## “Higher or Lower” – Part II

In Part I we calculated the probability of winning a single-suit version of the Higher of Lower card game. **The objective of this post is to find the probability of winning the full, unsimplified, 4-suit version of the game** played in pubs across the land.

I attended the pub quiz again this week out of curiosity and because I seem to have forged an attachment to this game having thought intensively about it for a fortnight. The manager of the bar was annoyed that I didn’t pay a quid to enter the competition, but I had to retain my dignity in the face of my own damning report on the game’s adversity. What they didn’t mention whilst rallying for entries though was their having changed the rules slightly to make it easier. I don’t know if they’d been reading my blog, doing their own calculations or simply applying common sense* but they announced that the player would have 3 opportunities to have a card changed, which they could use at will. **And to my surprise and delight, somebody won!** It was the first person to be picked to play and it was also her birthday. I love stuff like that.

*

Everyone has a sense of probability and some people have a real aptitude for this; it’s a useful skill in games like poker in which “hunches”, based on sub-conscious intuitive readings, allow a player to make good decisions without getting their calculator out.

So in this post we need to calculate the odds of winning 4-suit Higher or Lower and then make some allowances for these three opportunities to change a card.

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# Analysis

Let’s have a think about how the mechanics of the 4-suit version of this game differ from those of the single-suit version. In the single-suit version, if a particular value is revealed then that card cannot possibly turn up again – it’s already been used – but in the 4-suit game a particular value can be revealed up to 4 times. Importantly, as stated in the rules in Part I, an immediately repeated card constitutes loss of game (because it is neither higher nor lower than the previous card).

Example: Say the first card was a Jack and you said lower. In the single-suit game you’d lose if the next card were a King or Queen. However in the 4-suit version you’d lose if the next card were a King, Queen *OR a Jack*. Consequently the probability of passing each card is slightly lower than for the single-suit game because of this added possibility of an immediately repeated card. Hence the chance of winning the game overall will be even less than the 10.6% calculated in the last post.

**But by how much?**

I used similar methods to those described in detail in Part I to calculate exact probabilities for reaching the 2nd and 3rd cards, but for the 4th card, due to the added complication of “repeated cards”, I had to somewhat painstakingly work out the individual probabilities for all possible combinations of 3 cards and add them all up (that’s 13×13×13 = 2197 values to calculate, but there were patterns and “bounces” and symmetries again which made this practical and fairly speedy, and many of the combinations weren’t valid – hence probability zero – because they would have caused loss of game). The following graph shows these results:

And now for a concession: after many hours in pursuit, and even though I could describe qualitatively all the patterns and their origins, I simply found it too difficult to combine the results for reaching the 2nd, 3rd and 4th cards in the 4-suit game into a ɡeneral formula in the same way that it was done for the single-suit game in the last post. This is because the possibility of repeated cards makes it extremely complicated. ɪtʼs certainly possible thouɡh and ɪʼll have another crack at it in the not-too-distant future. **Fortunately however, in the meantime we can use everything we know to make a reliable estimate of the probability of winning 4-suit Higher or Lower.**

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## Estimating the probability of winning 4-suit Higher or Lower

Graph 4 includes exponential curves of best fit and those curves’ equations. To check the accuracy of the curve for the single-suit game, we can plug the number 8 into its equation: this gives 10.4% which is close enough to the 10.6% it should be. Plugging 8 into the equation for the curve for the 4-suit game gives 6.4%, which is approximately a 1-in-15.7 chance of winning. This is in line with what we should expect.

But we could do with corroborating this estimate, so **let’s also get there by another, more interesting route…**

At the bottom of the Part I post, Graph 3 shows the probability of reaching each card in the single-suit game (the same data used for the single-suit line in Graph 4) but also with a roughly horizontal line showing the average probability of reaching each card *once the previous card had already been reached*. We’ll take this idea a step further in pursuit of corroborating our estimate of the probability of winning 4-suit Higher or Lower.

