Am I the only one who sees it as 75%?
Never mind, upon further reflection I see why it is not, although the explanation is different and more complicated than the unsatisfying ones given upthread. (If there was a “Monty” who made sure there was a card showing with a black top, then it would be 75%, I am pretty sure.)
50% in that case, I thought.
I still get 2/3. There are six possible sides of a card that can be pulled from the deck:
Card W/W, side 1 (W)
Card W/W, side 2 (W)
Card B/W, side 1 (W)
Card B/W, side 2 (B)
Card B/B, side 1 (B)
Card B/B, side 2 (B)
If Monty made sure that any selection that showed up with a W was discarded you still have 3 remaining outcomes, each equally possible. If Monty threw away the W/W card there are four sides available but you only ever get to choose from one of the 3 that are black, so your odds remain.
How would Monty eliminating the W results change the odds?
You took Monty’s interference in a different way than I had intended or considered. I’m not talking about a random card peeking out but then Monty sweeps away any cases where it is a white card that is visible. I’m saying he looks in the bag while you are out of the room and chooses the closest card that has black showing to be the one that will peek out. There are thus no white results at all. Do you see why it would be 75% in that case?
I don’t understand what the bolded part means.
You’ll have to describe in greater detail how Monty picks and orients the card to be displayed. How is what you describe above different than just pulling out a card in the proper orientation and discarding any that show white on top?
Nope, not seeing it. Can you construct a probability table that describes your version of Monty’s process?
I think using those tables is obfuscating the issue. What I’m thinking of is more of a flowchart or decision tree. But I don’t know how to represent that in print, so I will just try to do a better job of describing it.
After the bag is “shaken about”, Monty peeks inside while the guesser is out of the room or has her back turned. He does not turn over any of the cards, and does not know which is which. He will see either two white cardfaces and one black, or two black and one white. If there are two black faces showing, he will determine which one happens to be lying closest to the mouth of the bag, and slide that one out. If there is only one black face showing, he will obviously just slide that one out, no matter where it is lying.
So here’s where I see it branching as a tree. Two of the cards will always look the same: one white, one black. They are almost like decoys in a sense. The third card, with black on one side and white on the other, is the only one that actually changes from getting shuffled. So there are only two possible “decks” (excluding their order for the moment): one (Deck A, let’s call it) in which the white side of the third card is showing, making two white faces and one black visible, and the other (Deck B) in which the black side of the third card is showing, making two black faces and one white visible.
(1) 50% of the time the cards are shuffled, you get Deck A. Monty will slide out the only available card with black showing face up, which is the card that is black on the bottom as well.
(2) The other 50% of the time, you get Deck B. Monty will slide out whichever of the two cards with a black face showing happens to be closer to the opening. So there is a 50/50 chance of a card slid out of a Deck B having a black bottom.
Take the 50% from (1) and add it to the 25% (half of 50%) from (2), and you get 75%. Does that make sense now? There are three possibilities, but they are not equal: one is 50% and the other two are 25% each.
I’m having a hard time accepting that those are a) equal possibilities because b) #2 and #3 are (to me) the same thing.
But when you draw the double white, it doesn’t apply to this thought problem. Its probability doesn’t seem to collapse into the B/B card like Monty’s goat does into the unchosen door.
This makes sense to me.
There are 8 possible card arrangements. In 4 of them, the B/W card is white, guaranteeing that “the flip” reveals a black side. In the other 4, you have a 50% chance of getting the B/B card, getting a black side on “the flip”. That’s 6 out of 8: 75%.
But that is the heart of the problem, and why it’s interesting. If you treat them as the same thing, you’ll see that it is twice as likely as the B/W card. Just because there are two options doesn’t mean they have equal probability. Try it with two cards, just a B/W and a B/B. Ignore the times W comes up. How often do you think you’ll get the B/B card and how often the B/W.
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It doesn’t need to, the W/W really doesn’t enter into the problem. The problem is exactly the same with just two cards.
Hrmm, it does make sense. I have to think if there are other methods of selecting that would change those odds, mainly sliding out random cards until he gets a black one. But I seems like it wouldn’t change the odds.
Back to the MH problem: there are a number of ways to explain this, and perhaps not one that works best for all people.
My favorite explanation is a variant of this:
Here’s the variant:
Restate the rules like this: You choose one door. Monty then gives you the option to stay with that choice, or exchange it for the chance to open both the other doors and select the best prize either one contains.
It’s fairly apparent that two doors are better than one, and that this restatement of the problem in fact matches the original.
