Let's make a deal! Can you solve the Monty Hall puzzle?

Yes. My high school math teacher gave us this problem one year. It's one of the few things I remember.

I wouldn't think so. There's still a 50-50 chance, right?

No.......... It does not in the same way that if you flip a coin 99 times and it comes up heads every time, it does not mean there is a greater chance it will be tails on the 100th flip

Yes.

No. He's either picked a car or a goat, so there is always going to be a door with a goat behind it that he didn't pick. Being shown that goat changes nothing. I know it looks like he starts off with one chance in three and then is offered "better odds" with a 50/50 chance, but that's just an illusion.

I was hoping for something better from you...

huh?.......... best analogy I could think of............its like that trick question:

If you have 50 white socks and 50 black socks in a draw and you are blinfolded:

How many socks would you have to remove to get one pair of the same color?

The amount of people who will answer.........51

Yes, you should switch. Initially, your chances of having the correct door are 1 in 3. After being shown that one of the other two doors is incorrect, the chances that the last remaining door is correct are the same as the chances that the door you already selected is wrong. Thus, your odds of winning by switching doors is 2/3.

I've heard the answer is "yes", but I don't understand why.

OK, here's what I think the reason is: The first time you pick, you have a 1 in 3 chance (33%). OK, so then in the second round, after a goat is revealed, you still have your 33% chance (I'm a little shaky on this point). But if you then change your pick, you chances start over and you're down to a 1 in 2 chance (50%). 50% is better than 33%, so it's better to change your pick in the second round.

The shaky part... I would think that when the goat is revealed, your chances just got better (upgraded to 50%), but apparently, that's not true?

I wrote a little computer program to run through this scenario over and over, so that I could get some stats. Sure, enough. Keeping your first pick yields a 33.3% change of winning overall. While, changing your pick in the second round, wins 66.6% of the time. This is actually better than the 50% I had originally thought. hm... I guess if the first pick is 33.3%, then the other pick has to be 66.6%. So, wow, you really should change your pick when they give you that option. It doubles you chances of winning!

They would now have a 50% chance of winning the car (instead of 33%), but they still might lose. I don't know. I stopped drinking caffeine about an hour ago.

and they say cheaters never win!

For anyone who would like a good mathematical explanation of the solution, I like this site, as it offers several:
http://exploringdata.cqu.edu.au/montyexp.htm

no, I know this for a fact

Yes. Switching improves the probability of getting the car.

then, Bob, i think you have your facts wrinkled...

No, it's 50-50.

I've seen this question create the most heated debates on the net and in person. It's kinda funny. Some people get so vehement in defending the wrong answer. At one time I knew three completely different ways to explain why the right answer was right and different people would respond to different explanations.

You started out right, when you first pick your chances are one in three of winning. After the goat is revealed, nothing has changed to change those odds; if you stick with your first choice your chances are still one in three. By revealing the goat, Monty Hall has showed you that if you chose incorrectly the first time (chances are two in three), which of the other two doors is the right one.

I think so because now he has 50% chance of winning when it was 33.33%. Anyway, I won't care about it unless I am the player and there is a real car behind that door!!!

No Gambler, there is a principle saying that "the more you try the higher your chance to win is". But the example you gave isn't the same of the question's one, because you are talking about how many you try (flip) to have a greater chance of winning (tails), whereas the question was about how many is your chance of winning if we reduce the number of choices!! Did you get it, this is not the same.

Sorry Zang, but the question was whether reducing the number of doors, and this will affect your chance of winning. you are talking about "Being shown that goat changes nothing" and this of course doesn't affect the %. (1/3<1/2 so the % will change I think).

Yeah it would change the percentage, but I am not sure if you have to change your choice, cause this depends on your chance and your "Luck"..

Actually, you are misunderstanding my point............ (and as a Gambler, I should know) I am referring to the ODDS............... even if you flip (as in my example) 99 tails on a coin, the odds are STILL 50/50 that it will be heads on the 100th flip.


In this example you have NOT reduced the chances, the ODDS are still 50/50, why because no matter what door you pick ..........there is ALWAYS gonna be a door, with a goat behind it, that the host can open to reveal......... therefore STILL making it 50/50 odds.

