r/theydidthemath • u/ThatsSo • 1d ago
[Request] Did Pokémon accidentally reinvent the Monty Hall problem?
The Pokémon TCG Pocket game has a mechanic where you are shown 5 cards face down, in which there is usually only one desirable card, and let you pick one at random to keep
That's all well-and-good, but recently they added an event where you get to "peak" at one of the cards before you make your choice.
Now - if the one you peak is correct, you can just pick it, so that math ends there. However, if the card you pick is incorrect, does the logic of the Monty Hall problem apply?
I, and I think most others, mentally decide a card to pick before it's even time (I always pick the bottom right). If I reveal a different card and it's incorrect, is it statistically probable for me to forsake my mental guess and pick one of the other 3 cards? This feels wrong, the game didn't know that was my choice, it feels like it should now be no less likely than the other 3 cards.
However, wouldn't the logic of the Monty Hall problem apply to this, and say it is incorrect? That, logically speaking, my initial probability doesn't change from 1/5 despite the fact another one was eliminate (this is, to be clear, under the theoretical that the 'peaked' card is wrong, as the peaked card being correct ends the scenario). If there was a 1/5 chance my initial guess was correct, there is a 4/5 chance it was wrong. If a card is revealed, there is still a 4/5 chance I was initially wrong, but if there are only 3 possible cards to switch to, they split that 4/5 3-ways, making them each 26.66% likely to be correct (as opposed to my 20%), no?
This is my way of understanding the Monty Hall problem but practically speaking I don't feel like that can be incorrect? The game doesn't even know my initial 'mental' pick, so how could there be a statistical difference if I choose it or swap it.
And if any of the 3 swaps are really 26.66% likely in that scenario, wouldn't that mean mentally envisioning one and then swapping it is actually (very slightly) more likely than peaking at one at random and then picking one of the remaining 4 (25%)? Again, in both these scenarios just not factoring in the possibility that the peak is correct, which should apply to both scenarios equally anyway and not change the end result.
I don't know what the flaw in my logic is but I can't imagine that 26.66% correct. If the 1/5 chance the "peaked" card was incorrect is distributed to all 4 cards then it's 25%, but wouldn't that mean the Monty Hall problem results in a 50% chance instead of a 66% chance?
Can anyone help me break down the probability in this scenario? Is there a flaw in my understanding of the probabiltiy or do you really increase your chances by mentally choosing one and then refusing to take it?
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u/UndeadCaesar 1d ago
I think Monty Hall does not apply here because you get to pick the card, so you have a possibility of being correct the first time and getting confirmation. In the Monty Hall a random incorrect option is removed from the pool, which is a different procedure.
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u/ThatsSo 1d ago
That tracks but I'm moreso thinking probablity after the point that card is confirmed incorrect
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u/mil24havoc 1d ago
Still no. Monty Hall works because the host knows which door has the prize. Monty would never randomly reveal the prize door.
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u/ThatsSo 1d ago
I don't think you understand what I'm saying, after the door is revealed to be false, it shouldn't matter if the host knew it was false or not, so I am asking the probability of switching or not after that reveal, as if that variable is constant
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u/Angzt 1d ago
after the door is revealed to be false, it shouldn't matter if the host knew it was false or not
It does. That is the entire crux of the Monty Hall problem. The fact that we know that the host will never reveal the main prize.
That's the one thing that fudges the probabilities in an unintuitive way.And going back to your initial question:
This isn't Monty Hall. Because there is nobody taking an action based on information that the player isn't privy to.
If you secretly pick a card, then peek (not "peak") at another, that choice was not based on additional info. It is pure chance.
As such, the probability that you were initially wrong is only adjusted from 4/5 to 3/4, given that the peeked card was a dud.If you draw out the probability trees for your scenario and Monty Hall, you'll see that all of your scenario's branches are reachable. Specifically, you can peek at the best card, the main prize. That is a possibility.
And exactly that scenario is not an option in Monty Hall; these branches of the tree are inaccessible. Because Monty has knowledge based on which he will avoid that scenario. That makes his choice non-random, which means a player can garner information from it that shifts the probabilities.6
u/jbrWocky 1d ago
try drawing a out two weighted decision trees representing the MHP with a random reveal and with a host reveal
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u/big_sugi 1d ago edited 1d ago
The Monty Hall problem only works if the host knows which door is false. If the door is opened at random, then you’ve gained no other information, and each of the remaining doors is equally likely to have the prize. Switching gives you a 25% chance of winning—but so does keeping your original choice.
