Monty Hall Problem

[Decider]

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

[matt674]

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

[LeKnave]

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...

[ACE2M]

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.

[Decider]

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.

[LeKnave]

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.

[Graham C]

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.

[ACE2M]

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.

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

[Decider]

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.

[Rod Paradise]

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

[Decider]

Heh, enough of the previous thread stuff  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.

[matt674]

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!!  

[Decider]

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.

[matt674]

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?

[Jon MW]

Heh, enough of the previous thread stuff  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.

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.