Open Mike: Valentine's Day (and Some Hard Math)

A cheerful plug—this is not objective or disinterested, because I love this book (I own a print from it), and Peter is a good friend, and Peter paid me generously to help publicize the book when it first came out. (It was a big success, thanks in part to me, and by "me" I mean you.) But—French Kiss is the perfect Valentine's Day gift. If you already own the book, you can probably attest to that. And if you order before the end of the day on Tuesday you can get it in time for Valentine's Day on the 14th. (International orders might not arrive in time.)

This is the print I own from French Kiss. This would be a great
shot even without the cat.

There are five hundred copies left of the second (and probably last) printing. It's a really nice book, suffused with a lighthearted spirit of romance and the affection that passes between lovers and friends. The book was made to be a gift, with a stout and handsome red cloth slipcase. If one friend doesn't like it, another will. Can't lose.

The real deal
On a different subject, what do you think the chances are of this?

This isn't a setup, this was a real deal, with a well-shuffled deck, dealing in rows from left to right. I once had my arithmetical skills derided by no less an eminence than Daniel Boorstin, celebrated historian and author and the Librarian of Congress at the time, and I doubt I could calculate the chances of one ace showing in a game of Klondike, much less all four, still less the chance of all four showing in a row. By the way, the idea of having to take a statistics class is a component ingredient in my idea of hell. I've never taken a statistics class—I only know just enough about it to know that it would torture me. Anyway, after some hard thought about my lack of mathematical aptitude and my ignorance of statistics, I've concluded that, in all mathematical likelihood, it's going to be a long time before I see this again.

Good Sunday to you. Gordon Lewis's thoughts about the new Olympus Pen-F on the morrow.

Mike

[UPDATE: As Gerry pointed out in the Comments, I should have said probability, not statistics. My bad. —MJ]

"Open Mike" is the Editor's page of TOP and is often off-topic or personal in some way. It appears on Sundays.

Original contents copyright 2016 by Michael C. Johnston and/or the bylined author. All Rights Reserved. Links in this post may be to our affiliates; sales through affiliate links may benefit this site.

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(To see all the comments, click on the "Comments" link below.)
Featured Comments from:

Patrick Perez: "I gave my copy to my girlfriend for Valentines Day two years ago. But now she's my wife so I still get to see the book."

Duncan: "Here's my back of the envelope calculation for the probability of getting four aces in a row. It is .000014775, or 1 chance in 67,681.25 deals of the cards. So unless you play Klondike something like 2,000 times a year (assuming you've got another 35 years left), it's unlikely you'll ever see this again. If you want the gory calculation details, let me know and I'll send them along."

John Camp: "You're a smart guy, so you probably don't lack an aptitude for math, but probably probably lack an interest in it. When I took ninth grade algebra, I got a C, and was lucky to get it—when my father pushed me on it, I told him that nothing I would ever do in my life would require me to know algebra. I was wrong about that, but something interesting happened in demonstrating my wrongness. When I was a newspaper reporter, I was packed off to a special intensive class at Northwestern to study statistics and probability and polling; I did quite well, because everything I studied had a practical application. Twenty years later, when I was involved in archaeology in the Middle East, I took a community college course in surveying, which included trig. I did quite well in trig, because again, it had a demonstrable relationship to something practical. So math without a demonstrable application was hard to me—but with an application, not really a problem. I suspect there's some math involved in lens calculations, and you're a lens guy. Does that math defeat you?"

Mike replies: In my very earliest exposure to arithmetic, I was quite put out by numbers because no one would tell me what. That is, when they said "what's two plus two?", my response was "two what?" It seemed obvious to my childish brain that "two" means "...of a kind," and you can't add two of one kind to two of another kind. For example, two sheep plus two sheep is four sheep, but two sheep plus two oranges is still two pairs of things and not four sheep or four oranges, and two sheep plus two days is not four of anything.

Elisabeth Spector: "Hmmm...my back-of-the-envelope calculation gives 1/67,659 (admittedly with some rounding error)—similar to Duncan's result. So, did you win the game with this auspicious start?"

Mike replies: I play Klondike allowing only two single-card-turn passes through the deck, so clearances are rare, but, yes, this one scored a 52.

