# Monty Hall Deniers?

My account of the big creationism conference will resume shortly, but I really must take time out to discuss this article by Brian Hayes of American Scientist. He is discussing the Monty Hall problem, you see.

The story begins with this earlier article by Hayes. He was reviewing the recent book Digital Dice: Computational Solutions to Practical Probability Problems, by Paul Nahin. Having enjoyed Nahin’s previous book Duelling Idiots and Other Probability Puzzlers, I suspect this new one is worth reading as well.

Hayes writes:

The Monty Hall affair was a sobering episode for probabilists. In 1990 Marilyn vos Savant, a columnist for Parade magazine, disยญcussed a puzzle based on the television game show “Let’s Make a Deal,” hosted by Monty Hall. The problem went roughly like this: A prize is hidden behind one of three doors. You choose a door, then Monty Hall (who knows where the prize is) opens one of the other doors, revealing that the prize is not there. Now you have the option of keeping your original choice or switching to the third door. Vos Savant advised that switching doors doubles your chance of winning. Thousands of her readers disagreed, among them quite a few mathematicians.

This is, of course, the way the problem is usually stated. I really must protest two things, however.

First, it needs to be stated clearly that Monty is guaranteed to open an empty door. This is plainly what Hayes has in mind in telling us that Monty knows where the prize is. The fact remains that this point is not completely clear in Hayes’ statement.

The more serious point is that we must have some piece of information regarding Monty’s method for selecting a door in those circumstances where he has a choice of doors to open (which happens whenever the contestant’s initial choice conceals the prize.) From Hayes’ statement we can conclude that if we play the game a large number of times, then a strategy of alwys switching will win two-thirds of the time, while a strategy of always sticking will win one-third of the time. The situation changes when we consider a single play of the game. Imagine that we initially choose door one and then see Monty open door two. What probability should we now assign to door one? We can conclude only that this probability is somewhere between one-third, and one-half, depending on Monty’s selection procedure.

The usual assumption is that Monty chooses randomly when he has a choice. Given this assumption we would say door one has a probability of one-third, and switching is plainly indicated.

So far, so familiar. Hayes begins his new essay with a story that will be sadly familiar to anyone who has engaged in serious discussion of the MHP:

In the July-August issue of American Scientist I reviewed Paul J. Nahin’s Digital Dice: Computational Solutions to Practical Probability Problems, which advocates computer simulation as an additional way of establishing truth in at least one domain, that of probability calculations. To introduce the theme, I revisited the famous Monty Hall affair of 1990, in which a number of mathematicians and other smart people took opposite sides in a dispute over probabilities in the television game show Let’s Make a Deal. (The game-show situation is explained at the end of this essay.) When I chose this example, I thought the controversy had faded away years ago, and that I could focus on methodology rather than outcome. Adopting Nahin’s approach, I wrote a simple computer simulation and got the results I expected, supporting the view that switching doors in the game yields a two-thirds chance of winning.

But the controversy is not over. To my surprise, several readers took issue with my conclusion.

Heh. The controversy is definitely over, the existence of a few hold-outs notwithstanding.

Hayes goes on to give some specific examples from his correspondence, and gives a wink and gives a shout-out to the importance of agreeing on common assumptions about the structure of the game. He writes:

A number of commentators–going back to the first wave of controversy in the early 1990s–have pointed out that certain assumptions are crucial to the analysis of the Monty Hall puzzle. In particular, it’s important that Monty Hall must always open one door and offer the option of switching, and the door opened can never be the one initially chosen by the contestant, nor can it be the winning door.

Leaving out the most important assumption, alas.

At any rate, Hayes’ essay is very interesting and worth reading. But I nearly fell out of my seat when I read this:

Making progress in the sciences requires that we reach agreement about answers to questions, and then move on. Endless debate (think of global warming) is fruitless debate. In the Monty Hall case, this social process has actually worked quite well. A consensus has indeed been reached; the mathematical community at large has made up its mind and considers the matter settled. But consensus is not the same as unanimity, and dissenters should not be stifled. The fact is, when it comes to matters like Monty Hall, I’m not sufficiently skeptical. I know what answer I’m supposed to get, and I allow that to bias my thinking. It should be welcome news that a few others are willing to think for themselves and challenge the received doctrine. Even though they’re wrong.

