blonde poker forum

Hello,

Dunno if this has been raised before, but I was reading Reds-thread about value and thought this may interest you.

Known as the Monty Hall or Three Doors problem, it was originally seen in a game show in the US many moons ago ( show was "Lets Make A Deal"). Please don't look up the answers, you'll enjoy this problem more if you work it out yourself. Lots of mathematicians at the time disagreed...

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat (he will never show the car if present, always a door with a goat). He then says to you, "Do you want to stick with your original choice or switch to door #2" (and he will always make this offer to stick or switch)

Is the probabililty of getting the car better if you stick to your original choice, or switch to door #2.

This is a difficult problem to describe without ambiguity, however that's my attempt. Have fun.

Cheers,

Eoan

The chance of you being correct is still 50/50.

Its either Door 1 or Door 2

The odds dont change if you change your mind and switch

Ohmy, is it like that cardtrick dan negreanu does on one of his video blogs...

il try find it and post the link when ppl have had a bash at this...

http://www.fullcontactpoker.com/video_diary/

It's been discussed before. Negreanu uses the same principle using cards and explains it very simply on his video blog. choose crazy neighbours and card tricks.

There's no trickery involved.

I don't think it's been discussed before on Blonde, I know it's been discussed elsewhere. I'll have a look at that link, but others should try the problem first.

yes, if u switch your twice as likely to be correct!


As, Say u pick #1, the car will be in #2+#3 twice as likely as it is in #1.

the odds have increased but to 50/50, when there was 3 doors, it was a 33% shot, now it's a 50% shot.

Doesn't matter which you pick.

It was discussed in the deal or no deal thread i think.

Ah ok, well it's got a dedicated home now.. I'll check out the other thread and see how far it got.

Cheers,

Eoan

found it... http://blondepoker.com/forum/index.php?topic=4121.msg115824#msg115824

Still think it needs it's own thread, it's kinda lost in there.

Saw it in previous thread - Car twice as likely to be behind door 2.

Heh, enough of the previous thread stuff  :redcard: Some folk may never have read page 6 and 7 of the deal or no deal thread ;) I know it's a disgrace if they haven't, but that's just the way it is.

Having now watched DN's blog i agree with the point that when you have a group of 2 cards and a 1 card alone then the chances of the ace of spades being in the group of 2 is twice as likely - but then once the card has been exposed you are back to being either/or a 50/50 shot.

If you are playing cards and you are up against two opponents and you all had the same hand AK o/s.

You are a 33.3% favourite to win therefore making the chances of either of your opponents winning as 66.6% - if you get one of your opponents to fold your chances of winning the pot dont stay at 33.3% - they rise to 50% as you now have an equal chance against just the one opponent.

but then i'd take the goats over a car every day of the week!!  :D

In this problem the past influences the future quite considerably. You know that he will always expose a door with a goat, and you know that he will always offer you the chance to stick or switch.

So he shows you the door with the goats - if they then mix up the remaining 2 doors so you cant remember which one you picked originally and ask you to pick what are the odds of you finding the car?

I hadn't read the other thread, but I have seen the Monty Hall problem before anyway. Gatso's explanation (#94) on the Deal or no Deal thread is a good straightforward explanation of the answer if you haven't seen it before and can't be bothered to work it out yourself.  8)

50/50

But they are not doing this. It is based on the fact that your first pick is 2/1 against.

Yes, if you change the problem like that then it does become 50/50 :)

Correct.. Here's my explanation:

Basically at the start of the show you know you can pick one door which has a 1 in 3 chance of being the car.

This leaves a 2 in 3 chance that the car is behind one of the other doors.

You know that he will always give you the option to stick with your original choice or switch.

The exposure of the door with a goat is a red herring, it's completely irrelevant other than to confuse you into thinking the problem is now 50/50.

He may as well have said at the start,choose a door, and then decide if you want to stick with your original choice or switch to the other 2 doors.

http://en.wikipedia.org/wiki/Monty_Hall_problem

For a very detailed explanation, it also points out that the monty hall advantage is lost in the deal or no deal game.

Another explanation :) I love this problem, when I first saw it it was clearly 50/50... :(

It's equivalent of having 1 million doors and having a car behind 1 of them and goats behind the others.

You choose 1 door a 1 in 1 million chance of getting the car. The host (who knows where the car and goats are) then shows 999,998 doors with goats and leaves just your original choice, and 1 other door. He offers you the chance to stick or switch.

Ok then by the same principle that you explain - how come in the example i have given your odds of winning the hand dont stay at 33.3%?

