Question for the Mathematicians

[Longsidebovril]

Fun puzzler for the maths lovers

Currently away for a few days and playing a lot of cards with the kids.
One of the solitaire games works like this..

Take the deck in your hand and turn over one card at a time, and as each card is turned you say Ace, two, three.. all the way up to king and then repeat four times.
The idea is to get through the whole deck without saying the name of the card you turn over.

Now, obviously the odds of the first card are 13/1.
And I’m guessing from then on it changes.

But the thing is it’s really really hard to do and in dozens and dozens of attempts we’ve yet to get through the whole deck. So the question is.. what are the odds of being successful in this game.

[ClaretTony]

I'm a mathematician but I'm not working that one out

[Rowls]

I think it's 13/12 odds on multiplied by 52.

Which is roughly 56-1.

I'm keen on maths but I often make mistakes. I believe CT is far more learned than myself on this subject, as well as a few others who post on here. So happy to be corrected if I'm wrong.

[Longsidebovril]

Wow. If it is in the region of 56/1 that explains why we never get it!

It’s funny cause it feels like it should be easy.

[Rowls]

Having thought about it, I believe it's 12/13 * 52.

12/13 = 0.92

Multiplied by 52 = 47.99

So the odds are 47.99/52 = 0.923

Which I think means there is a 92% chance you will lose and 8% chance of winning.

[dsr]

By my reckoning, each single card has a 12/13 possibility of being not the one you call. So to do this 52 times, the odds are 12/13 to the power 52, which my Excel spreadsheet has at 1.557%, or 1 in 64.

[distortiondave]

Quick edit so no one notices I'm a moron

[Rowls]

8% chance would be odds of 25/2.

Can you play the game 14 or 15 times and let us know how many times you win?

I'd like some real world evidence to support my maths.

NB: I've already confirmed i'm not all that good at maths.

[dsr]

Having thought about it, I believe it's 12/13 * 52.

12/13 = 0.92

Multiplied by 52 = 47.99

So the odds are 47.99/52 = 0.923

Which I think means there is a 92% chance you will lose and 8% chance of winning.

If you multiply by 52 and then divide by 52, you get the answer you first thought of. It needs to be by the power of 52, not a straight multiplication.

[Rowls]

By my reckoning, each single card has a 12/13 possibility of being not the one you call. So to do this 52 times, the odds are 12/13 to the power 52, which my Excel spreadsheet has at 1.557%, or 1 in 64.

What equation are you using to convert the fractional odds into a % chance?

I'm not sure I've got it right, even though we had the same basic formula.

[jdrobbo]

1/4329 apparently, if repeated for the entire deck (4 suits)

[Longsidebovril]

1/4329 apparently, if repeated for the entire deck (4 suits)

That’s mad if it’s true.

[ClaretTony]

It's a long complicated process

1st Card: you have a 1 in 13 chance of it being an ace
2nd Card: 51 cards left but you could have taken one of the twos with your first pick, so the chances of selecting a 2 are either 3 in 51 or 4 in 51.

And so on and so on.

[Longsidebovril]

Can you play the game 14 or 15 times and let us know how many times you win?

I'd like some real world evidence to support my maths.

Honestly.. done it 50 plus times over a couple of holidays and don’t think we’ve ever done it.

[Longsidebovril]

I had a feeling this wouldn’t be easy.

[dsr]

What equation are you using to convert the fractional odds into a % chance?

I'm not sure I've got it right, even though we had the same basic formula.

Chances of success = 12/13 x 12/13 x 12/13 x 12/13 x 12/13 etc etc, 52 times. I'm not using a formula, I'm just putting 52 lines of the figure 12 into box A, 13 into column B, and column C = A x B x the previous line's C.

[Rowls]

Yup, I'm satisfied that dsr has the correct equation and (this is where my maths is weak) the correct answer.

(12/13)^52 = 0.01557293512

ie. 1.5% chance of winning

Which is 63.23/1

[Rowls]

It's a long complicated process

1st Card: you have a 1 in 13 chance of it being an ace
2nd Card: 51 cards left but you could have taken one of the twos with your first pick, so the chances of selecting a 2 are either 3 in 51 or 4 in 51.

And so on and so on.

I think it's a lot simpler and that dsr has got it spot on. The cards don't 'know' which other cards have already been picked and the order is set in stone at the point the shuffle ends so each pick can be assumed as a simple 12/13 odds.

If somebody wants to play the game 70+ times and record the score we'd have a much better idea because if dsr is correct then there is a strong chance (funnily enough one that i can't quantify heeheehee hohoho) of winning once but if jdrobbo is correct the chance of them winning even a single game is minimal in the extreme.

[ClaretTony]

I think it's a lot simpler and that dsr has got it spot on. The cards don't 'know' which other cards have already been picked and the order is set in stone at the point the shuffle ends so each pick can be assumed as a simple 12/13 odds.

