If you are up at the Edinburgh Festival, I shall be talking about the strange science of sleep and dreaming on August 14th. Do come along, much fun will be had. Details here.

On Friday I posted this puzzle. Two cyclists are racing around a circular track. Jon can race around the track in 6 minutes and Eric can race round in 4 minutes. They start the race on opposite sides of the track. How many minutes will it take before Eric overtakes Jon?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

12 minutes – but how did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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I got six minutes. Not sure where I’ve gone wrong.

I think you were right – in six mins, the quick one does 1.5 laps and the slow one one lap.

Not sure you have gone wrong , Anne, but willing to hear what others say.

I suppose in theory it should be 6 minutes and a tiny bit, as in 6 minutes (I reckon) Eric has caught Jon up and if the race finished then it would be a dead heat. To actually overtake needs a little bit more.

You haven’t. In 12 minutes Jon does 2 laps while Eric does 3. But Eric only needs to gain half a lap, so 6 minutes is correct.

The visitors to this blog are never satisfied.

First no one must post an answer until after the weekend.

Second only answers arrived at by posters are allowed.Not even the puzzle creator is allowed to post a reply that the blog denizens disagree with.

As with so many things the true answer depends on how you read the question,

If it means “how long to reach the same spot” then the answer is three minutes as that is how long it takes to reach the other persons starting point.

If it means “how long so they have both moved the same distance” then then answer is obviously nine minutes.

Other possibilities were discussed (against the wishes of many blog posters) over the weekend and again below.

What this all tells us is the primacy of intuition over literalistic mathematical symbol manipulation in solving these puzzles. These puzzles are so much like real life.

If we can all simply be more intuitive and at one with the world around us then the questions of “wrong” answers will no longer apply.

Please, those of you obsessed with mathematics awake to a wider world. The “many answers” world of quantum cosmology is a far better place for us

@Barry Goddard. You know I love you and your stand-up routine, Barry, but I’m kind of disappointed with this response. It lacks your comic sparkle. It lacks your *cosmic* sparkle, come to that. I hope you don’t leave us with only this today.

The most reasonable and sensible post you have ever made, Barry.

Eddie

Maybe it was not Barry who posted this at all, but one of the imposters he has complained about in the past?

LMFC

@The. I thought about this myself, funnily enough — the impersonation thing. It certainly hasn’t got Barry’s usual joie de vivre… but to be fair, we’re all entitled to a bad Monday from time to time. Personally, I’ve got a head like a toyshop right now, but that’s from drinking too much wine in the garden.

I think Mr Wiseman made the mistake. The setup he describes sound like individual pursuit. In this discipline a rider wins when he or she overtakes his or her opponent on the track, not when one driver laps the other.

I cannot be sure, since track racing is an old sport, with many variants, and I have only some acquaintance with the sport. But if the point was lapping the opponent, starting at different places would make the race confusing for the audience, and lessen the excitement. If they start at the same position, the rider in the lead is always handicapped by the fact that the opponent benefits from the drag, making it necessary to break away from the opponent before lapping.

In 6 minutes, the 4 minute guy will have done 1.5 laps putting him at the point the 6 minute guy started. In 6 minutes, the 6 minute guy will be back where he started, so they meet in 6 minutes.

All answers/attempts are “allowed”, even your childish humerlous dribble, but that does not change what is correct and what is foolishness.

The puzzle creator has not been prevented from posting a reply, it’s simply that the people who are more switched on recognise that what it says is incorrect. I very much doubt that it is Professor Wiseman who actually puts the posts up. Perhaps it is someone in a bear costume.

I’m sure that in one of the many weird versions of reality that Barry’s consciousness dwells in the answer to this problem is 12 minutes but in the real world, a few minutes simple thought will show that the answer to the problem as originally stated is indeed 6 minutes.

You’re right. It’s 6 minutes.

I found the easiest was to consider angular velocity. Jon cycles at 60 degrees per minute (360/6) whilst Eric cycles at 90 degrees per minute (360/4). So Eric catches Jon at 30 degrees per minute. As they started 180 degrees apart Jon will reach Eric after 6 minutes (180/30).

angular velocity – that was neat, I like it

@Steve – yes, that’s an insightful method. There are many valid approaches, and none is necessarily better than another, although Barry has already demonstrated to us all that pulling answers out of your arse doesn’t work.

I think the teaser above was not copied correctly from the original source.

It would take 6 minutes for Eric to overtake Jon.

If the original wording of ‘ How many minutes will it take before Eric LAPS Jon?’

Then the answer would be 12 minutes, as in racing to lap someone you need to be a whole lap infront of them.

I’ve noticed that Barry seems to have difficulty when circles are involved. He confused diameter and radius in the puzzle involving the LP record, did the same with the rope around the Earth, and again comes up with an answer which is twice as great as it should be for this puzzle.