This is my reply to verouge:

Actually, you are misunderstanding my point............ (and as a Gambler, I should know) I am referring to the ODDS............... even if you flip (as in my example) 99 tails on a coin, the odds are STILL 50/50 that it will be heads on the 100th flip.


In this example you have NOT reduced the chances, the ODDS are still 50/50, why because no matter what door you pick ..........there is ALWAYS gonna be a door, with a goat behind it, that the host can open to reveal......... therefore STILL making it 50/50 odds.

because YOU KNOW every week that Monty will reveal a goat AFTER you have chosen , the ODDS remain at 50/50

Bill,

That is flawed, sorry Look at my responses to Verouge/Dab.

The difference here is YOU KNOW Monty is going to reveal a goat AFTER you have chosen?

therefore the ODDS remain at 50/50.......... there will always be a goat, you HAVE NOT chosen for Monty to reveal, so its an illusion.


To explain it easier:

You have three doors, One has a car behind it, the other two have goats..... one of the doors is made of glass and you can SEE the goat......what are your odds now?


Regards David

To explain it easier:

You have three doors, One has a car behind it, the other two have goats..... one of the doors is made of glass and you can SEE the goat......what are your odds now?


Regards David

I think you're wrong. But, I don't really care that much either. There is only one "coin flip" in this case. Only at the end, do they reveal the results. It's not 50/50... my computer program proved that empirically. It consistently produced the 33% vs. 67% results that the solution caviartaste's link gives as the answer.

> You have three doors, One has a car behind it, the other two have
> goats..... one of the doors is made of glass and you can SEE the goat......what
> are your odds now?

50/50 ... but, this is a different situation than Monty revealing a goat and you getting a chance to change your pick.

Yup, as a gambler you should know this. As I said, I've watched this debate on the 'net many, many times and it's kinda funny how vehemently some people will defend the wrong answer. You choose one door out of three so your odds are one in three. Those odds do not change if you stick with your first choice.

Yeah maybe, the odds will not vary, but here there is no odd or even, it's a percentage defined in how much you have doors. It's all three, so first it will be 1/3, but when you will have just 2, it will be 1/2. I know, it's a game, and I can see what you are talking about and I am with you, but when talking about that show-game it would be another thing.

If I can see the goat, the odd would be 1odd of course, bec I will have 2doors one of them it goat unless there are 2cars, one for me and another for you hehe, my chance of wining the car is 50%, but if I don't know which two of the three are the goats my chance would be 33%.

I like this question on the explanation page - that helps you understand why your choice after the door has been opened is not 50/50:
Imagine that there were a million doors. Monty knows which door conceals the prize, so he then opens 999,998 losing doors. You are now confronted with two doors, the one you chose initially and the one Monty has left. Do they each have a 50% chance of concealing the prize?

have you looked at the algebraic solution on the answer page?

AT THE END it will always be 50/50 and THATS what counts...........

Like I said.......... its AT THE END where it counts........ If you did not have the option to change, then it would be 1 in 3.....But because he shows a goat AND you have an option to chnage then its 50/50

Its at the END where it counts and 2 factors change it here.......

1) He shows the goat
2)You have the option to change


Not having either of these, then makes it 1 in 3

Nope...like i said , Monty giving you the option to change AND showing you the goat makes it 50/50

Yep, because Monty has made the odds so...................


I am talking at the END here, AFTER he has given you the option to change AND shown you the goat, your choice is 50/50..either you pick a goat or a car?

If neither of these are done , then the Odds remain 1 in 3...........


It matters not, if its three doors or a million at the beginning........... because MONTY adjusts the odds, by taking away your choices AND giving you an option to change

Seems like it should be and that's why so many people get this one wrong. Another way to examine the problem is to enumerate possibilities. Let's assume you switch doors and say you choose door A. The prize could be behind doors A, B, or C with equal probability. So if it's behind door A, Monty will show either B or C (doesn't matter), you'll choose the other one and lose. If the prize is behind B, Monty will show you the goat behind door C, you'll switch to door B and win. Similarly, if the prize is behind door C, Monty will show the goat behind B, you'll switch to C and win. So, by switching doors you win two out of three times. If you choose doors B or C to start, you do the same analysis and get the same two out of three times winning if you switch doors. With all possibilities enumerated and all the same probability, your conclusion has to be that switching doors increases your odds from 1/3 to 2/3.