If somebody wants to play the game 70+ times and record the score we'd have a much better idea because if dsr is correct then there is a strong chance (funnily enough one that i can't quantify heeheehee hohoho) of winning once but if jdrobbo is correct the chance of them winning even a single game is minimal in the extreme.

It's not simpler though, whether the cards know or not, which cards come out change the odds for subsequent selections.

[dsr]

It's a long complicated process

1st Card: you have a 1 in 13 chance of it being an ace
2nd Card: 51 cards left but you could have taken one of the twos with your first pick, so the chances of selecting a 2 are either 3 in 51 or 4 in 51.

And so on and so on.

I think that's one of the things about statistics that, counterintuitively, makes no difference in the end.

What you say is completely correct, of course, the probability of success if the first card is a 2 is different from the probability if the second card is a 2. But when you've worked out the two different probabilities, and then taken the next step of multiplying the first by 1/13 (probability of a 2 first) and the second by 12/13 (probability of no 2 first) and adding them together, you get back to the answer you would have got anyway.

It would take an uber-statistician to prove it, and that certainly isn't me, but I bet it has been proved somewhere. (But no guarantees!)

[ClaretTony]

I think that's one of the things about statistics that, counterintuitively, makes no difference in the end.

What you say is completely correct, of course, the probability of success if the first card is a 2 is different from the probability if the second card is a 2. But when you've worked out the two different probabilities, and then taken the next step of multiplying the first by 1/13 (probability of a 2 first) and the second by 12/13 (probability of no 2 first) and adding them together, you get back to the answer you would have got anyway.

It would take an uber-statistician to prove it, and that certainly isn't me, but I bet it has been proved somewhere. (But no guarantees!)

Once it all becomes about probabilities then it is no longer really maths but this is a complicated one.

[Longsidebovril]

Maybe it’s better to look at it as a full deck of cards and when shuffled what are the odds of say the Ace being in position 1,14,27 or 40. Or does that not change anything. That way drawn card doesn’t affect the outcome

[ollieclarets8]

Without access to specialized tools for combinatorial enumeration it's less than 1 in 1000.

[martin_p]

It's not simpler though, whether the cards know or not, which cards come out change the odds for subsequent selections.

While that is correct, before you have knowledge of the deck the odds are as dsr stated. The odds of course the change as the ‘game’ goes on.

[dsr]

Once it all becomes about probabilities then it is no longer really maths but this is a complicated one.

You're right. I tried working it out with a deck of 3 cards and the answer is different doing it your way from mine.

Basically, if you have 3 cards there are only 6 ways of shuffling them. 123, 132, 213, 231, 312, 321.

Of the six, only 231 and 312 will win, so it's 1 chance in 3. Working it out my way would be 2/3 cubed which would be 8 chances in 27. The difference is because the chance of success on the last card is so heavily dependent on what has gone before.

With a 52 card pack, I reckon the difference will be small so 1 in 64 is about right - ish. But I can't prove it.

[brexit]

As a model, it's a discrete markov chain as you are dealing with three variables card chosen, instances of current card stated already found and number of cards left in deck. Because you know how many cards have been drawn you can skew your answers, i.e if you know 4 aces have been drawn you can call aces every time.(assuming the rules allow you to have the same guess every time)
To me, the game reduces to knowing which cards have gone before which means that the probability is variable.

[steve_f]

This is why we miss spades claret the bridge wizzo, she would work it out I’m sure

[ollieclarets8]

This is why we miss spades claret the bridge wizzo, she would work it out I’m sure

English teacher, but Gemma worked it out. These puzzles are now a thing of the past.

[Rowls]

To me, the game reduces to knowing which cards have gone before which means that the probability is variable.

Yes but it surely still be expressible as a chance of winning at the outset, before we know the order of any of the cards?

[distortiondave]

I thought the rules were to count up from ace to king, rather than trying to bluff the cards.
If you're allowed to know what cards have already been drawn before declaring a card then that changes everything.

[whiffa]

Just asked AI to explain it because I couldn't be bothered typing it out, but basically agree with CT:

This is a fascinating card game! Let's break down the odds of success.

First, let's clarify the game's mechanics:

You have a standard 52-card deck.

You turn over one card at a time.

As you turn each card, you count: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

You repeat this sequence four times (since there are 4 suits, and 4×13=52 cards).

The goal is to go through the entire deck without the rank of the card you turn over matching the number you say. For example, if you say "Ace" and turn over an Ace, you lose. If you say "Two" and turn over a Two, you lose, and so on.

You correctly stated that for the first card, the odds of not matching are 12/13 (not 13/1, which would be the odds of matching if there were only 13 cards in total). If you say "Ace" and turn over any card other than an Ace, you continue.

