the goat behind it when you choose wrongly/correctly then I still believe it makes no material difference whether you switch or not.

Culchie

I'm in the same place as you Iago, but await with interest explanation of 'switching' advantages.

Imposter

As has been already said:
You first choose a door with a 33% chance.
Now you can stick with this or choose to open both of the other doors (as the host opening the other door effectively means you get to see behind 2 doors).
Therefore by switching you get a 66% chance.

NickyOD

There are 3 possible outcomes if you always switch.

you pick goat #1 and host reveals goat #2. You switch and WIN
you pick goat #2 and hoat reveals goat #1. you switch and WIN
you pick the car. host reveals a got and you switch and LOSE

Therefor switching gives you 2/3 chance of winning

There are 3 outcomes if you always stick

you pick goat #1 and stick and LOSE
you pick goat #2 and stick and LOSE
you pickt he car and stick and WIN

so sticking is always 1/3 chance of winning.

Culchie

Imposter wrote:

As has been already said:
You first choose a door with a 33% chance.

OK, but there are 3 doors to choose from ...33% , and I haven't made a decision, because Host has now stepped in and given further info.

Imposter wrote:

Now you can stick with this or choose to open both of the other doors (as the host opening the other door effectively means you get to see behind 2 doors).
Therefore by switching you get a 66% chance.

Decision to be made, 2 doors left, 1 has a car behind it, 50% chance.

I hated Maths

Imposter

Culchie wrote:

Decision to be made, 2 doors left, 1 has a car behind it, 50% chance.

I hated Maths

The host has given you info. If the host did not know the correct door and he was opening a random door (which happens to have a goat behind - if it was the car he opened you'd pick that) then it's 50:50. But the host knows where the car is, meaning with his help you are seeing 2 doors if you switch and not one as happens when you don't switch.

PPP-Pit Boss

You're all mental. I still say oddswise it makes no difference the only difference being the overwhelming regret factor. No hint to host participation we given in the question so it cannot be taken into consideration as it is merely hypothetical..

On another note .... I trust we shall be watching the heats of the WSPO on Poker channel tonight and tomorrow night? I and X and X are going back for the semi's today and final today ;0)

Culchie

I won't pretend "I get it" .

This is where I ask teacher if I can drop down to the Pass Maths class and not cause a fuss, because I think teacher is going to burst a blood vessel cos I'm so thick.

To me they are independent decision points to be taken based on known information (1 out of 3 chance initially) then 1 out of 2, once the host has eliminated a booby prize.
No matter how many times I have each decision to make, I will have the same odds every time.

I won't frustrate you any longer, I'm off to sulk.

Iago

A = Car

B = Goat 1

C = Goat 2

I choose A host opens B, I switch to C I lose
I choose C host opens B, I switch to A I win
I choose B host opens C, I switch to A I win
I choose A host opens B, I stay with A I win
I choose C, host opens B, I stay with C I lose,
I choose B, host opens C, I stay with B I lose

50/50

1. No matter what I choose initially I get to see behind two doors

2. No matter if I choose rightly or wrongly the host is going to show me one of the wrong answers

3. There are only 6 decisions I can make, 3 will see me win, 3 will see me lose

NickyOD

Ah lads come on now. :rolleyes:

frodi

Iago wrote:

A = Car

B = Goat 1

C = Goat 2

I choose A host opens B, I switch to C I lose
I choose C host opens B, I switch to A I win
I choose B host opens C, I switch to A I win
I choose A host opens B, I stay with A I win
I choose C, host opens B, I stay with C I lose,
I choose B, host opens C, I stay with B I lose

50/50

1. No matter what I choose initially I get to see behind two doors

2. No matter if I choose rightly or wrongly the host is going to show me one of the wrong answers

3. There are only 6 decisions I can make, 3 will see me win, 3 will see me lose

You switch, win twice lose once 66/33
You stay win once lose twice. 33/66
Overall win thrice, lose thrice 50/50

Winning play is to switch, twice as likely to win car.

Fatboydim

Iago you've answered it yourself - Looking at your ABCs when you switch you can win 2/3 times. When you don't you can only win 1/3 times. Stick and you trust to luck. Switch and you improve your odds. But it only works if the host knows the answers.

The simple way to test it if you like is to do what I did. Get a deck of cards and pull out two queens and a king. Shufle the cards and get another person to try and pick the K. Do this say 10 times. Each time they stick on their choice. Therefore the cards you pull away are irrelevant. They have stuck. Now do it where they switch each time and you will clearly see that there is an advantage. Because the amount of times they get it right in the first instance will be significantly lower than the amount of times they get it wrong. Therefore switching makes sense.

PPP-Pit Boss

My head just exploded
thanks for that Fatboy
:0)