Oh for heaven’s sake! The mathematical community is unanimous and the problem is settled. Given the usual assumptions of the problem the wise course of action is to switch. Period. That is a fact, not an opinion. Anyone who says otherwise is wrong, wrong, wrong! Wrong in the same sense that it is wrong to say that 2 and 2 make 5. As for dissenters, well, they should not be stifled (because stifling people is rude), but they should definitely be treated as people who are confused. The whole point of mathematics is that problems get resolved to a deductive certainty. It is the great appeal of mathematics over other branches of science that when one of our problems is solved, it stays solved.

Believe me, I understand the problem is difficult and counter-intuitive. There are many plausible sounding arguments that can be made to defend different conclusions, and it can sometimes be difficult, even for mathematically savvy people, to distinguish the wheat from the chaff. But for all of that there is no controversy. There is no consensus opinion on the one hand with a handful of plucky dissenters on the other. There are only people who understand the problem and those who do not.

1. #1 Luke O'Dell
August 20, 2008

2. #2 ctw
August 20, 2008

I think a possible reason for the result appearing to be controversial is that among the many proposed “solutions”, a majority always seem to be “informal” (essentially verbal reasoning) as opposed to formal (conditional probabilities or event counting). On one of the longer threads on this blog, I recall that even after 200+ comments among which were numerous formal proofs, most commenters were still offering variations on the many more numerous informal “proofs”, even those which had by then been proven unequivocally wrong.

For those who (like me) initially get fooled by the informal “proof” but (unlike me) can’t construct (or even follow) a formal proof once doubts have been raised about our “solution”, perhaps it’s hard – maybe even impossible – to get beyond that initial error.

But why the apparently mathematically savvy continue to doubt the formal proofs is mysterious.

– Charles

3. It’s funny that even people with degrees in mathematics can sometimes get the Monty Hall problem wrong. Those are usually the types that didn’t take an introductory course in probablity. But they can always be set straight.

Another problem that most people who consider themselves to be rational often stumble on is the four-card problem also known as the Wason selection task. I’ve have this problem as part of a five-question quiz on my business website which is frequented by computer science types. These guys live and breath logic and you’d be suprised at how many of them get the problem wrong.

4. #4 j a higginbotham
August 20, 2008

I thought I understood this when it came out, but I am confused now.

1) “The situation changes when we consider a single play of the game. Imagine that we initially choose door one and then see Monty open door two. What probability should we now assign to door one? We can conclude only that this probability is somewhere between one-third, and one-half, depending on Monty’s selection procedure.”

a) I don’t see why a single play strategy is different from multiple play.
b) I don’t see how the probability for door 1 can be anything other than 1/3 given that Monty chooses to open an empty door.

2) “Leaving out the most important assumption, alas.”
Which I guess I missed as well.

5. #5 Dan
August 20, 2008

j a h:

1) I think you are correct on both counts. Monty gives you information by opening an empty door, and hence the probabilities, which are conditional, change. Consider 1000 doors. Pick one. M opens 998, all empty. Obviously switch.

2) I also missed that. Jason? ๐

6. #6 Blake Stacey
August 20, 2008

Sure, he’s a Radiohead fan, he just prefers “Paranoid Android” to “Idioteque”.

When I am king, you will be first against the wall
with your opinion which is of no consequence at all

Obviously.

7. #7 Blake Stacey
August 20, 2008

I presume the “most important assumption” refers to this bit:

The more serious point is that we must have some piece of information regarding Monty’s method for selecting a door in those circumstances where he has a choice of doors to open (which happens whenever the contestant’s initial choice conceals the prize.) […] The usual assumption is that Monty chooses randomly when he has a choice. Given this assumption we would say door one has a probability of one-third, and switching is plainly indicated.

8. #8 Dan
August 20, 2008

Ah, the assumption that the doors are identical. Fair enough.

9. #9 wazza
August 20, 2008

You choose a door, Monty opens another door, and now you have the choice of switching or staying. But the choice of switching or staying isn’t between a door that you chose at a probability of 1/3 or a door that you choose at a probability of 1/2. It’s a choice between two doors. The prize could be behind either of them.

It’s like rolling a triangular prism with a circle, a square and a pentagon on its faces, getting the pentagon, and then getting a coin toss between the circle and the pentagon. The probability is 1/2.