Yonatan Katznelson: "The back of my envelope agrees with the back of Duncan's envelope. Explanation (even though you didn't ask, and which you can ignore): To deal the pictured Klondike hand the four aces have to be in positions 8, 14, 19 and 23 of your well-shuffled deck. There are three other sets of four positions in the deck for the four aces that result in seeing them displayed in a row like that—so four total 'good' ways of positioning the aces in the deck so they come out in a row. There are 270,725 (52C4) possible placements for the aces in the deck, so the probability of seeing 4 in a row like that is 4/270,725 ~ 0.000014775."

Rube Redfield: "Statistics has nothing to do with math (neither does Sudoku), but everything to do with probability. I teach statistics (along with a basic 'learn your digital camera' course) at university in Japan and I am sure you would love it. I do it in a computer lab; the computer does all the math, everything is hands on, and the graphics are both beautiful and instantaneous. All you have to do is learn how to interpret them."

mao: "Given that you have two sets of answers from your readers, and you claim to be bad at math, what are the odds that you picked the right one for the Featured Comments? My reading of the problem agrees with the low probability camp, but to achieve better agreement on answers you receive, you may need to improve you skill at writing math problems  ;-)  "

Mike replies: ...Which deficiency certainly stands to reason. Also my lack of concern about discrepancies in the answer.

For the record, the problem, I think, is four aces next to each other but in any order of suits and anywhere in the array (four possible positions). Reason? Well, I dunno. But I think the fact that they were all in a row increased my sense of wonder at the occurrence, whereas I don't think I would have been any less amazed if the row of four was in any different position in the array (i.e., all the way to the right or all the way to the left or with two other cards to the left of the aces instead of one). Which is a totally subjective and impressionistic reason for formulating a probability problem, but then, the only college-level math course I ever took was called "Math for Poets."

Comments

[Alex Buisse]

Duly ordered. I can't help but notice the irony of paying an extra $45 for shipping the book back to its French origins, though!

[Hugh Smith]

I bought French Kiss when Peter first published and it holds a special place in my library.
Thanks Mike and Peter.

Mon deux Euro!

[Fred]

Yeh, but did you win?

[dd-b]

A generation or so ago I did take the two-term mathematical probability and statistics course ("Papa stats"; not to be confused with the one-term "mama stats" and "baby stats" versions). What I remember is that it's surprisingly hard to figure out anything like this one.

Part of the problem is defining exactly what you mean. I'm confident you don't care about the order of the suits. I'm reasonably confident you don't care if it's in the first 4 positions or the second four (as pictured) or either of the other two possible positions for four-in-a-row. But it sounds like you *do* require them to be adjacent, not just all 4 showing at once?

But, now that you have the photo, you can see it again whenever you want!

[John Igel]

Obviously the odds of four aces coming up in a row are much better than the odds of winning the Powerball Lottery!

[Moose]

The odds are just the same as any specific four cards showing up in the initial spread twice. 7♥, J♦, 3♣, A♥ have just the same probability of showing up in any initial spread as your A♦, A♠, A♥, A♣.

All in sequence, longer odds.

In the same sequential order each time, longer odds yet.

In the same spots, in this case 2 through 5, even longer odds.

We, in our minds, are the ones who put special significance on particular cards. Chance doesn't.

I made the mistake, near the middle of the last century, of taking a statistics class from George Kuznets, brother of Nobel laureate Simon.

Wow was he brilliant. Wow did he cover a lot of material in each class. And boy did I come close to drowning. ººº{:~(

[Edd Fuller]

I am not going to attempt to calculate the odds on this, but "all things being equal" meaning that the deck as shuffled is in truly random order, this arrangement is no more or less probable than any other arrangement. It's just that we notice four aces.

[Gerry]

It's probability, not statistics.

Assuming that the important thing is the four aces and that any other three cards would be equally good, here is how to work it out:

There are 133,784,560 seven-card hands in a 52-card deck (assuming that we do not care what order we receive the cards in).

The formula for this is 52! / (7! * 45!)

"52!" means 52 factorial. In other words, 52 x 51 x 50, etc, all the way down to 1.

Next, we have to work out how many three-card combinations there are in the remaining 48 cards (because, once we have the four aces, we need three more cards from the remaining 48 cards). Again, we do not care what order we get the cards in.

