Probability 101: Monty Hall

[OPFOR]

Very very, simple math.

Suppose there were fifty doors. You pick number one. Monty then opens number three through forty nine revealing 48 goats (pass the curry please).

There are now two unopened doors, number one and number two, and again you have chosen number one.

You stick with number one and your chances are 50/50.

If you switch to number two your chances are...........guess what, 50/50.

It does not matter how many doors you start out with.

Gunny, do you understand why it's not 50/50 now? If, in your example, those other doors were opened RANDOMLY, and by some very small chance you managed to open 48 of the remaining 49 and still have the "winner" left unrevealed... then, yes, it would be 50/50 between the originally chosen door and the one remaining door.

But Monty isn't opening a door randomly - he knows where the prize is, and MUST open a door that DOESN'T reveal the prize. The video explains this very well...

[DefensiveCarry.com]

[Hopyard]

10 years in college, a Master's in Public Health Epidemiology, and a Certified Payroll Professional.

You were right, I was wrong. I'm sorry. Seriously - apologies for pushing the wrong way. Sigh.

The key here is that it is NOT a second problem: the car stayed put. Good YouTube explanation - thanks for the link.

Don't feel bad. I'm in a similar boat. This one aggravated me all night.

If you do JoininNY's coin experiment the experiment proves the outcome.

The next issue, why?

It has nothing to do with Memory or Information or what you learn when the goat or penny is revealed.

On step one the chances of picking wrong are 2 out of 3.

On step two, if you do nothing at all the chances of being wrong remain the initial 2 out of 3

If you were dealing only with step 2 your chances of being wrong are 0.5

The overall chance of being wrong if you change your choice in step two is the product of 2/3 * .5 or 1/3 So by switching you change your chance of being wrong from 2/3 to 1/3.

It is totally counterintuitive but that is how probability works when you have a series of things happen each with its own probability.

The key to the puzzle is that by not switching you are retaining your initial high probability of having been wrong. By not switching you have taken only one toss of the dice not two.

Where you and I went wrong wasn't our initial view that there were only two choices behind each door (though that was part of it) it was our failure to recognize that by staying with the initial pick we had not tossed the dice twice. At least that is how I think we went wrong.

This statement by OPFOR isn't quite right al though I see what he is getting at.

"Paymeister, as has been said, you are correct in that any series of random events is random, and does not depend on preceding events (roulette spins, coin flips, etc). This, however, does NOT hold true here: there is only one event,; you are just given more information about it DURING the event. "

There is ONLY one event ONLY if you DON'T switch. If you switch there are two events. And that is the key to making the probability end up in your favor. Switching creates two events so the probabilities ofan outcome from the two independent events can be multiplied. 2/3 * 1/2= 1/3 and the odds of your having picked wrong change from 2 in 3 to 1 in 3. I do not think the workings of this puzzle have anything to do with "you are given more information about it during the event." The revelation of where one penny or one goat happens to be doesn't in anyway change the odds or the necessary strategy. The key to the correct strategy is recognizing that you are being given a SECOND CHANCE with a second independent outcome from the first, and if you do nothing at all (don't switch) you are retaining the initial unfavorable odds.

You guys and my own slowness caused me a sleepless night. ACH!

Two quick points: 1) JoininNY said "do the experiment." He was right. That is how you get to see the truth.
2) JoininNY said this isn't a trivial matter. He is right again because we all make decisions with insufficient information all the time and often these
decisions are matters of guestimating the odds. If we don't come close, as some of us in this example, we can lose a lot of money, lose
an airplane, lose a war.

[JonInNY]

Where you and I went wrong wasn't our initial view that there were only two choices behind each door (though that was part of it) it was our failure to recognize that by staying with the initial pick we had not tossed the dice twice. At least that is how I think we went wrong.

Well done Hopyard!

Another way to look at it is to think that you have a "free pick", which indeed you do, when Monty opens the door with the goat. You have a choice of 1/3 if you stay with the one you initially chose, or 2/3 if you choose one of the other two! Since you know that the one he shows you is not the winner (unless you want a goat), then switching is the best choice!

[Hopyard]

Gunny, do you understand why it's not 50/50 now? If, in your example, those other doors were opened RANDOMLY, and by some very small chance you managed to open 48 of the remaining 49 and still have the "winner" left unrevealed... then, yes, it would be 50/50 between the originally chosen door and the one remaining door.

But Monty isn't opening a door randomly - he knows where the prize is, and MUST open a door that DOESN'T reveal the prize. The video explains this very well...