The key is that the Monty Hall problem conceals the actual choice. It’s not one door out of three; it’s one door or all the other doors.
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u/Willeth 1d ago
That's the common explanation that makes it make intuitive sense. But the probability does not depend on the host's knowledge.
Even if Monty's choice was entirely random from the doors the player did not choose, the probability would still work out. In fact it would be even better odds, because he would also have the option to open the door with the car and offer the switch.
The bigger confounding factor here is that there are more than three doors in the Pokémon example. It still makes it make sense to switch, but the advantage you get by switching diminishes the more doors/cards are in the problem.
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u/Quiet-Mango-7754 17h ago
You contradict yourself. You say the probabilities don't depend on the host's knowledge, and then that it would yield better probabilities if the door opened was random. You are right in the second part tho. But I would add that in that case, you don't need a specific strategy to reach maximum expected outcome.
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u/Quiet-Mango-7754 16h ago
The difference is that in the Monty Hall problem, there is a malicious agent (the host) that has knowledge you don't and is changing the odds by opening only wrong doors. That's why you need to come up with a specific strategy to achieve maximum chances of winning. When everything is random like with sneak peek, there is no strategy you could devise that would yield better odds than picking at random. When in doubt, just draw the probability tree. Let's do a simpler version with 3 cards. Wrong card 1 W1, wrong card 2 W2, right card R.
Strategy 1 : keep the same card
Choice r, sneak peek r, result : 1
Choice r, sneak peek w1, result : 1
Choice r, sneak peek w2, result : 1
Choice w1, sneak peek r, result : 1
Choice w1, sneak peek w1, result : 1/2
Choice w1, sneak peek w2, result : 0
Choice w2, sneak peek r, result : 1
Choice w2, sneak peek w1, result : 0
Choice w2, sneak peek w2, result : 1/2
Total expected gain : 6/9Strategy 2 : switch card
Choice r, sneak peek r, result : 1
Choice r, sneak peek w1, result : 0
Choice r, sneak peek w2, result : 0
Choice w1, sneak peek r, result : 1
Choice w1, sneak peek w1, result : 1/2
Choice w1, sneak peek w2, result : 1
Choice w2, sneak peek r, result : 1
Choice w2, sneak peek w1, result : 1
Choice w2, sneak peek w2, result : 1/2
Total expected gain : 6/9
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u/blacksteel15 1d ago
No. In the Monty Hall problem, the door that's revealed is guaranteed to be wrong. The reason it works the way it does is because while we don't know which door is correct, we can infer information from the actions of the host, who does know and who decides which door to reveal. The host's actions differ depending on whether or not you picked the correct door the first time. In you did, he opens a random wrong door. If you did not, he opens the only remaining wrong door, which is not a random choice.
In this case, you're choosing a card at random, then revealing a different card at random. It has a chance to be the card you want, giving you a 100% chance to win, which cannot happen in the Monty Hall setup. But because it was selected completely at random from among the unchosen cards, if it isn't the one you want then it tells you nothing about the others except that the card you want is one of the remaining 4.
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u/superheltenroy 1d ago
To be fair, the Monty Hall problem still works brilliantly if the host randomly reveals the other door, you will still win by switching in 2/3 of the picks, only you now must remember to sometimes pick the revealed prize.
The significant part is that the host doesn't reveal the door the player picked.
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u/blacksteel15 1d ago
I mean, that's a completely different setup than the Monty Hall problem. You're right that the probability of winning by switching is still 2/3, but for completely different reasons.
In the Monty Hall problem, we can increase the odds of winning to 2/3 by noting that which door is revealed to be incorrect is only random in 1/3 of cases. In the 1/3 of cases where it is, you have a 100% chance to lose by switching. In the 2/3 of cases where it isn't, you have a 100% chance to win by switching. The significant part is that we know the rules the host uses to select the door to reveal, not just that he doesn't pick the initially selected one.
In your problem, assuming you choose rationally, the chance to win is inherently 2/3. After the door is revealed, in 1/3 of cases you have perfect information (the cases where you picked the wrong door and the correct one was revealed) and have a 100% chance of winning. In the other 2/3 of cases, the odds are 50/50. As long as you pick the right door when you know which one it is, your win rate will be 2/3 regardless of what you do in the other cases.
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u/Gravbar 20h ago
sure but only if you're allowed to switch to the revealed door (not an option in monty hall). Otherwise it's back to 1/3.