The blue line in Graph 5i is the same data as in Graph 3 for the probability of reaching each card in the single-suit game *once the previous card has been reached*. We described this line before as being “approximately horizontal” but, as you can see on this vertically stretched graph, it’s anything but horizontal, rather it oscillates but increases on the whole as the game progresses. This is not surprising, since as the game progresses there are fewer and fewer cards remaining and the chance of making a correct guess will increase.

Does the same happen in the 4-suit game? That is, **does the chance of making a correct guess gradually increase as the game progresses?** Again, the answer is YES, but it only makes a weeny bit of difference. But we’ll factor this in because it’ll make the estimate more reliable. The green curve in Graph 5 is a quadratic trendline (curve of best fit) with equation 0.0011x^{2}-0.0095x+0.7659. In 4-suit Higher or Lower we’d expect the rate of this increase to approximately quarter, because there are 4 times the number of cards (52 instead of 13).

Jumping briefly back to Graph 4, if we sample the 4-suit trendline at each integer on the horizontal axis, the results form **a geometric sequence with common ratio 0.709**. This is the estimated average probability, in the 4-suit game, of correctly guessing a card once the previous card has been reached. Compare this value with that for the single-suit trendline: 0.754 (and with the “real” average for the single-suit game calculated from the actual results from the last post, which give 0.756 as the average).

So we can take the equation of the green curve, flatten it out by a quarter with a little calculus, and position it vertically with the 0.709 average in mind, and we have an estimated trendline for the 4-suit game: 0.000275x^{2}-0.002375x+0.7122

Now we simply sample the purple line in Graph 5ii at 2,3,4,5,6,7,8 and 9 and multiply all the resultant figures together (because we’re effectively calculating the probability that the player reaches the 2nd card AND reaches the 3rd card AND reaches the 4th card AND etc… up to the 9th card).

Our survey says?

## 6.4% (EXACTLY WHAT WE CALCULATED EARLIER BY A COMPLETELY DIFFERENT METHOD!)

Consequently we can be confident that **a reliable and fairly accurate estimate of the probability of winning 4-suit Higher or Lower is 1-in-15.7**

…which goes some way to explaining why, years ago, I sat and watched people play this game on repeat for about 45 minutes before somebody was successful.

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## What about the “Changing a Card” rule?

We certainly need to discuss this because it improves the odds radically. We know that sometimes this game can leave us with a tough decision, for example if a 7 comes up making it 50/50 for the next card to be higher or lower. That’s when you’d use up one of your “change card” opportunities. But because a 7 will only turn up approximately once in every 13 cards, I’d also suggest using up a “change card” opportunity if a 6 or 8 is revealed. It’s likely, after chanɡinɡ that card, you’ll be left with a value closer to one of the higher or lower extremes, increasing the probability of a correct guess on the next card significantly.

Ignoring this rule for a moment, if a 6 or 8 comes up you have roughly a 7/12 chance of being correct on the next card; and if a 7 comes up it’s roughly 6/12. The average of 7/12, 6/12 and 7/12 is 5/9 or 0.556. That’s the average chance of being correct on the next card if you get a 6, 7 or 8. But remember that figure we calculated a few minutes ago of 0.709? That was an estimate of the average probability of making a correct guess at any time in the game. Therefore, on average, if you use up one of these opportunities on a particular go, you’ll improve your chances by a factor of 0.709/0.556 which is equal to about 1.28. So we take that figure of 15.7 (from the calculated estimate of “1-in-15.7”) and divide it by 1.28 three times, (*three *because youʼre allowed up to three opportunities to change a card) **giving a final estimated probability of success of approximately 1-in-7.5 or 13.4%**.

So birthday girl got a bit of luck to walk away with the £40! But since about 30 people entered the competition and they would only have allowed up to 2 people to play the game (before rolling the money over to next week), **I’d still only have entered if there’d been £120 or more in the pot***. AND THAT’S THE CASE REGARDLESS OF WHETHER SOMEBODY WON THE COMPETITION OR NOT!! It’s easy to assume you should regret not entering a competition when you see someone win, but technically if it wasn’t *worth* entering, you shouldn’t have entered. Inevitably, some ‘told-you-so’ turned to me and said something like, “You said the probability was 1-in-16 but she won!” because to that person, seeing someone win was “proof” the calculation was irrelevant. But independent events that occurred in the past prove nothing about the likelihood of future recurrence or worthiness. At the risk of being radical, it could be said that someone who plays Lotto and wins is just a* lucky fool*, because if you played Lotto every week for eternity, you’d lose far more than you’d win back.