Yes, Monty isn't opening a door randomly. But that is irrelevant to the outcome of the game, I think. It is just a means to provide the player with a second opportunity to make a choice where the probability of picking the loser is different from the first "toss." *

You can think of it this way. Monty randomly opens a door. If he finds the winner the game is over; if he finds a goat you get another chance to pick. If you don't switch you are foregoing your second chance to pick and retaining your initial unfavorable odds. If you do switch then the probability of being wrong for BOTH tries changes from 2/3 to 1/3.

The whole key to the puzzle is you must take the second chance or you retain your initial losing odds.

[OPFOR]

Small typo, Hopyard: 2/3 * 1/3 doesn't = 1/3, it equals 2/9, but I understood that you meant 2/3 * 1/2...

In any case, I see your point. You can look at it is not getting more information, because you already knew that your chance of being wrong the first time was 2/3. When Monty reveals one of that 2/3 and asks if you want the other, he is in effect just asking if you want to switch your original 1/3 for "his" original 2/3. He could simply say, without revealing anything, "do you want to stick with your door, or switch to BOTH the other doors."

However, if he opened randomly, you wouldn't even GET a second chance 1/3 of the time, because he would randomly open on the winner. It has a certain sense to it (your theory), but I don't think it holds up...
Glad everyone is beginning to see the logic in it, even if the logic here isn't obvious at first glance. I'll post another one right now...

[Hopyard]

It has a certain sense to it (your theory), but I don't think it holds up...
Glad everyone is beginning to see the logic in it, even if the logic here isn't obvious at first glance. I'll post another one right now...

OPFOR, this is the part of your explanation where I am having a problem and think you are mistaken.

"you are just given more information about it DURING the event."

I don't think you are being given more information. You are being given a second chance in which the odds are different from the first chance. The key to losing the game is not recognizing this fact and retaining your original wrong choice.

Unless what you mean by being given more information plays out as follows:

If the game were played where monty, knowing where the prize is, MUST tell you to open the door when you pick correctly the first time, then yes, you would be getting more information when monty shows you a goat, as he would effectively be telling you that you picked incorrectly the first time and he is going to give you a second chance to pick. You would be getting the message that you picked wrong. You would know that you must make a second different choice to stay in the game.

But if monty has no obligation to stop the game if your initial pick is correct, then I think information has nothing to do with how you play the game. It is pure probability of two independent events being multiplied. But you must recognize your need to toss the dice that second time, and also recognize that not changing is not making a second choice.

[GunnyBunny]

Yes, Monty isn't opening a door randomly. But that is irrelevant to the outcome of the game, I think. It is just a means to provide the player with a second opportunity to make a choice where the probability of picking the loser is different from the first "toss." *

You can think of it this way. Monty randomly opens a door. If he finds the winner the game is over; if he finds a goat you get another chance to pick. If you don't switch you are foregoing your second chance to pick and retaining your initial unfavorable odds. If you do switch then the probability of being wrong for BOTH tries changes from 2/3 to 1/3.

The whole key to the puzzle is you must take the second chance or you retain your initial losing odds.

In my scenario, Monty know the goat is behind number one or number two. After that it doesn't matter how many doors are added or tossed out. When you're down to two doors, you have a 50/50 chance, period.

If you toss a coin in the air and it comes up heads 100 times straight, it's still 50/50 that it'll come up heads on the 101'st toss too.

That professor dude reminds of my university days where they mess with your head. I bet he voted for Obama too!

[HKinNY]

I bet he voted for Obama too!

I say we toss Obama in the air and see if he lands heads or tails.

After Obama took office who has a coin?

[highvoltage]

But, but, I WANT a goat...

Then stay with your original choice. You'll have a 66% chance of getting the goat.

[OPFOR]

In my scenario, Monty know the goat is behind number one or number two. After that it doesn't matter how many doors are added or tossed out. When you're down to two doors, you have a 50/50 chance, period.

If you toss a coin in the air and it comes up heads 100 times straight, it's still 50/50 that it'll come up heads on the 101'st toss too.

That professor dude reminds of my university days where they mess with your head. I bet he voted for Obama too!

Monty does indeed know where both goats are, but they (and remember there are two) aren't restricted to doors one and two. They could be anywhere, but that has no bearing on the game.

This isn't a coin toss, as there are more than two possible outcomes...

[OPFOR]

OPFOR, this is the part of your explanation where I am having a problem and think you are mistaken.

"you are just given more information about it DURING the event."

I don't think you are being given more information. You are being given a second chance in which the odds are different from the first chance. The key to losing the game is not recognizing this fact and retaining your original wrong choice.