The challenge in calculating the exact odds for the entire deck is that it's not a simple multiplication of probabilities for each step, because the events are dependent. The cards you've already turned over affect the probabilities of the remaining cards. This type of problem is best approached using concepts from combinatorics or by simulating the game.

Why it's "Really, Really Hard"

You've observed that it's "really, really hard," and that's absolutely true. The probability of success is extremely low. Here's why:

Fixed Sequence: You're forced to say the numbers in a fixed order (Ace to King, repeated). You can't adjust your call based on the cards you've seen.

Four Suits: Having four of each rank significantly increases the chances of a match.

Dependence: As you draw cards, the composition of the remaining deck changes. If you manage to avoid matching for the first 20 cards, you've removed 20 potential "losing" cards, but you've also reduced the total number of cards, potentially making matches more likely for certain ranks if many of that rank are still in the deck.

Estimating the Odds

Calculating the exact probability analytically is a complex problem, often tackled using advanced combinatorial methods or generating functions. It's similar to calculating the probability of a "derangement" (a permutation of a set of items where no item appears in its original position), but with the added complexity of multiple identical items (the four suits of each rank) and a fixed sequence of "called" values.

While I can't give you a precise, simple fractional answer without performing a very elaborate calculation, I can tell you that the probability is exceptionally small.

Studies and Simulations:

Problems like this have been studied in recreational mathematics. Simulations are often used to estimate the probability. When mathematicians or statisticians simulate this game thousands or millions of times, they find that the success rate is incredibly low.

Without performing an extensive simulation or diving into very deep combinatorial calculations, a precise numerical answer (e.g., 1 in X attempts) is not readily available as a simple formula. However, it's generally understood to be significantly less than 1 in 1,000, and potentially much, much lower (e.g., 1 in 10,000 or even 1 in 100,000).

In summary:

It's a very low probability event. Your experience of never succeeding in dozens of attempts is entirely consistent with the mathematical odds.

The exact calculation is complex. It's not a straightforward probability problem due to the dependencies between draws and the fixed calling sequence.

It's akin to a "matching problem" in probability, but with specific constraints that make it unique.

If you were to set up a computer simulation, you would likely find that the number of successful runs is tiny compared to the total number of runs. This is why it feels so "hard" to accomplish!

[Longsidebovril]

Wow.

[Rowls]

I just asked Grok (Twitter/X) and ChatGPT.

Grok said the odds of winning were 7/2. Ho hum.

ChatGPT came up with 62/1.

I vote we discount the "thoughts" of artificial intelligence.

[Prof]

IMG_6415.jpeg (1.71 MiB) Viewed 1676 times

It’s still early here but it should be something like this. Will check my maths once I’ve had a coffee.

If you assume that the deck has been shuffled and that there is a random permutation of the 52 cards at play then you can calculate this.

I saw some comments about awareness of the cards turned over and that affecting the probability, but you can calculate that too if you’re trying to figure out the probability before the game starts. If you’re trying to do it during the game, then it’s a different thing altogether and it becomes dynamic at that point. In this model I am using the former which is the unconditional probability.

Stick the formula into a computer and it will give you a percentage.

You could use a poisson to get a close answer too. It wouldn’t be exact because you aren’t replacing the cards but the difference would be small.

[mdd2]

I'm a mathematician but I'm not working that one out

Guess that sums you up Tony.

[ClaretTony]

Guess that sums you up Tony.

Sums? I’m good at sums.

[Longsidebovril]

IMG_6415.jpegIt’s still early here but it should be something like this. Will check my maths once I’ve had a coffee.

If you assume that the deck has been shuffled and that there is a random permutation of the 52 cards at play then you can calculate this.

I saw some comments about awareness of the cards turned over and that affecting the probability, but you can calculate that too if you’re trying to figure out the probability before the game starts. If you’re trying to do it during the game, then it’s a different thing altogether and it becomes dynamic at that point. In this model I am using the former which is the unconditional probability.



Stick the formula into a computer and it will give you a percentage.

You could use a poisson to get a close answer too. It wouldn’t be exact because you aren’t replacing the cards but the difference would be small.

When I posed the question I most definitely didn’t expect this!
Brilliant!

[ClaretTony]

When I posed the question I most definitely didn’t expect this!
Brilliant!

Has it put you off playing though?

[Longsidebovril]

Ha. No way. I’ll crack it one day.

[dougcollins]

I'm relieved that factorials eventually entered the field of play.

I wasn't comfortable reading a probability conversation without them.

[Claretprinter]

I just put that into chat gpt to work it out.

It replied: There's more chance of Brownhill re-signing for Burnley

[Tricky Trevor]

Odds change for each card drawn.
If the first 4 cards are 2s then the odds on the next card drop to 1 in 12.
If they were 2x2s, a 3 and a 4. Different again.
3x2 and a king. Different again.
I can’t see anyway there could be one answer before the first card is drawn but I’m not a mathematician.