I thought of another way to think about this problem that might make it more clear. If there were 1,000 doors to choose from instead of 3. You picked one, then Monty revealed 998 goat doors and asked you if you wanted to change your pick. How would you feel about it then? Does it still seem like 50/50? What are the chances that you picked the right door out of 1,000 the first time?

Yes, because the odds go from 33.33333% to 50% being right.

Other.

The 'right' answer to the Monty Hall problem is that switching doors gives you a 66.66% chance of winning the car.

However, this is only the case if the wording of the puzzle is right. Monty *has* to know where the car is in order to pick either one of the two goats (if, by fluke, you picked the car first time) or the one remaining goat if you picked the other goat. Motivation is a factor and the puzzle only works if we know that Monty wants to open a door with a goat behind it in order to help the contestant.

Another key to why so many people get this one wrong is that Monty doesn't have a free choice about which door he opens. If you pick a goat (which you will 2 times out of 3) he *has* to show you the location of the one remaining goat. By definition, this means that 2 times out of 3 you'll win the car if you change your mind.

Another reason why many people can't get their heads around the fact that it isn't 50/50 is that probability is relative to the observer insofar as it is relative to the information the observer has.

If you have 1000 doors and pick a door (which 999 times out of 1000 will be a goat), Monty is then duty bound to open all 998 remaining goat doors and leave behind the one door that is almost guaranteed to be the car (at odds of 99.9%).

However (and this is where it gets confusing) if you brought somebody off the street into the studio and asked them to pick between the two remaining doors (so long as they didn't know which door the contestant had picked first), his chances of picked the right door would be 50/50. The reason is information. When Monty opens the door to reveal a goat he is giving you information. Somebody walking off the street has no additional information.

Yeah, in the beginning it's like that: 1goat (a), 1goat (b) and 1car (for me and not you!!).
But when there will be just 2doors, it would be: 1goat and 1car, so what is the option???
(If you are bored from me, just leave the discussion, but I like it cause I want to see who is the one who will give in, hehe..)

Let me read the question..

"""However, after the player selects a door but before opening it, the game host opens another door revealing a goat. The host then offers the player an option to switch to the other closed door. Does switching improve the player's chance of winning the car?""", you have seen one of the three, and you still confused where is the 2nd goat (2-un-open-doors!!), isn't the percentage now 50%???

we did not destroy a door. The first door is still there. Therefore, if you stay with your first choice, your odds remain at 1 in 3. If there were a million doors, and only MONTY knew which door were the right one - and he opened 999,998 losing doors, you would be lucky that you were down to the last TWO doors, but your ODDS would still be a MILLION to one, gambler!!!!!

NO

Door_1 and Door_2 is a goat. Door_3 is the car.

Player_1 randomly picks Door_2 or Door_3. He then finds out that Door_1 is a goat. Switching between Door_2 and Door_3 won't matter in terms of odds.

If you ain't cheating, you ain't trying.

Let me ask this,

He gives you two doors and says choose one......one has a goat, one has a car...what are your odds then ?

mmmmmm makes no difference AT THE END, because Monty will remove 998 doors and still leave one with a goat behind it

not bored at all.......I know alot of people are disagreeing with me......I might well be wrong?,

But IMHO..... it matters not HOW MANY doors there are........ because at the end of the day you are left with only TWO choices.........................,


Regards David

Yep 50% not 33.3333333333%

In the beginning YES,................ But NOT in the END.............(which is where it counts, because this choice decides whether you win the car or not, NOT the 1st choice).......... Million doors?....its an illusion..........OK,

assuming a million doors..you will either choose a goat or Car right?........ your chances of getting the car are 999,999-1 right?.........THERE WILL ALWAYS be 999,998 doors that MONTY can remove to leave you with ONE goat AND ONE car..................

Now, your left with two doors and you change your choice and gosh darn it, if you do not win the Car!!!!!!!!..........

Now did you just hit a 999,999-1 shot or did you just win a 50/50 toss-up?


Let me know,


Regards David

That's 50/50. But that's not the situation presented. The difference between the two is that in the Monty Hall situation, when your initial choice was a goat (though you don't know it yet), Monty then always chooses the remaining door with a goat. That's a piece of information not available to you in the two door case. In essence, what Monty's doing is telling you that if you change your selection, which door you should change to. If you chose the car initially (1/3), then he tells you nothing and changing will get you a goat. If you chose a goat initially (2/3), then he tells you, by showing you one of the goats, which of the two remaining doors is the right one.