You are right, Anne. 12 minutes is the answer to a different question, namely: If Jon and Eric start out together, when are they next together? Note that they would be opposite each other halfway through–after six minutes–which is the position they start from in the Friday puzzle. So they have 6 minutes to go from this position.

This may depend on the definition of overtaking.

After 6 minutes Eric and Jon can shake hands, as they are on the same spot. After 12 minutes Eric has completed 3 rounds, while Jon completed only two.

@Roland. I take your point but I hope it’s not this answer. It seems like a cheat!

I agree Anne. Surely it’s 6 minutes?

I drew a diagram consisting of two circles. One dived by 4 pieces the other divided by 6 pieces. I also got 6minutes. What have i done wrong?

Nothing. It’s Richard that is wrong.

I got 6 minutes too. Jon is ahead Eric with a half lap at the start because they start at the opposite sides of the circle. The difference between them decreases by 1/4 – 1/6 = 1/12 part of the lap. To the overtake, the difference must decrease by 1/2 = 6 * 1/12. So it’s 6 minutes.

Definitely takes 6 minutes for Eric to catch up to Jon if they start on opposite sides of the track. If they both started on the same spot, then Eric would lap Jon after 12 minutes.

Definitely takes 6 minutes for Eric to catch up with and overtake Jon if they start on opposite sides of the track. If they both started at the same point, Eric would lap Jon after 12 minutes.

Somehow managed to post that twice, one anonymously. Many apologies!

@Colonel Wiseman. You have a mutiny on your hands, sir!

Will this be another ’22’ moment?

22/7 = 2

actually, it’s 22/7 = pi and 22 = 11

I agree with A.E., after 6 minutes Jon will have done a lap, but Eric will have done a lap and a half, therefore will have reached Jon as he started half a lap behind. If one considers 12 minutes, Jon will have done 2 laps and Eric 3 – so Eric is 1/2 a lap ahead. I may have missed something though.

A mutiny indeed. We need our captain to tell us what went wrong.

Definitely 6 minutes.

The answer given is wrong – no doubt we can expect the post from Friday and the post from today to both be silently edited in the next few hours. Perhaps is will change from “overtake” to “lap”, or perhaps it will change from 12 to 6. The fun is in finding out which!

@Dan A. I was just thinking the same thing! The question posted is going to change at some point.

For future reference, as of Monday morning the question still reads:

“Here is this weeks puzzle. Two cyclists are racing around a circular track. Jon can race around the track in 6 minutes and Eric can race round in 4 minutes. They start the race on opposite sides of the track. How many minutes will it take before Eric overtakes Jon?”

We’ll see if/when it changes.

Definitely 6 minutes.

Lets assume the track is 300 meters long, so Eric is 150m behind Jon. (about right for a velodrome)

Jon travels the 300 meter lap in 6 mins, so a speed of 50 m/min. (very slow!)

Eric travels the 300 meter lap in 4 mins, so a speed of 75 m/min. (still very slow)

This means Eric gains 25 m every minute. So a gap of 150 m, takes 6 minutes to close.

It doesn’t matter what the length of the track is, let it be L.

Jon has head start of L/2 and travels at L/6

Eric travels at L/4

Eric gains (L/4) – (L/6) m per minute, this equals L/12.

To close a gap of L/2m at a speed of L/12 m/min takes 6 minutes.

Lets assume the track is 300 meters long, so Eric is 150m behind Jon. (about right for a velodrome)

Jon travels the 300 meter lap in 6 mins, so a speed of 50 m/min. (very slow!)

Eric travels the 300 meter lap in 4 mins, so a speed of 75 m/min. (still very slow)

This means Eric gains 25 m every minute. So a gap of 150 m, takes 6 minutes to close.

It doesn’t matter what the length of the track is, let it be L.

Jon has head start of L/2 and travels at L/6

Eric travels at L/4

Eric gains (L/4) – (L/6) m per minute, this equals L/12.

To close a gap of L/2m at a speed of L/12 m/min takes 6 minutes.

Richard is quite simply wrong on this one.

Agreed, Paul H. It’s six minutes.

Slow is putting it mildly. Even a moderately slow walker will travel at about 80m per minute. The track would have to be about 2km in circumference for even a very modest cyclist to take 6 minutes to circumnavigate it. Even the 4 minute cyclist would only be doing 30km/hr on such a track. (The world record for the individual pursuit would be equivalent to circulating such a track in fractionally over 2 minutes).

I’m not quite sure where Richard would find a circular track over 600m in diameter.

@Steve, it’s not circular but the distance is about right – en.wikipedia.org/wiki/New_Hampshire_Motor_Speedway

Apologies if this duplicates, the comments system seems to be having issues (overloaded maybe!).

Lets assume a simple example where the track is 300 meters long, so Eric is 150 m behind Jon. (about right for a velodrome).