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u/superheltenroy 20h ago
Really not an option? In that case it would be absolutely cruel to reveal the prize.
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u/Gravbar 19h ago
The way we usually pose the problem is that the open door is eliminated, but I guess they could just give it to you if we wanted it to work that way. I think irl it wouldn't be that cruel if it worked like that because you've already picked your door after all so you're getting your fair shot
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u/superheltenroy 20h ago
Thanks for explaining. This assumption wasn't explained to me when I learned about the Monty Hall problem, so I thought it was unnecessary.
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u/JupiterRai 1d ago
Everyone has answered the Monty hall part of the question but I want to point out that wonder pick cards are picked the moment you open select it and you decisions do not matter.
In a snack peek pick the both cards are generated right at the start. Your sneak peak will always be the same no matter where you pick and after that if you pick any of the 4 remaining cards they will always be the same card that was generated at the start.
It’s an illusion of choice that protects the game from exploits to guarantee a card
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u/DonaIdTrurnp 21h ago
If you select the card that you peeked, you get that one, right? So either you keep the sneak peek or reroll it. (Regarding the computer science of when the seed is selected, it is useful to describe the roll as occurring when the user gains any evidence of what the roll is, rather than when the entropy used to seed the RNG is collected or at any other stage in the process, and you don’t need the strong form of timeless decision theory to notice that a die already rolled in a manner where the outcome is hidden from everyone is the same thing as a die that will be rolled later.
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u/Umicil 1d ago
No, because the revealed card might be the desirable card.
The reason the Monty Hall problem works the way it does is because the host knows where the good prize is and always makes sure to reveal a bad prize, and in doing so gives you information.
Since the Pokemon version can reveal any random card, including the desirable prize, it's distinctly different.
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u/DonaIdTrurnp 21h ago
Well, the logic of the Monty Hall does apply, but not in a way that gives you an advantage.
The Monty Hall paradox relies on the choice about which card is revealed leaking information, since Monty is specified to only reveal a lower value prize.
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u/hilburn 118✓ 1d ago
To look at the full decision tree:
1st pick (secret): 20% chance of being good, 80% bad
Peek at a different card:
- if 1st pick was good, 100% chance of being bad
- if 1st pick was bad, 25% chance of being good
So we have 4 cases
- P(reveal good | secret pick good) = 0% * 20% = 0%
- P(reveal bad | secret pick good) = 100% * 20% = 20%
- P(reveal good | secret pick bad) = 25% * 80% = 20% - game ends
- P(reveal bad | secret pick bad) = 75% * 80% = 60%
3. is where the "information" sneaks in that makes switching better - the only way for the game to end on the "peek" is if your secret guess was bad - so the only way you hit the chance to switch is when you chose bad to begin with.
So let's consider switching in the case of 2 and 4:
Case 2: Switching always provides a bad result as the good card was our secret guess.
- P(switching good | 2) = 0% * 20% = 0%
- P(switching bad | 2) = 100% * 20% = 20%
Case 4: Switching has a 33% chance of being good
- P(switching good | 4) = 33% * 60% = 20%
- P(switching bad | 4) = 66% * 60% = 40%
So what about the probability of switching in the abstract, as we don't know if we're in 2 or 4 when we are considering switching
P(switching good) = (P(switching good | 2) + P(switching good | 4)) / (P(2 or 4) = (0% + 20%) / 80% = 25%
P(switching bad) = (P(switching bad | 2) + P(switching bad | 4)) / (P(2 or 4) = (20% + 40%) / 80% = 75%
Note that this is exactly what we would expect from a completely random choice of the 4 remaining cards once you have eliminated one of the bad cards in the peek - a switch is no different.
Compare this to 5 door Monty Hall where the 1 door reveal is always bad
- P(reveal good | pick good) = 0% * 20% = 0%
- P(reveal bad | pick good) = 100% * 20% = 20%
- P(reveal good | pick bad) = 0% * 80% = 0%
- P(reveal bad | pick bad) = 100% * 80% = 80%
Again, we are always in 2 or 4, but unlike previously - the P(2 or 4) is 100%, not 80%
P(switching good) = (P(switching good | 2) + P(switching good | 4)) / (P(2 or 4) = (0% * 20% + 33% * 80%) / 100% = 26.67%
P(switching bad) = (P(switching bad | 2) + P(switching bad | 4)) / (P(2 or 4) = (100% * 20% + 66% * 80%) / 100% = 73.33%
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u/ThatsSo 1d ago
Actually, another question, and bear with here in case I am just an idiot
After you have been told the 'reveal' is incorrect, what makes that situation different than the 5 door example at the end? After gaining that knowledge, wouldn't you then be in a situation where there is a 20% chance your initial guess was correct, and 80% it was wrong, thus bringing us back to the same probability as your ending example of a 5 door Monty Hall?