*sigh* But, you're first pick is a very unlikely 1 in 1000 chance. Him remobing 998 doesn't change that... Anyway, I give up.

we are not solving between two heads of a coin toss. we are solving for the probability of one of three or two of three.
And the math doesn't lie. "Since the sum of the three probabilities is 1, the probability that the prize is behind the other door is 1 - (1/3 + 0), which equals 2/3."

???? Forgive me......... Are you saying that, depending on whether you choose the goat or car in your initial choice, Monty then decides what to do?

I was assuming that NO MATTER what you chose he would always expose the 2nd goat and give you a choice, No?

I think I do too............... Lol.......... gonna post this on my poker site let them figure it out

Dear Caviar taste,

I posted this to my ......... resident Poker Guru:





Re: I have a problem............
« Reply #5 on: Today at 03:05:27 PM »

--------------------------------------------------------------------------------
David,

You should take his offer and switch. The mistake is assuming that the prize is equally likely to be behind the remaining two doors. This isn't the case.

If you swap, you win when your initial guess was wrong. The probability of being wrong is 2/3. If you stick, you only win when your initial guess was correct. The probability of being correct is 1/3. So you're twice as likely to win the prize if you swap.

This one is hard to believe I know, but if you don't believe me try it yourself. Take three playing cards, one of which is a queen and play 'find the lady' with a friend. shuffle the cards face down and ask your friend to choose one (without looking at it). You then show him a different card (which is not the queen) and ask him if he wants to swap. Do it 100 times where he swaps and 100 times where he doesn't, and look at the results. He'll win about twice as often when he swaps.

Ain't probablility wonderful? .




Now as to Maths not lying:


Three old ladies go to buy a TV set, the TV set is $30......... each pays $10 and goes home.


The manager says to the clerk, That the TV set is only $25, here is $5 go give the Old ladies their money back.


On the way there he puts $2 in his pocket.


When he gets to the old ladies house he gives each old lady $1, so now instead of $10 they have now paid $9.......... 3 x $9= $27 + $2 the clerk put in his pocket is $29 ..where is the other $1.

Pffft ............. and Maths doesnt lie

Dear Bill,

It seems I was wrong..............

I posted this to my ......... resident Poker Guru:





Re: I have a problem............
« Reply #5 on: Today at 03:05:27 PM »

--------------------------------------------------------------------------------
David,

You should take his offer and switch. The mistake is assuming that the prize is equally likely to be behind the remaining two doors. This isn't the case.

If you swap, you win when your initial guess was wrong. The probability of being wrong is 2/3. If you stick, you only win when your initial guess was correct. The probability of being correct is 1/3. So you're twice as likely to win the prize if you swap.

This one is hard to believe I know, but if you don't believe me try it yourself. Take three playing cards, one of which is a queen and play 'find the lady' with a friend. shuffle the cards face down and ask your friend to choose one (without looking at it). You then show him a different card (which is not the queen) and ask him if he wants to swap. Do it 100 times where he swaps and 100 times where he doesn't, and look at the results. He'll win about twice as often when he swaps.

Ain't probablility wonderful? .

Dear Dab,


Seems I am wrong,

It seems I was wrong..............

I posted this to my ......... resident Poker Guru:





Re: I have a problem............
« Reply #5 on: Today at 03:05:27 PM »

--------------------------------------------------------------------------------
David,

You should take his offer and switch. The mistake is assuming that the prize is equally likely to be behind the remaining two doors. This isn't the case.

If you swap, you win when your initial guess was wrong. The probability of being wrong is 2/3. If you stick, you only win when your initial guess was correct. The probability of being correct is 1/3. So you're twice as likely to win the prize if you swap.

This one is hard to believe I know, but if you don't believe me try it yourself. Take three playing cards, one of which is a queen and play 'find the lady' with a friend. shuffle the cards face down and ask your friend to choose one (without looking at it). You then show him a different card (which is not the queen) and ask him if he wants to swap. Do it 100 times where he swaps and 100 times where he doesn't, and look at the results. He'll win about twice as often when he swaps.