[dougcollins]

You could use a poisson to get a close answer too. It wouldn’t be exact because you aren’t replacing the cards but the difference would be small.

Those French fish are damn clever.

[Goalposts]

I’m starting to think that the odds on this are more favourable than Trafford going to Newcastle at the moment

[Prof]

Odds change for each card drawn.
If the first 4 cards are 2s then the odds on the next card drop to 1 in 12.
If they were 2x2s, a 3 and a 4. Different again.
3x2 and a king. Different again.
I can’t see anyway there could be one answer before the first card is drawn but I’m not a mathematician.

What you are describing is conditional probability, which is different to the question being asked. In the original question, unless I am getting this wrong, you have a deck that has been shuffled, and therefore we calculate the probability based on a randomly shuffled deck and the probability of the cards falling where they will within that randomly shuffled deck. We assume there is a deck that is equally likely to occur having been shuffled as any other deck that might occur.

Either way, it’s approximately a 1.6% chance of it happening. Run it about 60 times and let us know if you win one. Hopefully, our win percentage this season is slightly higher!

[dsr]

Odds change for each card drawn.
If the first 4 cards are 2s then the odds on the next card drop to 1 in 12.
If they were 2x2s, a 3 and a 4. Different again.
3x2 and a king. Different again.
I can’t see anyway there could be one answer before the first card is drawn but I’m not a mathematician.

There are 80658175170943878571660636856403766975289505440883277824000000000000 different orders of a pack of cards. Of course, if we aren't bothering about suits, it's less, But of that number of ways of shuffling, only a certain number will win the game as described. That certain number, divided by the big number, is the odds of winning.

[Oshkoshclaret]

ChatGPT:

You're absolutely right to suspect that the odds are extremely low — the game you’re playing is often known as "Clock Patience" or "13-Card Solitaire", and it’s famously difficult to win. But your version — saying "Ace, Two, Three... King" repeatedly while flipping through the entire shuffled deck, and trying never to say a card name that matches the one flipped — has a precise probabilistic interpretation, and we can work it out.

Game Summary
52-card deck.
You flip one card at a time.
As you flip each card, you say: "Ace, Two, Three, ..., King" in repeating cycles — 13 values, repeated 4 times = 52 calls.
You lose if at any moment the card you call out matches the card you flip over.
You win only if you complete the deck without a single match.

This is a matching problem — a kind of "derangement" problem, where you're asking for the probability that no item matches its position in a fixed cycle.

The Core Idea
You're comparing two sequences:

The fixed spoken pattern: Ace, 2, 3, ..., King, repeated 4 times
The random shuffled deck

You want to know: what’s the probability that no card matches the called card at that position?

This is equivalent to asking: What’s the chance that a random permutation of the deck produces no matches when compared to a fixed 52-card sequence with four of each rank, repeated in order?

This isn't quite the classical derangement problem (which involves unique items), but a variation known as the Montmort Matching Problem with repeated values.

Simulated Estimate
Since there's no simple formula for this exact setup due to the repeated values, the best approach is a Monte Carlo simulation. But we can shortcut to an empirically known result.

People have simulated this game extensively, and the odds of success are about 1 in 13,000 to 1 in 15,000.

So, roughly:

Your odds of winning are approximately 1 in 14,000.

Why So Hard?
Because there are:

52 cards,
Only 13 different ranks,
And 4 full repetitions of those ranks in the called sequence.
Each card has a 1 in 13 chance of matching the spoken word — and those risks are independent but cumulative, which quickly makes the game unwinnable in most cases.

Even if each individual chance were independent and 12/13 safe (which it's not exactly), you'd have:

(12/13)^52 ≈ 0.0175
Which is ~1.75%, or about 1 in 57 — and that's way too optimistic, because duplicates make matches more likely than pure independence suggests.

Bottom Line
The game is not broken — just very, very unlikely to win.
Odds are about 1 in 14,000 per attempt.
It's a perfect example of a game that feels like you might eventually win, but is nearly impossible in practice.

[Prof]

And this is why we tell our students to not trust ChatGPT. It can convince you it is right when you don’t know the answer.

The Monte Carlo would result in about 1.6% and give a close enough response.

The clock patience is an error but it is another form of solitaire. In this problem, it is just a fixed sequence against a shuffled deck so you can use rook polynomial and inclusion-exclusion.

Can we get back to the football now?

[Longsidebovril]

Genuinely interested… are you actually a maths professor… you must be!!

[Prof]

Strategy professor. 2 maths A levels, undergrad degree in maths. MBA in finance, and PhD in strategy. Lots of maths game theory, and stats!