Probability is probability, regardless of the result.

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* *Calculated by taking the reciprocal of the following:
P(I am first person to be picked)×P(I win) + P(Someone else is picked first)×P(They lose)×P(I am picked next)×P(I win)
=(1/30×0.134) + (29/30×(1-0.134)×1/29×0.134)
=1/120
Hence the competition is only worth entering if the pot contains at least 120 times the entry fee. (ʙecause then, if you played on repeat for eternity, youʼd win more money than youʼd lose.)*

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### Recommended Reading

1. Poker Pot Odds – Deciding whether or not to play http://www.thepokerbank.com/strategy/mathematics/pot-odds/

Wow, interesting! Has anyone else come across the same thing compared to this?

Hello from Russia!

Can I quote a post in your blog with the link to you?

hi, sorry for the late reply! of course you can, I’d be honoured 🙂

Yes the card could be thought of as being put back into the deck and reshuffled. ‘replacement’ I think it’s referred to as?

I said infinite deck as the game I played was electronic. There was no mention of number of decks etc so I have just assumed the program picks any random card independently of the proceeding card. I have not played it long enough to notice a repeated card. I hope you have fun with it!

Hi. This article made a good read.

I came across it when looking for a similar game.

Are you interested in another challenge?

In this game you pay 8 tokens to play.

You are presented with a card from an infinite deck.

Higher or lower?

If you get it right you get +6 tokens, a second card presented to you

and the chance to “let it ride, higher or lower?”

It then repeats like this until you either lose or take your winnings.

you then must pay 8 tokens to play again.

The probability of winning each round is easily calculated.

Whether or not it’s the best decision to take your winnings under each

circumstance is what I have been trying to calculate.

My instinct has told me to take the winnings whenever I am risking 18

tokens and receive something close to 7 and to only play on when risking

24 tokens if a 2 or queen is returned. It has been my undoing to play on

risking 30 tokens for a queen before.

In this game if the same card number is drawn in succession you do not

lose, you are asked to pick again. (effectively you can assume that the

same card never appears twice in succession in the maths).

If you would like to take on this challenge, I would be very interested

in your findings.

The game just doesn’t feel like the house has the edge…

-Scott

Very interesting problem that Scott, thanks. By an “infinite deck” I presume you mean a normal deck but, effectively, that between each revealing of a card the previous card is returned to the deck and the deck shuffled? I’ll have a go at this at some point! Mick

Hi Micky,

Well, i am interested in Maths and Physics but i am not a mathematician, so i recommend you better refer to the Wiki Page for elaboration. I became kinda obsessed with the DA since reading about it in Stephen Baxter’s Book ‘Manifold : Time’.

Maybe you post your take on it on your intereting site when you have studied it properly, as it seems the opinions about it’s validity differ greatly.

Hi Malenfant,

I’ll have to study the Doomsday Argument properly as it’s not something I’ve been exposed to before. On first consideration though, the only similarity I can identify between it and this post is that they are both calculated predictions of probabilities. I’ll study further when I have a moment, but in the meantime perhaps you’d elaborate on your own thoughts! Thanks for the heads up. Micky

Hmm, is this a variation of the Doomsday Argument ?

“magine two urns, one containing 10 balls numbered 1 thru 10; and a second urn containing a million balls, numbered 1 thru 1,000,000. If you pull a ball at random from one of the urns and you obtain a “3”, the likelihood, according to Bayesian prior/posterior probabilities, is much higher that you have pulled from the 10-ball urn than the 1,000,000-ball urn.

http://en.wikipedia.org/wiki/Doomsday_argument

Very nice site!