The formula for this is 48! / (3! * 45!) = 17,296

Finally, we divide 17,296 by 133,784,560, giving a result of 0.00129282

So the probability is 0.00129282, or one in approximately 774.

[Ken]

Gerry and Duncan came up with wildly differing numbers. Assuming one is correct, which of them got it right? Anyone?

[Gerry]

Sorry -- I was off by a factor of 10 in my final calculation. So the correct probability is 0.000129282, or one in 7,740.

[Elisabeth Spector]

Forgot to add: my calculation was: 4*(4*3*2*1)/(52*51*50*49) = 0.00001478, or 1/67,659. This assumes the 4 aces can be in any starting position (1-4) along the face-up row.

[Joe Kashi]

You could have taken a statistics class from Tom Lehrer. That was fun.

[Stephen Gilbert]

The formula for this is 48! / (3! * 45!) = 17,296

Finally, we divide 17,296 by 133,784,560, giving a result of 0.00129282

So the probability is 0.00129282, or one in approximately 774.

What?

[Steven Ralser]

The cards reminds me of a day when I was doing postdoctoral research up in the Canadian Arctic. One day we were were playing bridge, while waiting fro the helicopter to come into camp. I was dealt a hand with 4 kings and 4 aces. Of course, just after the cards were dealt, we heard the helicopter coming in.

[psu]

"That is, when they said 'what's two plus two?', my response was 'two what?'

This means that you wanted to be a physicist.

[Bob Cook]

What's the probability that I knew the game is called Klondike? I thought it was just your basic solitaire.

[I thought so too, Bob, but I have to research stuff before I write about it here. Turns out there are lots of Solitaire games, and Klondike is only one. A lot of people call it "Patience," too, and I'd never heard that. --Mike]

[Jonathan]

I agree with Elisabeth Spector.

Which works out to 96/6497400 or 1/67681.25

[Gerry]

To explain the card problem in words rather than numbers, I assumed that the problem can be reduced to: "What is the probability of receiving four aces among seven cards dealt from a deck of 52 cards?"

Finding the solution involves calculating 1) the total number of combinations of seven cards, 2) the total number of combinations that contain four aces, and 3) the probability of getting one of the four-ace combinations given the probability of getting any individual seven-card combination.

First we need to find the total number of unique combinations of seven cards in a deck of 52. The answer is 133,784,560 combinations.

Next we have to calculate how many combinations of seven cards there are that include all four aces. This is the same as removing the four aces from the deck, leaving us with 48 cards (52 - 4 = 48), then calculating how many combinations of three cards (7 - 4 = 3) exist in our deck of 48 cards. The answer to this is 17,296 combinations.

In this type of probability problem, the probability of Thing 1 given Thing 2 is equal to the probability of Thing 1 divided by the probability of Thing 2. So the probability of getting all four aces, given a hand of seven cards is equal to the probability of getting all four aces divided by the probability of getting any particular set of seven cards. In other words, 17,296 divided by 133,784,560. The result of this deivision is 0.000129282.

Probabilities are expressed as a number between zero (no chance) and one (absolute certainty). Our probability is 0.000129282, or approximately 1 in 7,740.

Alternative summary: it's not very likely. I hope you bought a lottery ticket when you saw that combination, Mike!

[Duncan]

The difference between Gerry's probability calculation and mine (and others') is that Gerry does not require the 4 aces to be next to each other while my calculation does. For example, Gerry's aces could appear on the first, third, sixth and seventh piles. Since there are many more ways in which 4 aces may appear in any order than there are when they must be adjacent to each other, Gerry's probability is much higher -- 1 out of 7,740 rather than 1 out of roughly 67,680.

[Bill Allen]

Having seen his work in two different Leica stores (I just happened to be passing by; I'm not *that* guy), I was kicking myself a bit for not ordering a print of the photo that appears on the cover of French Kiss. I told myself I could wait and order the book any time. With only five hundred copies left, "any time" appears to have arrived. Paid in full.

[Kjell]

So, if I get it right, agree that the chance of seeing 4 aces in a row is 1/67681, but I guess you would be almost equally amazed if there had been any other 4 cards of the same value in a row. This would increase your chances quite a lot. Inutition tells me by a factor 13 (but intuition is rarely correct when working with probability, so I may be wrong).