At least, that’s the way I’ve always figured it. Correct me!

10. #10 j a higginbotham
August 20, 2008

Thanks Blake. I was just going back and reading that. It’s an assumption I made implicitly (always the kind which cause trouble). I still don’t see how it makes any difference unless I have watched the show long enough to figure out how he chooses the door. So as a one off problem, how can I figure out anything from his selection of doors? I can’t see that his method of selection has any effect on my decision unless I know what his method is. From the original comments, I should be able to figure out from Monty’s door choice something that raises the odds to 1/2.

11. #11 j a higginbotham
August 20, 2008

wazza,

the simple way i think of it is

1) If we know nothing (easy for me to say), there is a 1/3 chance of picking the door with the prize.

2) Therefore there is a 2/3 chance the prize is behind one of the other two doors.

3) Since there is only one prize, one of the two unchosen doors must NOT have the prize.

4) Given two doors, Monty can always open one that does not have a prize, which he does.

5) But the chance that the prize was behind one of the 2 doors is 2/3 so by being able to eliminate one door, the chance that the prize is behind the remaining door is 2/3.

And there is pparently something I am doing wrong in 4.

12. #12 Jeff Chamberlain
August 20, 2008

Dan said: “Consider 1000 doors. Pick one. M opens 998, all empty. Obviously switch.”

Not obvious to someone who has trouble with the normal 3-door version of the MHP. For that person, the 3-door problem and the 1000-door problem end up the same: 2 doors remaining, behind one of which is the prize and “thus” a 50-50 chance. A person who makes this kind of mistake on the 3-door problem makes exactly the same mistake on a 1000-door version.

13. #13 Mark
August 20, 2008

So the real problem becomes the psychology of understanding mathematics – why is this problem hard to understand for many people? (possibly because it contains some subtly not normally found in probability problems)

14. #14 Shawn
August 21, 2008

I would wager that people have difficulty understanding the problem because they are not used to solving probability questions by computing the negative outcome (in the case of a binary situation like Monty Hall). It’s fairly clear with the standard Monty Hall assumptions that switching is the same as betting that you initially chose the wrong door, and that probability is 1-(1/n) where n is the number of doors.

15. #15 Dan
August 21, 2008

Sorry, I was too terse. ๐

Let’s try this: I’ve hidden money in something in my apartment. You get to choose one item. I then remove everything in my apartment except the item you chose and one other item. Do you switch? As Shawn says, your probability of winning by switching tends to 1 as the number of items in my apartment increases.

As to the question of randomness Monty’s choice: if you know no better, then the uniform distribution has the fewest assumptions (maximum entropy). If you think there are correlations between Monty’s choice and the prize then you should take that into account.

16. #16 ephant
August 21, 2008

@Jeff Chamberlain

For me, at least, the 1000 door version of the puzzle was the one that switched the light in my head. I couldn’t see how the probability wasn’t 1/2 until I considered the 1000 door version and then it suddenly made sense.

17. #17 j a higginbotham
August 21, 2008

The other interesting observation was that when this was contentious, people were arguing for both numbers. But whenever someone said ‘Aha, I see it now,’ they always went from 1/2 to 2/3.

18. #18 NM
August 21, 2008

“You choose a door, Monty opens another door, and now you have the choice of switching or staying. But the choice of switching or staying isn’t between a door that you chose at a probability of 1/3 or a door that you choose at a probability of 1/2. It’s a choice between two doors. The prize could be behind either of them.”

The important thing is that Monty doesn’t pick the door he opens completely randomlu — he chooses one without the prize. If he indeed chose at random, he would in one of three runs open the door with the prize!

Think about it this way, there are three possible combinations, with the player picking the first door:

1. (P) (-) (-)
2. (-) (P) (-)
3. (-) (-) (P)

In case 1, you first pick the prize door, Monty opens one of the other 2 doors. If you stick, you win; if you change, you lose.
In case 2 and 3, you pick a door without the prize. Monty can only open 1 door, the remaining one without prize. If you change, you win, if you stick, you lose.

19. #19 Blake Stacey
August 21, 2008

So, I just woke up with the burning desire to try solving Monty Hall Classic with a genetic algorithm. (Why my brain operates like this, I have no idea.) Ten minutes of Python fiddling later, most of which was spent puzzling over the sort of problems you get when coding in a just-woken state, and I get a nice graph of increasing switching probability. The damn thing converges to the “always switch” strategy so directly that I’m going to have to complicate it somehow to get any interesting dynamics.