Just having trouble wrapping my head around the difference of knowing the leak was wrong vs it being guaranteed to be wrong
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u/EGPRC 1d ago edited 1d ago
I will try to explain the difference adding a coin flip, but with the standard Monty Hall problem (three doors, two goats and one car), to have less cases to count.
Remember your chosen option can never be removed, that's the rule; the revealed wrong one must come from the rest. Now, when he knows the locations and your choice is already wrong, only a wrong one remains in the rest so he is 100% forced to take it, while when your is the winner, the other two are wrong so he is free to reveal any of them, meaning that neither is guaranteed, each has 50% chance.
In this way, the host could secretly flip a coin to decide which of the two bad options he will discard when yours has the prize. For example, if you pick #1 and it is the winner, he could open #2 if the coin comes up heads, and open #3 if it comes up tails. (You don't see the result of the coin, the host flips it for himself.)
However, remember he only has one possible goat door to show from the rest when yours already has a goat, so in such cases he must specifically open it regardless of the result of the coin, completely ignoring it.
Then we have 3x2=6 equally likely scenarios depending on the location of the car and the subsequent result of the coin. If you start selecting door #1, those cases will turn out to be:
- Door #1 has the car. Coin=heads. He reveals #2.
- Door #1 has the car. Coin=tails. He reveals #3.
- Door #2 has the car. Coin=heads. He reveals #3 anyway.
- Door #2 has the car. Coin=tails. He reveals #3 anyway.
- Door #3 has the car. Coin=heads. He reveals #2 anyway.
- Door #3 has the car. Coin=tails. He reveals #2 anyway.
Suppose he shows a goat in door #3. That leaves available the three bolded cases, that are 2), 3) and 4). You only win by staying if you are in case 2), so the coin must have come up specifically tails, while you win by switching both in cases 3) and 4), as the coin could have come up either heads or tails, it does not matter. Therefore it is one case versus two, and that's where the 1/3 vs 2/3 comes from.
So the issue with him knowing the locations is that when he sees that you choice is wrong, he ignores the result of the coin and always takes the same one from the rest.
But if he didn't know the locations, he would have always followed his algorithm of revealing door #2 if the coin comes up heads and revealing #3 if it comes up tails, regardless of where the car is:
- Door #1 has the car. Coin=heads. He reveals #2.
- Door #1 has the car. Coin=tails. He reveals #3.
- Door #2 has the car. Coin=heads. He reveals #2 ---> Car is shown.
- Door #2 has the car. Coin=tails. He reveals #3.
- Door #3 has the car. Coin=heads. He reveals #2.
- Door #3 has the car. Coin=tails. He reveals #3 ---> Car is shown.
So, if he randomly reveals door #2 and just by chance it results to have a goat instead of the car, the possible remaining cases would only be 2) and 4), because in order that he opened #3, it was required that the coin came up specifically tails, not heads. Either staying or switching would be correct in only one case, so neither would be more likely than the other.
The conclusion is that the host that knows the locations would have made the current revelation even in cases that the host that does not know wouldn't have.
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u/Willeth 1d ago
we don't know if we're in 2 or 4 when we are considering switching
I'm getting hung up a bit on this part of your working. This is also true in the original Monty Hall problem. What is different here?
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u/glumbroewniefog 1d ago
That part remains the same. The difference is:
Again, we are always in 2 or 4, but unlike previously - the P(2 or 4) is 100%, not 80%
To explain this with three doors, for the sake of simplicity:
You have 1/3 chance of picking the right door initially. If you do not pick it, it's left with Monty. If Monty always eliminates a bad door, then he is guaranteed to end up with the right door every time you don't pick it. Thus there is a 100% one of you ends up with the prize.
If Monty opens a door at random, then there's a 1/3 chance you pick the prize, 1/3 chance Monty's door has the prize, and 1/3 chance he reveals the prize by accident. So if the revealed door turns out to be bad, then you and Monty have equal chances of winning.
