It looks to me as if our odds are the same.
I agree that when you look through the cards, exposing 98 and leaving one hidden, that implies a high chance that the hidden one is the winner - my card retains its original odds of 1%.
But (as noted) the problem at hand involves chosing the cards to be exposed randomly, which produces a very different result.
Just trying to make sure I’m on the same page here…Xema are you talking about only those cases where Monty opens doors/flips cards at random and Monty happens to reveal only zonks? In other words, in the majority of cases the prize is revealed before we’re down to just 2 choices and the trial ends, but we’re not considering those outcomes. Do I have that right?
I’ve just gotta’ make sure I understand the question before I can even think about considering the answer. 
Actually I like the explanation that kind of reverses the logic. You pick a door, say 1. Before revelaing the door Monty asks if you would like to switch to having both the other doors. Of course you switch. Now MOnty says,'Ok one of your doors is a zonk and it is door number 3!" He opens door niumber three and then asks, “Would you like to switch back to 1?” Of course not, why would you. Since there is no possibility that Monty would reveal the prize and you knew at least one of your doors was a zonk already. You gained no new information.
Right. The rules for the case with three doors were outlined in post #48. Since then we’ve taken up the case with 52 “doors” (a deck of cards).
The fundamental point is that if the selection is random, things are quite different from when Monty knows what’s behind the doors and can choose based on that knowledge.
You’re close to the main point. The 50 non-winning cards are not selected completely at random. The person who is revealing the cards knows which is the winning card, and deliberately doesn’t reveal it. This is different from the case where the person revealing the cards picks them completely at random, and though it’s possible that he could reveal the winning card, he just happens not to reveal the winner.
Suppose the original puzzle were stated this way: after you pick one of the doors, Monty flips a coin to decide which other door to open. The one he opens happens not to have the big prize. Then he gives you a chance to switch to the other unopened door. Should you switch? In this case it makes no difference, since the chances are 50 : 50. The difference from the original puzzle is that in the cases where the prize is behind one of the doors you didn’t initially pick, Monty actually takes a chance on revealing the prize to you when he opens a door picked at random.
Here’s another thought experiment. Suppose after picking a door, Monty puts a blindfold on you, then announces that he will open an empty door that you didn’t pick. He opens it, then offers to let you swap the lone door you picked for the two you didn’t. One of those doors is open, and you know its empty, but you don’t know which door it is because you can’t see. Should you switch? You knew in advance that at least one of those two doors was empty, since there is only one prize, so when Monty opened an empty door you didn’t learn anything you didn’t know before. For you, the situation is the same as it was before, so switching will double your odds of winning. Now, how would things be different if the blindfold were removed before you made your choice? How would knowing which door had been opened change the odds?
For the problem we’ve been considering (under rules described in post #60), they are.
I think the vast majority here are in agreement with the standard answer to the standard problem.
I’m finding the modified version much more interesting, where Monty randomly opens all but one of the remaining doors, tossing out all scenarios except the few where Monty happened to pick only zonks.
I simply cannot wrap my brain around it. The annoying part is that Cecil included the same example in his penultimate post in the linked article, and didn’t explain it.
Let’s see if I can brute-force the three doors version. I always pick door #1, the prize is the letter o, zonks are the letter x, and Monty’s reveal is the capital letter.
Okay, so there’s six possibilities when I always pick door 1. The two where Monty reveals a prize get removed. That leaves 4 possibilities, two of which win. This gives the 50% chance many are advocating. Now, the heart of my question is this: How do we reconcile this state of affairs with the following exchange from earlier in the thread?
It really seems to me that we are, in effect, counting a “Bar 12” as action. If we are, then we would be in the wrong.
Explanation of Bar 12:
There are two “line bets” in craps: the Pass and the Don’t. Of the 1980 total possible craps scenarios:
976 pay the Pass and lose the Don’t
949 lose the Pass and pay the Don’t
55 lose the Pass and count as “no action” (push) for the Don’t. This is the Bar 12.
When comparing odds, one could do the simple:
976/1980 = 49.29% to win the Pass, which is correct.
949/1980 = 47.93…% to win the Don’t, but this would be wrong. One must exclude the “no action” on the Don’t to calculate the true EV:
949/1925 = 49.30% to win the Don’t, which is correct.
The problem I’m having is considering scenarios 1 & 2 from my previous post as distinct, when I feel they should be collapsed into a single scenario.
Note that BwanaBob’s post applies to the original Monty Hall problem, not the random choice variant. His error is regarding one position of the prize as two cases, and the other two possible positions as one case each.
The simplest notation lists 3 possible starting positions (P=prize, Z=zonk):
P-Z-Z
Z-P-Z
Z-Z-P
A more detailed notation (which isn’t necessary) regards the two zonks as distinct - zonk1 and zonk2. We then get 6 possible starting positions:
P-Z1-Z2
P-Z2-Z1
Z1-P-Z2
Z2-P-Z1
Z1-Z2-P
Z2-Z1-P
BwanaBob is mixing these two notations. From the list of six, he regards #1 and #2 as separate, but then merges #3 and #4 into one case, and #5 and #6 into another.
Ran 1,000,000 trials, took 9 minutes. Results: You picked the ace 19, 268 times, the ace is the last card turned 19,236 times, and the other 961, 496 times the card is somewhere in the middle of the deck.