Or, you know, I could go back to sleep like a sane person and in the rationality of daylight wonder why I ever bothered. . . .

20. #20 James W
August 21, 2008

Blake: “Ten minutes of Python fiddling later”…

21. #21 Dan
August 21, 2008

All this discussion of Monty’s problem (my fault too) misses the big point in Jason’s post: that Mathematical and other Scientific proofs are different.

In math, like in the MHP, the world of outcomes is strictly controlled, and all hypotheses are enumerated. A mathematical proof is not a proof unless it is a logical necessity, a deduction. Disagreements, like those alleged over climate change, cannot occur in math, because you are either a competent mathematician or you are not.[1]

Other sciences (and everyday life in general) are not like math, because we cannot possibly hope to enumerate and test all hypotheses. It is simultaneously the tragedy and beauty of Natural Science that inferences are uncertain, and depend both on your data and your assumptions. It’s as if you had to solve the MHP without knowing the rules.

The triumph of Science, however, is that for a given set of assumptions (hypothesis/model) and data we can do something like a mathematical logical deduction of the answer. It is the “rational” answer.

[1] I should add that mathematics, being an evolving human endeavour, does have its disagreements, which extends to questions of standards and presentations of proof etc. But that’s beside the point here.

22. #22 kiwi
August 21, 2008

Another way to look at … Monty is essentially giving you the option of keeping your original choice or having BOTH the other doors.

23. #23 Jeff Chamberlain
August 21, 2008

Ephant: I’m interested in the question of “how come” some people don’t “get” the MHP. A “1000-door” example may make the light come on for some, but not for others. These latter, I think, view the circumstances as “resetting” once a door has been opened (revealing no prize). They think that once a door has been opened, it becomes irrelevant; the initial problem (a 3-door or 1000-door problem) is transformed into a new problem when/as doors are opened (changed from a 3-door problem to a 2-door problem, or from a 1000-door problem to a 1000-n door problem). On that view it doesn’t matter how many doors there are at first, since as each no-prize door is opened it essentially goes away and is not seen as having any further connection to the now-reset — different — problem. (This, by the way, would respond also to Dan’s apartment reformulation of the MHP — again, on the subject of “how come” some people have trouble with the MHP.)

24. #24 --bill
August 21, 2008

j a higganbotham–

here’s an argument for why Monty’s method of choosing doors affects the probabilities
suppose there are three doors, marked 1, 2, and 3.
if you pick a door without the prize behind, Monty’s choice is determined for him.
if you pick the door with the prize behind, Monty has two choices. Suppose he always picks the door with the lower number (of the two he has available to him).
How does this change the probabilities?

25. #25 --bill
August 21, 2008

oops…i think i posted before thinking…
i don’t think that what I said above changes the probabilities at all….
so a question: is there a selection procedure for Monty to choose a door that changes the probabilities? upon further reflection, I don’t think there is…

26. #26 Tony Jeremiah
August 21, 2008

1. (P) (-) (-)

Pick(1),Stay(1),Switch(-),Outcome(W)
Pick(1),Stay(-),Switch(2),Outcome(L)
Pick(1),Stay(-),Switch(3),Outcome(L)

Pick(2),Stay(-),Switch(1),Outcome(W)
Pick(2),Stay(2),Switch(-),Outcome(L)
Pick(2),Stay(-),Switch(3),Outcome(L)

Pick(3),Stay(-),Switch(1),Outcome(W)
Pick(3),Stay(-),Switch(2),Outcome(L)
Pick(3),Stay(3),Switch(-),Outcome(L)

2. (-) (P) (-)

Pick(1),Stay(1),Switch(-),Outcome(L)
Pick(1),Stay(-),Switch(2),Outcome(W)
Pick(1),Stay(-),Switch(3),Outcome(L)

Pick(2),Stay(-),Switch(1),Outcome(L)
Pick(2),Stay(2),Switch(-),Outcome(W)
Pick(2),Stay(-),Switch(3),Outcome(L)

Pick(3),Stay(-),Switch(1),Outcome(L)
Pick(3),Stay(-),Switch(2),Outcome(W)
Pick(3),Stay(3),Switch(-),Outcome(L)