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u/Gravbar 21h ago edited 20h ago
Bad explanations to the Monty Hall problem lead to misapplication to other problems. your mental choice of a card prior to the reveal is irrelevant, monty hall only works because the host knows which doors are incorrect, can only reveal incorrect doors, and can't reveal your door. here none of those assumptions are met. The question you should be asking, is "What new information do you know after the reveal?", and the answer will be only the value of the card you chose to reveal at random. It doesn't tell you anything about whether your mentally selected card is more or less likely than the remaining 3.
tldr; odds are the same as randomly selecting 2 cards.
Here, we have the conditions
5 cards, 1 good 4 bad
C1 C2 C3 C4 C5 where C1 is the good card. C2-5 are interchangeable, doesn't matter which is which.
you pick one at random to look at, C
Assuming C is bad, we're left with
C1 C2 C3 C4
And now we have to pick a card at random.
In Monty Hall, we have already selected a card, which cannot be revealed, but here the card we selected was revealed, and will always be revealed
From the remaining there's no longer a concept of switching, so our probability of selecting C1 is simply calculated as random for each of our two choosing events E1, E2. note that E1 and E2 are independent events when they occur, so no need to use conditional probability notation.
the overall probability is
P(E=C1) = P(E1=C1)+P(E1≠C1) * P(E2=C1)= 1/5 + 4/5 * 1/4= 2/5.
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u/atomicsnarl 1d ago
Hmmm. 1/4 good bad. One good card, and four others. One of the five is in your hand
You see one of the 4 cards. Now still 1/4, but good card is yours if exposed, else bad. Good card - test ends.
Good card not seen. Now 1/3 good/bad. Odds are now 25% you have the good card and 75% one of the other 3 is the good card. You have no particular advantage to choosing one of the other cards.
Now the Monty Hall problem.
3 cards you see none. You hold one. Odds 1/2 good/bad. You see one of the two cards. It's bad. Odds still 1/2 good/bad, but 33% your card and 66% other two cards. But one of which is bad, so 66% the other card! Best solution -- pick the other card for a 2:1 advantage over your card. Note: revealing one of the two cards does not change the initial odds, which were 1/2 good/bad.
The 5 card version does not give you and advantage. The 3 card version does.
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u/ThatsSo 1d ago
By this logic the issue is strictly the number of cards, am I understanding that correctly?
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u/atomicsnarl 1d ago
Yes. If it was 100 cards, you have 1/99. Showing one of the 99 still gives you 1/99, but of the 99 you have 1/98, so no significant improvement. 1/5 goes to 1 + 1/3 so not much of an improvement.
People argue about the Monty Hall problem because the confuse the subset odds with the overall odds. The overall is 1/3, but the subset is 1/2. But since you know that since half of the 2 is revealed, the overall 1/3 is now 33% your card and 66% the other two cards (the original odds), one of which is bad. That makes the remaining card 66% odds.
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u/Crazed8s 1d ago
It should. Lot of ranting about the specifics of the Monty hall problem setup but something doesn’t have to be exactly the same to be applicable. I mean, if the host in the Monty hall problem had a chance to show you the winning door you’d want to change even more.
Changing the specifics of the setup simply changes the bonus you receive but the basic elements are there. You have an initial random choice, you gain information about the other doors, and are given the chance to switch.
With 5 items it’s not nearly as significant a boost as the standard 3 door variety, but yeah you’d want to switch.
Initial selection: 1/5 chance. This doesn’t change.
Peak at winning card: obviously switch to win.
Peak at losing card: The other 3 remaining cards have to share the 4/5 odds to win. So switching gets you from 20% -> ~27%. Not really a gain you’d be able to feel on as small sample size.
Also this is assuming binary outcomes. Since we don’t have that it’s a bit muddy in that determining what is a loser door vs winner door isn’t quite as clear. But at a conceptual level it should apply.
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u/glumbroewniefog 1d ago
You can see that switching doesn't provide any benefit in Monty Hall if the bad door is revealed randomly.
Say that Monty is just another contestant hoping to win the prize. You pick a door at random. Monty picks a door at random. You each have 1/3 chance to win. The third door is opened, and turns out to be empty.
Are you both supposed to switch with each other now? Why? You had the exact same chance to win.
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u/Gravbar 20h ago
damn I'm gonna use this next time someone tries to argue random is the same problem. That shows why so succinctly and doesn't require me to explain math they don't understand.
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u/glumbroewniefog 1h ago
Eh, it doesn't always work:
https://www.reddit.com/r/trolleyproblem/comments/1jjhekw/comment/mk0jo89/?context=3
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