Ain't probablility wonderful? .

Dear Verouge,




It seems I was wrong..............

I posted this to my ......... resident Poker Guru:





Re: I have a problem............
« Reply #5 on: Today at 03:05:27 PM »

--------------------------------------------------------------------------------
David,

You should take his offer and switch. The mistake is assuming that the prize is equally likely to be behind the remaining two doors. This isn't the case.

If you swap, you win when your initial guess was wrong. The probability of being wrong is 2/3. If you stick, you only win when your initial guess was correct. The probability of being correct is 1/3. So you're twice as likely to win the prize if you swap.

This one is hard to believe I know, but if you don't believe me try it yourself. Take three playing cards, one of which is a queen and play 'find the lady' with a friend. shuffle the cards face down and ask your friend to choose one (without looking at it). You then show him a different card (which is not the queen) and ask him if he wants to swap. Do it 100 times where he swaps and 100 times where he doesn't, and look at the results. He'll win about twice as often when he swaps.

Ain't probablility wonderful? .

I don't think so !!!

Ahh... that's pretty much the kind of explanation I had going on in my head. Kind of a reverse of a lot of the other explanations I read. I summarize it this way:
You are more likely to pick a goat the first time than the second time.

So we are OK now, we have the same result!!! (But showing the goat behind the first door doesn't really push you to change your choice, it only changes your % of winning but not your choice, cause you still confused!!)

IT'S PERFECT!!! How did he think about this???? I was very "impressed"!!! Although I believe in what I said, but this way was "Damn, where did he get it??"
Thanks about this!

I would like to meet him, can I have free cards for the casino? I think he's working in a casino, isn't he??? hehe

LOL...........He plays at much higher levels than I, and when it comes to Odds and probabilities etc.............he knows his stuff,

Best Regards David

God bless him and you!!

(I like this Best Regards David), Best Regard Marwan.. (but I prefer it with the last 2pts "..")

30-5=25+2=27+3=30
If you work the problem backwards, you come up with the correct answer.
Math doesn't lie.

> Now as to Maths not lying:
> Three old ladies go to buy a TV set, the TV set is $30......... each
> pays $10 and goes home.
> The manager says to the clerk, That the TV set is only $25, here is
> $5 go give the Old ladies their money back.
>
> On the way there he puts $2 in his pocket.
>
> When he gets to the old ladies house he gives each old lady $1, so
> now instead of $10 they have now paid $9.......... 3 x $9= $27 +
> $2 the clerk put in his pocket is $29 ..where is the other $1.
>
> Pffft ............. and Maths doesnt lie
>
As caviartaste says, Maths doesn't lie. Only people do.

This puzzle is as old as the hills and the reason why it works so well is because of misdirection and clever wording.

By adding the $27 the old women have paid to the $2 that the clerk has in his pocket, you are treating the clerk as if he is one of the customers. This is the central lie open which the puzzle is built.

To put it another way. You can't add up money someone is up to the money someone is down as if they are two positive numbers. Maths doesn't work that way. The old ladies are all $9 down whereas the clerk is $2 up.

You can either express this as:

-27 + 2 = -25 (plus the 25 in the till = 0)
or
27 - 2 = 25 (minus the 25 in the till = 0)

What you can't do is express it as 27 + 2 = 29. The clerk works for the store, not the old ladies.

don't know!

Thank you

Hee hee........... I know just teasing you.......... strange though?........ you pay $10 fact......... you get back $1 fact......... means you have only paid $9 fact........... 3x9 = 27...........fact..........

I know (look at my post to CT) still makes me think......... like I said ...... You pay $10 get back $1........ means u only paid $9?............ its counter-intuitive

You are welcome
Best Regards Marwan..

I like the number 5, wanna be the five one in the chat room right now??

now you would be the sixth, hehe, i like it too..

............ The firewall at work will not let me in there......... and its too much hassle to re-configure....but thanks

You are welcome

I love you gambler.....
when are we going to Vegas?

It's a cool problem huh? Glad you got it.

........Yea, In the end

talk about living life on the edge

After opening the goat door, there's a 50-50 chance of getting the prize.

Guess it depends on the left 50 or the right 50.

Actually, with gas prices like they are today, I'd rather have the goat!

Classic problem, love it.

Increases it.