There’s nothing to reconcile, since these two setups are different. In the “classic” Monty Hall problem, scenarios 3 and 6 never happen, Monty doesn’t let them happen, they get converted to 4 and 5 by MH because he never wants to reveal the prize. So, 1 and 2 happen 1/6 of the time each, 4 and 5 happen 2/6 of the time each. That means you want to switch when given the opportunity.
In the “new” Monty Hall problem you never get the chance in situations 3 and 6. Those two cases represent the advantage in switching. When those cases are removed (Monty opens those doors by chance) then the incentive for winning goes away.
MH’s knowledge of where the prize lives changes the scenarios between “classic” and “new” and make all the difference in the world. With the “new” problem, in 1/3 of the cases, you never get the chance to switch, and that 1/3 of the time was your advantage.
To put it another way, let’s slightly restate the problem (without in any way changing the setup):
We are considering two scenarios, one in which Monty knows where the prize is and deliberately opens a non-prize door, and one where Monty randomly opens either of the as-yet-unchosen doors. The fact is that under both scenarios, changing your choice of door gives you a 2/3 chance of winning on average. Not 1/2 for the random scenario, 2/3, because 1/3 of the time Monty will randomly open the prize door (which, presumably, you will choose).
But if, in the random scenario, we exclude the cases where Monty opens the prize door, your chance of winning is unchanged by switching to the other unopened door - half the time you win, half you lose.
I think the confusion arises from the fact that it is implicit in the first scenario that we exclude cases where the prize door is opened, because it cannot be opened - the rules prohibit it. In the random scenario, it needs to be made explicit, because otherwise we don’t know whether to include certain cases. This is what leads to the apparently contradictory results.
The best explanation I’ve come across for this simply lists all of the possible outcomes:
Option 1: The prize is behind door number 1
You pick number one, Monty reveals another door
If you switch, in this case, you lose.
You pick number two, Monty reveals three
Switching wins
You pick number three, Monty reveals two
Switching wins
So out of Option 1, in one outcome switching loses, and in the other two it wins.
Option 2: The prize is behind door number 2
You pick number two, Monty reveals another door
If you switch, in this case, you lose.
You pick number one, Monty reveals three
Switching wins
You pick number three, Monty reveals one
Switching wins
So out of Option 2, in one outcome switching loses, and in the other two it wins.
Option 3: The prize is behind door number 3
You pick number three, Monty reveals another door
If you switch, in this case, you lose.
You pick number two, Monty reveals one
Switching wins
You pick number one, Monty reveals two
Switching wins
So out of Option 3, in one outcome switching loses, and in the other two it wins.
These are all of the ways it could go, and 6 of the 9 possibilities win if you switch.
The most interesting aspect about this problem is that it gets rehashed every year or so on this board. The participants are different, but the arguments are the same. There will be some people who get it right away after hearing one of the many valid solutions that depend on math or logic. There are others who will have to write a computer simulation in order to convince themselves of the right solution. Then there are others who get it after enumerating all the possible outcomes. Inevitably someone will change the problem slightly and the discussion goes off on a tangent. And, of course, there are those who just don’t get it, but think they got it. And you will always have some who will likely never get it. Just fascinating.
Some people just have a hard time internalizing the fact that having additional information can change the odds. To others, that fact is intuitively obvious.
This arguement had me going each way at one point or another. I’d read one side and go ‘Hey, that make sense! That’s got to be right!’ then read the other side and go ‘That make even more sense!’.
Instead of continuing to confuse myself even more I decided to ask an expert. That expert, who also happens to be my Dad*, sent the following response. The problem as I outlined it to my Dad was the classic Monty Hall problem where Monty never picks the door with the prize and the contestant picked door #1. My Dad responded with the following:
So the answer to the question is that switching doors does nothing to impove the odds. It’s 50/50 once Monty opens a door.
Slee
*My Dad has a PHD in math(with a 4.0 GPA). His dissertation was on statistics. He spent most of his career(nuclear reactor safety) doing studies on the probability of certain events taking place.
Go and play the game that Triskadecamus outlined earlier and see if the odds are 50/50.
Just for curiosity, I worked out the conditional probability here. P(1|not 3) = P(1)P(not 3)/P(1) = P(not 3). If we assume that the prize is distributed at random, P(not 3) = 2/3.
You can find the conditional probability formula here, or check any introductory probability textbook.
I take that back. The condtional probability is 1/2. Ask your dad about dwalin’s listing of all possible outcomes.
No offense to your Dad, sleestak, but he’s wrong. I’m decidedly not an expert on probabilities, but my guess is that he has correctly solved the wrong problem.
Ask him this:
If there are three doors, and behind one is a prize, you will get the prize about 1/3 of the time by randomly selecting a door, and lose the prize about 2/3 of the time. No one disputes this.
Monty then opens one of the other two doors (making sure that the door he opens doesn’t have the prize) and gives you the chance to switch to the third door.
If you switch, you will lose the prize in every case where you would have won the prize if you had not switched. In other words, if you already picked the door with the prize, you obviously lose by switching. According to our first statement, this will be about 1/3 of the time.
Conversely, you will win the prize in every case where you would have lost the prize by not switching. In other words, if you originally picked a door without the prize, switching will get you the prize, every time. Again, according to our first statement, this will happen about 2/3 of the time.
How does he explain this?