I may have misread the scenario. I don’t know.
I took it to be equivalent to the following:
There are just two cards–a double black card, and a black-white card. I am allowed to pick one at random without looking at it. Monty then reveals that card, being sure to place the black side face up. What is the probability of the other side being black?
There’s a 50% chance I initially chose the all-black card. And in that case, there’s a 100% chance the other side is black. So there’s 50% of the total probability space filled in as “the other side is black.”
There’s a 50% chance I initially chose the all-white card. And in that case, there’s a 0% chance the other side is black. So there’s 0% of the total probability space filled in as “the other side is black.”
Add the two up, and you get a total probability of 50% that the other side is black.
ETA: I think this is the key. If a card is picked first, then a side revealed (guaranteed black), then the probability is 50%. If, on the other hand, a side is picked initially (i.e. of the three black sides, each has the same probability of being the initial reveal), then the probability is 66%.
I don’t play Bridge, but I have played Hearts. I assume “Contract Bridge” means played in teams, opposite sides of the table. Cards dealed around, 13 per player. Each round the teams try to take the points similar to Hearts, i.e. first card played sets the suit for that round, highest card that suit takes the win, and you must play in suit if you can, otherwise you dump anything you want. I assume no crosstalk or pre-planning, so I couldn’t ask if you have the trump before I play. But then how do you know which cards your opponents have vs your partner? Does Contract Bridge allow teams to share knowledge? Is that what the “contract” part means?
Again, to clarify, N/S pair are trying to run hearts to collect all the points. You want the high card every round. Ace and King are in your collective possession, but Queen and Jack are not - Q and J being two high cards that could spoil the run.
First round your partner drops Hearts, West drops low card 2, you play Ace, East drops Jack - that makes East’s lowest heart the Jack. That’s how you know he can only have Q or nothing, because you hold the K and played the A.
Next round, partner leads a low heart again, West plays the 3. Now it’s your turn. Your options - do you play your King now or later?
If East has the Queen they must play. But if they don’t, then West is holding the Queen and will take the next Heart round. That’s the game, either one or the other case. So do you bet the odds or play against?
However, it might involve gamesmanship if the player is trying to make you think they have or do not have the card to affect your decision. That makes it a poker game. Ignoring attempts at bluffing, mathematically the odds favor you dropping the King and assume East has to play Queen. The odds difference is small, but that’s what it favors.
But you asked “which do you play for?”, and to me that puts bluffing etc on the table, and that involves knowing your opponents and strategizing around the straight odds.
I’m assuming that rule is straight odds based. If you have 8 cards, odds align slightly that the opponent can take a round with the Queen, 9 cards the odds align with a slight favor that you will get the Queen with the King or Ace.
As for the rule, it only applies if you are strict odds. It doesn’t apply if you’re looking at bluffs. As they say in poker, you’re not playing your cards, you’re playing the other player.
GuanoLad, you understand. The host did not actually reveal anything that changed the odds because he is always going to show an empty door and never reveal a prize.
Because you are past that part of the odds, the W/W card is irrelevant. At the point you are considering, you know it can’t be the W/W card, so your odds have shifted to the following cases.
Card B/B side 1 (B)
Card B/B side 2 (B)
Card B/W side 1 (B)
Those are the only three cases that can remain. Either the card has a black back or a white back, but the face is a given black.
The reason why it isn’t 50/50 is because there are two sides to the B/B card. Each side of the card has an equal chance of being the one drawn.
The act of drawing eliminated 3 equal shares of the original 6 equal possibilities, what remains is 3 equal shares. Each side of the card must be treated independently as far as possibility of being the one that is face up.
But they are not for the purposes of probability.
Consider if you have a fair coin, heads or tails, 50/50. Then you have a 2-headed coin, odds are 100% heads will turn up.
If you flip both coins together, what are the odds you will get 2 heads? 50/50. What are the odds you will get a tails? 50/50. The extra coin is irrelevant because it is a given.
Now what if I take and hold 1 coin in each hand. You can select the coin, but do not know which is which, so you pick left or right. It is 50/50 which is the fair coin. Now I flip the coin you selected. What are the odds you get Heads?
Case 1: you pick 2 heads coin (50%), I flip and get heads (100%).
Case 2: you pick fair coin (50%), I flip and gets heads (50%).
Case 3: you pick fair coin (50%), I flip and get tails (50%).
Those are not equal sized categories. Case 1 has twice the number of throws as Case 2, because the 50% shown for fair coin in Cases 2 and 3 is the same set of results.