Unless what you mean by being given more information plays out as follows:

If the game were played where monty, knowing where the prize is, MUST tell you to open the door when you pick correctly the first time, then yes, you would be getting more information when monty shows you a goat, as he would effectively be telling you that you picked incorrectly the first time and he is going to give you a second chance to pick. You would be getting the message that you picked wrong. You would know that you must make a second different choice to stay in the game.

But if monty has no obligation to stop the game if your initial pick is correct, then I think information has nothing to do with how you play the game. It is pure probability of two independent events being multiplied. But you must recognize your need to toss the dice that second time, and also recognize that not changing is not making a second choice.

I get your point. I relate it to cards, where each card revealed gives you more information about what cards are still un-played. That analogy doesn't work exactly in this event - here you are (at the most basic level) being given the opportunity to switch from your one door to Monty's two doors after the "reveal."

[Pistology]

A goat is a goat. If I put three marbles in a can - identical except two are black and one is white, and, blindfolded, you pick a marble and hold it in your fist, and I pull a black marble from the can and show it to you and offer you to exchange the marble in your fist for the one in the can and tell you that I will put $5 on the table if you will put $6 on the table with the total $11 going to you if you hold the white marble and to me if you don't, would you be willing to play the game?
After you choose - whether you chose a goat or the BP - and Monty shows you a goat, by your Youtube / Wikipedia understanding, you give me odds of $6:$5 that you win by switching and play the game 1,000 times. Over the long run, I am up and you are down $500.
If your claim of a 16% (2/3-1/2) advantage holds, then the results for the same long run are that you are the one up $1,320.
I'd be willing to take odds of 6:5 if you're willing to give 'em.

[Hopyard]

In my scenario, Monty know the goat is behind number one or number two. After that it doesn't matter how many doors are added or tossed out. When you're down to two doors, you have a 50/50 chance, period.

If you toss a coin in the air and it comes up heads 100 times straight, it's still 50/50 that it'll come up heads on the 101'st toss too.

That professor dude reminds of my university days where they mess with your head. I bet he voted for Obama too!

There is a distinction to be made between the probability calculation based on whether one bet or two bets are made successively. If you toss a coin two times the chances of a tail are 50:50 each time. But if you want to know the chances of coming up with two tails on two tosses, you have to multiply .5 *.5 to get .25 That is what is going on here. If you switch you make two bets. If you don't switch you have made 1 bet. Intuitively, you know that if you gamble long enough (in a fair game) you will eventually have an outcome where you win. All OPFOR is doing is showing that you can calculate your chance of winning this game on two successive tosses, and if you do that (make two successive bets), your chance of losing goes from 2/3 to 1/3.

These statements are correct: 1) the probability of being wrong and getting the goat for the first bet is two out of three

2) the probability of being wrong on the second bet by itself is 50:50

3) the probability of an outcome on a series of two bets overall is the product of the two probabilities: 2/3 * 0.5= 1/3

Don't worry, I was up half the night till I figured this one out. It was intended as a brain teaser and it is.

It is hard to see though until you actually sit down and either play the game out in your mind or better yet at a table.

Do the experiment, as JoininNY suggested.

There are only a limited number of ways this game can be played out, and if you sit down with a piece of paper a quarter and two pennies and look at the individual outcomes which happen, and keep track of them on paper as you try different options, you will see that OPFOR is right in fact.

No one is messing with your mind, or anyone else's mind. There is no trick here. OPFOR posted this because it is intriguing and informative, and FUN to play with. (IT wasn't fun being in bed awake half the night trying to see what was going on--now I'm tired and more grumpy than usual.)

JoininNY explained it because he wanted to make the very valid point that lots of times really important decisions get made erroneously as the concepts of probability are difficult to comprehend and don't always play out intuitively. That is scary. It is also why I was obligated (when I still worked) to review what I did with a mathematical statistician. And even with the statistical consult, you can still mess up a probability calculation, easily. (Especially if you are relying entirely on software instead of brains.)

Paymeister explained that he does have the appropriate credentials (degree in epidemiology --i.e. almost always a math/ statistics intensive program-- and he as I missed this at first.

No need to let the fact that you didn't see it bother you.

[Pistology]

I just read the Wiki, and now know that I'll never get a job working for the house at a casino.
First pick chances of picking the door with the car are 1 in 3.
Monty will always show a goat, and this does not effect the outcome. The information given is worthless to Monty and the player and may as well be stated before the first pick, above - "For your original choice of one door I'm going to offer you the other two doors; and I'm going to show you a goat behind one of them."
Switching gets 2 doors for 1.
Who knew? Thanks, OPFOR.

[GunnyBunny]

Someone pass the goat please.