3. (-) (-) (P)

Pick(1),Stay(1),Switch(-),Outcome(L)
Pick(1),Stay(-),Switch(2),Outcome(L)
Pick(1),Stay(-),Switch(3),Outcome(W)

Pick(2),Stay(-),Switch(1),Outcome(L)
Pick(2),Stay(2),Switch(-),Outcome(L)
Pick(2),Stay(-),Switch(3),Outcome(W)

Pick(3),Stay(-),Switch(1),Outcome(L)
Pick(3),Stay(-),Switch(2),Outcome(L)
Pick(3),Stay(3),Switch(-),Outcome(W)

Total Wins: 9
Wins due to switching: 6
Wins due to staying: 3

27. #27 Dan
August 21, 2008

Jeff:

Yes, my apartment example is the same as the 1000-door example. I just rephrased it because usually when money and gambling are involved, people are a little more rational. Or am I wrong? ๐

{Afterall, probability theory as used in science grew out of its success in the world of gambling.}

28. #28 Dan
August 21, 2008

I hope this is not considered spam, but this is a great short page telling the history of probability theory:

http://www.cc.gatech.edu/classes/cs6751_97_winter/Topics/stat-meas/probHist.html

although I would argue that you don’t need all the weight of measure theory to have the most useful parts of probability theory.

Well, I found it interesting… ๐

29. #29 drdave
August 21, 2008

Pick a door. It has 1/3 chance of being right.

Monty picks a door. Your door still has a 1/3 chance.

The remainder of the chances add up to 2/3.

Which belong to the unopened door.

30. #30 bmkmd
August 21, 2008

Empty Door Steps of Evolution:

If cutting down the number of options in Monty Hall doesn’t change the original odds (2/3 stays 2/3, therefore switching works), why does each step of selected evolutionary change make the result less random, i.e change the odds?

With naturally selected steps of evolution, the creatures with apposing thumbs aren’t selected from all the creatures with thumbs, but from the prior generation’s choices of thumbs. Mount Improbable isn’t the evolution of apposing thumbs from all creatures, nor from all primates, nor from all apes. Each step which selects out branches of the bush, increases the odds of getting apposing thumbs in humans.

The odds change when you select out options. That’s why its not random.

Do the odds stay 1/1000000000000000 (apposing thumbs vs all creatues? from the beginning of evolution?)or decrease down to 1/2 (apposing thumbs vs. not apposing thumbs, after the empty doors are selected out, sorry after the less selectively advantageous mutations are selected out)?

Don’t the odds change when you select out all those other possibilities (empty door steps in evolution)?

31. #31 j a higginbotham
August 21, 2008

bill

Thanks for the ideas. I see two problems though:
1) The information about Monty choosing the lower number wasn’t in the original question so I don’t think it is relevant. Why not assume I bribed some underling to tell me what door the prize is behind? That would also affect the probability.
2) If we did know Monty chose the lowest number empty door (another scenario is that he stands in the center and just picks the closest empty door) and I chose door one and Monty opens door 3, then there is a 100% probability the prize is behind door 2. But Jason said the odds are for door one, 1/3 or 1/2 so that doesn’t satisfy his conditions either.

32. #32 ctw
August 21, 2008

“We can conclude only that [the conditional probability that the car is behind door 1 given that Monty opens door 2] is somewhere between one-third, and one-half, depending on Monty’s selection procedure.”

I don’t see this. If when Monty has a choice (ie, when the car is behind door 1) he chooses door 2 with probability p, I calculate the conditional probability that the prize is behind door 1 (RV C=1) given that Monty opens door 2 (RV O=2) to be:

P{C=1|O=2} = p/(1+p)

which ranges from 0 to 1/2 for 0 &lt p &lt 1.

As a check, we can evaluate the conditional probability for several values of p and make sure the results seem reasonable:

– for the usual assumption of a fair toss (p=1/2) the result is 1/3 as we know it should be.

– for p=0, Monty will only open door 2 when the car is behind door 3, in which case it clearly isn’t behind door 1, consistent with the result being 0.

– for p=1, Monty will open door 2 whenever the car is not behind door 2. But then it can be behind either door 1 or door 3 with equal probability, consistent with the result being 1/2.

Am I missing something?