I think you need to look at numbers rather than percentages. Take 100 throws.
You pick the fair coin half the time (50 throws), the 2 headed coin half the time (50 throws). I know that is not reality, but we’ll do an ideal case so I don’t have to speculate on millions of throws, because who has that time?
The half time you pick fair coin, you get fair split on H/T.
That means you end up with 25 Heads, 25 Tails.
Half the time you pick the 2 headed coin. How many throws is that? 50 throws. So you have to account for 50 throws coming up Heads.
That means you had
25 T, 25 H, 25 H, 25 H, = 75 Head to 25 Tails.
In other words, despite the fact that the 2 headed coin looks like the same results (Heads), each side has to be treated separately as a possible result to reflect actual numbers of results.
Same thing for the cards from the bag. The B/B card has two sides, each side is a different possibility for whether it is up or down.
Think of it differently, number each card face. You have B1,B2, B3/W3, W1/W2. Now what are the odds the other side is black? You don’t know, they aren’t visibly numbered, but think of them as independent options.
If you see B, is it B1, B2, or B3? Three equal probability options you are looking at B, what are the odds the other is W?
If you are looking at B, that eliminates 1 B from the list. You have 3 Bs and 1 W, minus 1 B = 2 B versus 1 W.
66% B, 33% W, 1% the universe explodes. (Okay, not really, there’s a rounding error there. 
)
Different point in the problem. This isn’t Monty revealing the empty door, this is you walking in to the room after the door has been revealed and removed. You have different odds than the original contestant. Just like the odds are different for the guesser before/after the W/W card is eliminated from the choices.
Yes, that is a good explanation. But then you have to argue that this restatement is a correct reflection. It’s all in justifying that Monty can give you the best of 2 as opposed to only 1 but he picks the best of the two you didn’t already pick.
That’s another reason to favor empty doors over goats. You pick a door, then Monty offers to trade his 2 doors for your door. An empty door is an empty door, but you still get the best prize of those 2 versus your original pick.
Yes, exactly, if you treat the card as the decision point, then the results are different than if you consider each side as a separate option in the decision point.
Just like Monty’s doors. If you pick from three and Monty reveals a dud, you have different odds than if I walk in the room after the door has been revealed and removed. I am picking from 2 even percentage doors, but you are picking from 2 uneven percentage doors.
In your example, you picked a card that was guaranteed to have a black side, then you look at odds on that card whether the other side is white. The example stated was to pick a possible side from a bag with 3 black sides possible, then you look at the odds the other side is white.
If East started with the QJ, then on the first round East could have played either the Q or the J, and an expert player would often play the Q to try to deceive South, Therefore, I think it’s more likely that East only has the J: South should take the finesse.
But if everyone playing is expert and East knows this, they could play the Jack as an even higher level bluff.
(I said the problem required almost no knowledge of cards, but I overlooked that you need to know Queen and Jack are equal – there’s no strategic reason for East to play one rather than the other.)
Just as in rock-scissors-paper where you play each with 1/3 probability if you have nothing else to go on, the default assumption is that holding QJ and no low cards East will play the Queen 50% of the time and the Jack 50% of the time. Any other strategy, if visible to the opponents, would put himself at a disadvantage.
The section you quoted was actually about a different version of the problem. We’re all in agreement about the 2/3 in the random case, we’d moved on to the case where Monty makes sure that a black card is always shown and his methods of selecting that card/side to display.
If you pull the cards at random from the bag, and if a white face shows up you put all the cards back in the bag and try again then your odds are 2/3.
But, if you flip all three cards, and have Monty look in the bag and always pull out the closest card with a black face your odds change to 3/4. It’s an interesting variation. My initial analysis was wrong, I thought it stayed at 2/3.
At least I think I’ve summarized the variant correctly.
Yes, except “flip all three cards” might mislead some people into thinking you are alternating each time (not that this would actually change the odds, but it is certainly less random). “Shuffle” would be a more precise word there, or maybe “randomly shake up”?
Trying to remove Monty from the equation, what are the odds here:
Case A:Shuffle the cards. Slide a card out of the bag. If it shows white, put the card back in the bag, re-shuffle, and repeat until you pull out black. This appears to be the classic 2/3 scenario.
Case B:Shuffle the cards. Slide a card out of the bag. If it shows white, discard that card and reach in for a second card from the two remaining. Repeat if needed with the last card. Does this map to Monty looking in the bag and pulling out the black card closest to him?