If the difference between a “normal deck” and a “climate changed deck” is just one card in 52, or one year in 30, then it’s going to take an awful lot of draws before you could be at all confident in any sort of conclusion (though precisely how many that “awful lot” will be will depend on the prior you assign to tampering).
But on the other hand, if the difference between the decks is in something like 80% of the cards, then it’s going to be much easier to tell the difference.
You won’t know that it’s been tampered with, but you can have well-defined probabilities for how likely it is that you’re drawing from the tampered deck after even a single card. The more cards you get, the better your estimate will be.
The easy way to see this is to imagine that the tampered deck is much more tampered, and is like 50 black cards and 2 red ones. If you draw one card and it’s red, it is very likely you are drawing from the non-tampered deck.
The math still works with only slight tampering, it just means the probability will be closer to 50%.
What is the well-defined probability you aren’t drawing from the 50B2R deck after you’ve drawn one red?
Sure, but unless I go in with the understanding that the deck might be tampered with, the probabilities are so small as to go unnoticed and without arousing suspicion until I draw the 27th black.
Now that I’ve posted that, I think I’m missing some detail here. @am77494, are we given a deck of cards with no indication that it could be tampered in any way and in a manner not to arouse suspicion, or is it randomly chosen from a pile of card decks, of which we have knowledge that half of them have been tampered, or something completely different?
I’m not 100% sure I have the math right, but here’s a stab at it. Note that even if the below is wrong, the math to solve this problem exists and is quantifiable.
If we assume that you’re either drawing from a fair deck or an unfair deck with equal probability, then there are 28 ways to draw a red card (26 in the fair-deck universe, and 2 in the tampered-deck).
If you draw a single red card off a deck, then the chance that the deck is fair is 26/28, and the chance that it’s the tampered deck is 2/28.
As you draw more cards (or if you have different starting assumptions) the math gets more complicated, but it is well-defined.
Not necessarily. The black card must duplicate another so suppose the first two cards are queens of spades?
As a follow up to above, I think rather than counting to the 27th black card, the question boils down to given two matching black cards, what is the probability that the nth card matches its pair already drawn.
Clearly p(n=1) = 0; p(n=2) = (2/52)*(1/51) or 0.000754
Just reread those (only 1 cup of coffee so far). I would say they agree with me.
Then we’re at an impasse. I read post 4 as saying the same thing you are and post 5 saying that that is not correct. As stated earlier, I don’t think we can answer the OP until they come back and give additional detail.
It seems to me that the OP in #5 specifically said that he’s ignoring the characteristics of the card other than the colors, so the probability of finding a duplicate card is not the solution.
Ok got my 2nd cup of coffee and you are right according to Post 5 but not the OP. Now I’m imagining 25 Aces of Hearts and 27 Aces of Spades. If am77494 sees this can you confirm that is your scenario and not a standard deck with one duplicate? Because according to
Suppose I have a deck of cards that is tampered by removing one red card and replacing it with a black card from another deck.
Seems like a standard deck with a duplicate while
I am looking for just the colors.
Is more like all reds the same card and all black the same card and looking for the 27th black card.
@am77494, can you please answer the following:
As you can see, it might be simpler to just ask your climate change question.
Here is my thought. The question can be restated as: given that you have drawn n cards what is the probability that the rest are red? After n cards and assuming the last one is the 27th black, there are n-27 red cards showing so 25 - (n - 27) = 52 - n red cards left. So for example the 30th card is the last black so we have 27 black, 3 red with 22 red cards left.
I believe this means that that works out to (looking at the stack of red cards at the bottom) is the permutation P(25, 52-n) so to answer the OP the probability is P(25, 52 -n) ÷ 52! but I need to think about that a bit.
If you assume the last one is the 27th black, you’re assuming a tampered deck in advance. Am I understanding you correctly?
No it is assuming no prior knowledge and after the nth card you go, “Wait, that’s 27 black cards”
Then we’re right back to my scenario in post 9. I still don’t have confirmation from the OP that that is the situation.
I think we can all agree the question needs to be written more clearly. But if we know a priori the deck has been tampered with then p(tampered) is already 1 so I assume we don’t know.
Although the odds are extremely remote, you could draw 26 black cards consecutively. This still wouldn’t tell you anything and is potentially the same even when the 51st card has been drawn. It’s just that in real life, the consecutive order would look very strange, perhaps in some context, suspicious.
But in this scenario, the odds are great.