But if I am right, your chance at being amazed is about 1/5206, which is much better than the lottery

[Bob Rosinsky]

Trig and physics play a role in billiards. Although you may not realize it, you've been acquiring skill in those areas.

Many years ago, I read studies of animals who possess keen number sense. Believe it or not, Fido knows trig. Throw a ball, and he/she will find the shortest path to retrieve it. Now take it a step further--throw a ball into the sea and Fido will swim the shortest route, taking drift into account.

[Rich]

I think I prefer your Pool/Snooker posts. At least I understand the Comments on those....

[ctein]

Dear Gerry, et.al,

I think** the discrepancy between you guys is that Mike implied two different problems:

1)What is the likelihood of all four aces turning up in the seven top cards?

2) What is the likelihood they'd all turn up in a row?

Gerry, I believe you're solving the first problem, while others are solving the second problem.

Does that make the results match up? Someone else do the number crunching.

I'm guessing that everyone has taken into account that Mike did not specify a particular order to the aces, just that all four turn up.

pax / Ctein

**I'm not actually doing the maths, just reading the logic. It's very busy here right now. Just spent the last five days prepping for and having a Janis Ian Living Room Concert (you can Google that), plus two girlfriends staying over the weekend (finally, the new guestroom gets field-tested), plus one visiting for the rest of this week.

Hence, kinda overscheduled.

My life truly does suck.

[Bruno Masset]

Here's an explanation of why Duncan's calculation is the correct one.

Mike thought the pattern he got was unusual because he got four contiguous aces.

We'll make the important assumption that Mike would have been surprised by the occurence of four contiguous aces anywhere in the seven cards that are face up.

That is, in the 52-card stack, the four contiguous aces could have been at positions:

{1,8,14,19}

or

{8,14,19,23} (as actually happened with Mike)

or

{14,19,23,26}

or

{19,23,26,28}

and presumably create a similar amount of surprise.

There are 52! ways to order a 52-card stack.
Assume the aces occupy the positions 8,14,19,23.

There are 48! ways to order the remaining 48 cards.

Furthermore, there are 4! = 24 ways to order four aces:
A♦, A♠, A♥, A♣
A♦, A♠, A♣, A♥
A♦, A♥, A♠, A♣
etc.

There are thus 4! * 48! ways in which a 52-card stack can be ordered such that the cards at positions 8,14,19,23 are aces.

The probability of having four aces at positions 8,14,19,23 is thus (4! * 48!) / 52!

Given that we assume that four contiguous aces anywhere among the seven face up cards would have been surprising, and given that there are four possible such position combinations — i.e. {1,8,14,19}; {8,14,19,23}; {14,19,23,26}; {19,23,26,28} — the final probability of a contiguous four-ace pattern appearing anywhere among the seven cards is simply "four times (4! * 48!) / 52!" = 1 / 67681.25

[Gerry]

To Duncan, Elisabeth, Bruno and others...

You are quite right -- I was solving a different problem than you were. As Ctein pointed out, my solution was for getting four aces (in any order) when dealt seven cards from a 52-card deck, whereas your solution also took account of the order in which the cards appeared.

It's unclear to me whether having the aces appear next to each other was an important part of the problem but I'm happy to consider myself outvoted in that regard.

[Bob Blakley]

"I assumed that the problem can be reduced to: 'What is the probability of receiving four aces among seven cards dealt from a deck of 52 cards?'"

But it can't be reduced to that problem, which is one of the reasons probability and statistics are hard. Both fields are "just counting", but figuring out what to count is very difficult.

The wrinkle here is that it's not 7 cards, being dealt; it's all 52. And to get exactly the result Mike got, you need to deal aces in positions 3, 6, 10, and 15.

If you care about getting the suits of the aces in the positions Mike got, the odds will be higher.

It's also not clear what question Mike is asking. Is it:

(1) "What's the probability that I will get all four aces dealt face-up in a Klondike spread?"

(2) "What's the probability that I will get aces as the second, third, fourth, and fifth face-up cards in a Klondike spread?"

(3) "What's the probability that I will get the ace of diamonds as the second face-up card, the ace of spades as the third, the ace of hearts as the fourth, and the ace of clubs as the fifth, in a Klondike spread?"

These all have different probabilities.

[Maggie Osterberg]

Valentne's Day is my 52nd birthday.

Just putting that out there. ;^)