– Charles

33. #33 Foggg
August 21, 2008

Well, Charles, you are missing the fact that Monte’s p for choosing the empty door 2 is dependent on the success of the contestant’s choice, so your equation is false.
If the prize and contestant’s pick is door 1, Monte’s p for door 2 is 1/2. The only way p could ever be be 0 is if the prize is door 2 – Monte will never reveal it, but you’re considering the prize door 1, so p can’t ever be 0.
If the contestant’s pick is door 1 and the prize is door 3, Monte’s p for door 2 is p=1. Since the a priori prize prob for door 1 was 1/3, that now leaves the prize prob for door 3 as 2/3.

34. #34 Pierce R. Butler
August 21, 2008

The controversy is definitely over, the existence of a few hold-outs notwithstanding.

A controversy is generally defined as a public disagreement. So long as those few hold-outs continue to speak up – regardless of whether they have any factual, mathematical, logical, mythical or esthetic basis for their arguments – controversy (literally, “against-turning”) persists.

35. #35 Pierce R. Butler
August 21, 2008

BTW, has anyone ever sacrificed the brain cells necessary to tabulate the respective scores of stickers & switchers on tapes of the game show?

36. #36 ctw
August 22, 2008

j a higginbotham:

“Jason said the odds are for door one, 1/3 or 1/2 so that doesn’t satisfy his conditions either.”

What he actually said is that it can be any value between 1/3 and 1/2, and I am suggesting that this may be wrong. According to my analysis, the range is actually 0 to 1/2. I.e., I think you are right.

Your example of MH always opening the lower numbered empty door (either 2 or 3) and opening door 3 is equivalent to the case p=0 in my comment (because you assumed MH opens door 3 instead of door 2 as stated in my comment, that may not be obvious, but I think it’s correct). And as you noted, that yields a conditional probability that the prize is behind door 2 equal to one, ie, a conditional probability it’s behind door 1 equal to zero – as does the calculation for p=0 in my comment.

– Charles

37. #37 AJS
August 22, 2008

OK, I’ll admit: The first time I heard of this one, I got it wrong. But it wasn’t hard to prove it to myself by drawing diagrams.

The important point is, the prize stays wherever it was before you picked a door. There are two ways you can get the winning door: either pick the winning one first time and stick, or pick a losing one and switch. You are twice as likely to pick a losing door initially — and therefore be in the position to benefit from switching.

I think bmkmd is missing that the benefits of evolutionary adaptations are not necessarily constant from generation to generation. And hence the apparent fixedness of kinds: the better adapted an organism is to its environment, the smaller the proportion of useful mutations among the available ones. Unfortunately, it’s next to impossible to put actual numbers to this.

It’s rather like the fact that prime numbers get progressively more sparse the higher you count: as n increases, there are more small numbers of which n could be a multiple. But just try proving it mathematically!

38. #38 Mike
August 22, 2008

2+2=5 for large values of 2.

39. #39 ctw
August 22, 2008

“2+2=5 for large values of 2.”

Actually, even acknowledging that our choice of symbols for integers is arbitrary, if you restrict the assertion to integers and retain the feature of the entity we label “5” that it is a prime, the assertion is wrong no matter what entities “2” and “5” are meant to represent.

Of course, 2+2=6 is another matter – although it might be better to say for “appropriate mappings” of “2” (and “6”). The phrase “for large values” usually implies asymptotic behavior, which is inapplicable here.

– Charles

40. Of the first 100 people who came to my website from Evolution blog and took the five-question quiz, here are the results:

The five highest scores were 428, 393, 378, 376, 364. That’s impressive. Actually any score over 300 is impressive. Others had higher scores the second time around but I didn’t count those.

The first person to take the test scored 364. I wonder who that was.

The average score of those that finished the quiz was 152. Many gave up after missing the first question. This would probaby make the real average much lower.

77% of you got the four-card problem wrong. If you were one of them it suggests that you are not a logical thinker.

41. #41 jo5ef
September 28, 2008

Dear Randy
A correction:
77% of you got the four-card problem wrong. If you were one of them it suggests that you are not a (logical thinker) “maths puzzle obsessed nerd”. For the record, i still dont see why turning over the card with the odd number is the right answer. Supposing there was a consonant on the other side, how would this prove the statement?

42. #42 jo5ef
September 28, 2008

I just noticed it said card(s) not card, oops, my bad.