The odds are only increased because of the way they choose to explain it, it's 50/50 once one door has gone.

Take 3 from 6 counting backwards, 6 - 5 - 4.  6-3 isn't 4 but the way you do it makes it 4.

Its all so mathematical with you internet players!!!!!

If its not poker tracker its something else.......sigh!! :D

This is a LIVE situation!!! you have to incorporate whether or not the gameshow host has a well trimmed beard or is clean shaven, this is integral to making the right choice.

Needless to say now Ive highlighted the beard aspect your all able to very easily work out whether or not the correct move is to stick or change, i dont have to tell you that if he has a well trimmed beard........well, its just soooooooo obvious ::)

A simple and easy way of looking at it is that the ONLY way of losing by swapping is if you picked the winning door in the first place which was a 1 in 3 chance. So the chance of NOT losing by swapping is 2 in 3.

What if you want the goat?

:redcard: Pervert.. 8) :D

The example you give matt, bares no resemblence to this problem.

As pointed out earlier this is a problem of prior action and knowledge affecting the choice.

If you choose 1 from 3 there is twice as much chance that you did not choose correctly than you did choose correctly.

The host is always going to expose a goat and offer you a change.

The intial odds now affect the choice as the chance of your intial choice being correct remain at 1/3 whilst the section with 2/3 chance of containing the car still exists but with one of the options exposed.


or that which is very clear

Oi!! Where i come from a goat is far more practical than a car!!  :D

Goats + boxes usually = indignant Sark!!

They're stuffing goats in boxes for the purpose of entertainment over here Sark, get over here!

http://blondepoker.com/forum/index.php?topic=11985.0  ;D

;D

You've missed a  factor here - Monty Hall KNOWS the winner & will expose 1 loser.

I know - there is always a 100% chance that one of the "losing" hands/doors will be in the group of 2 then removed from the equation.

I got the problem after reading about it but it seemed that the only people posting answers on the thread were those who had seen and read about this "problem" before - and therefore knew the answer and explanation.

By asking other problems and throwing different examples into the equation it is easy to see who knows more other than just what they read from the original problem in the first place. ;)

I avoided your question as it had too many incorrect parts that i assumed you knew were incorrect anyway.   ;)

I did the same when I first got asked about it. 50/50 seemed obvious. When this program was on TV, a lady wrote in and suggested that players should always switch as it doubles there chances of winning. She was then rebuffed by a few top mathematicians in the country and told she was daft. Eventually they worked out she was right. It's not an easy problem even after it being explained as its counter intuitive.

This has been doing my head in.

if we pick a door at random we will pick the Car .333333 of the time and hence .66666 not.

if we always swap
.3333333 (when we pick the car first) we lose 0.0
.66666666(when we pick a goat)  we swap and win 1.0

net +ev= .33333*0 + .6666666*1=.6666666


If we always swap
.3333333 we win 1.0
.6666666 we lose 0.0

net +ev=.3333333*1+.66666666666*0= .333333333

if we randomly swap or stick on say a coin toss

.3333333 we win half the time .5
.666666666 we win half the time .5

Net +ev .3333333*.5 + .66666666*.5= .5

ie if we always swap we win 66% of the time
if we always stick we win 33% of the time

and if we swap 50% of the time we win 50% of the time.

I can't believe 'top mathematicians' could get this wrong! It's just plain obvious!

I have no verification of this, this was the original story I had been given.

Having read the wikipedia article, it appears that in fact, the game show host, Monty, didn't offer anyone a chance to switch. :)  But he did expose the goat to make it more interesting.

The person who wrote in to say always switch, was in fact writing into a magazine which had updated the problem to include the switching part and hence to start the ball rolling on the discussions thereafter.

This link explains who the top mathematicians were:

http://www.letsmakeadeal.com/problem.htm

surley the garage door has the car behind it, the barn door the goat?

Where's Sark when you need im

I remeber Sword, Rod and myself trying to explain this to tigscoco one night, and giving up after an hour and a half.  ;frustrated;

The odds of you choosing the correct door in the first place, is 33%. 1 in 3 doors will contain a car.

Now that there are two doors, the odds of you having the correct door is 50%, 1 in 2 doors will contain the car.

But we pick the car only 33% of the time, and we pick a goat 66% of the time (as there are two goats!) so by always swapping we win a car 66% of the time making swapping teh +EV decision.

What if the goat's driving the car?

Theres only a 33% chance of that!

Yeh, and you know what, I still dont get it!