Jon travels the 300 meter lap in 6 mins, so a speed of 50 m/min. (very slow!)

Eric travels the 300 meter lap in 4 mins, so a speed of 75 m/min. (still very slow)

This means Eric gains 25 m every minute. So a gap of 150 m, takes 6 minutes to close.

It doesn’t matter what the length of the track is, let it be L.

Jon has head start of L/2 and travels at L/6

Eric travels at L/4

Eric gains (L/4) – (L/6) m per minute, this equals L/12 m/min.

To close a gap of L/2 m at a speed of L/12 m/min takes (L/2)/(L/12) = 6 minutes.

Richard is quite simply wrong on this one.

Paul, any chance you could post the whole thing again? I seem to have missed the first three.

Apologies, I got an error message saying that my comment could not be posted due to maintenance. Then they all appeared a while later.

Barry, what do you think? I note you’re not shedding any light yet.

Barry has commented above, although I’m not sure he’s shedding any light!

I think Barry sounds rather tired and bored this morning. I was looking forward to his take, as I do every Monday. I’ve grown very fond of him over the months. Today’s reply was disappointing, to be honest.

Goddard very rarely sheds any light on anything and when he does it is but the light from a tallow candle in the midday sun.

His sole worth is to sporadically and briefly entertain those who are easily pleased.

@Starman. If that’s an insult aimed at me, you’re wasting your breath, wee man.

@Mr Anagram – I hadn’t even noticed you, wee wee man.

@Starman. Mine was the comment directly above yours. You’ll be able to identify the similarities in the spelling of the first and second names.

Good use of ‘wee wee’. You must be very proud of that razor-sharp wit of yours.

@Mr Granama, thank you for taking time out of your busy schedule to reply.

I made you a card: tinypic.com/r/swftk5/8

You’re very brave, wee man. Well done. You clearly put a lot of thought into that. Keep working on amusing variations of my name while you’re at it. I bow to your superior intellect.

6 minutes is the correct answer and Professor Wiseman is smiling now

I felt a bit panicky when I saw Richard’s answer. I’ll stick with 6 after reading the comments though.

Wiseman, you are an idiot

If the track length is x units in length

Jon’s speed is x/6 units/min, then in 12 mins he would have travelled 2x units

Eric’s speed is x/4 units/min, then in 12 mins he would have travelled 3x units

Likewise in 6 mins:

Jon would have travelled x units and Eric would have travelled 1.5x units.

If Eric started on opposite sides of the track then that is 0.5x units away, thus it takes 6 mins to meet Jon and just about to overtake.

If overtaking implies that Eric meets Jon for a 2nd time, then by inference that would take 18 mins

Mathematically the exact time of overtaking cannot be defined as a definite. It’s the old hare and the tortoise conundrum.

Maybe they stopped for a tea break, or to adjust their go-pros or they were inebriated and kept on falling off their bikes.

the only way 12 minutes works is if we are talking completed laps, at 12 minutes E has 3 full laps J has 2, both are at their original start point, however this is inferred by the answer, not the question

Yeah, setting Eric’s start position at position 0, and Jon’s at 1/2 track, the formula for their traveled distance measured in track lengths at time t minutes is:

Jon(t)=1/6*t+1/2

Eric(t)=1/4*t

The intersect of the two lines will be their first meeting point.

Which is

1/6t+1/2=1/4t => 2/3t+2=t => 2=1/3t => 6=t

6 Minutes

Now Eric finishes a lap when Eric(t) is integer, and Jon does it when Jon(t) has a fractional remainder of 1/2.

So Eric will lap Jon the first time when the Eric(t)-Jon(t)=1/2

So

1/4t-1/6t-1/2=1/2 => 1/4t-1/6t=1 =>1/12t=1 => t=12

This is an effect of me choosing the formula for Jon’s distance as if he started at the distance of 1/2 track. We could have just looked for driven distances in which case the formula for their driven distance would be simply 1/6*t and 1/4*t, and the lapping would happen when 1/4*t-1/6*t=1, which is what we found above to be at 12 minutes, This makes sense since lapping does not depend on the riders position on the track, but rather the distance traveled..

In fact if overtaking is meant to be when the riders are on the same position on the track, then we have to use the fractional part of my formulas to decide when they are in the same position, which would make the formulas cyclical.

Jon/t)=frac(1/6*t+1/2)

Eric(t)=frac(1/4t)

So overtaking would happen whenever

frac(1/6*t+1/2)=frac(1/4t)

But as were shown above, the first this happens is at time t=6 minutes

and Eric laps Jon every 12 minutes, so Eric would overtake Jon for t in (6,18,30,42,…,TOs) Where is the maximum number in that series where TOs <= the total time of the race, and he will lap him for t in (12,24,..,TLs), again where TLs is the maximum number of the series where TLs <= the total time of the race.

Given that the setup sounds as the track discipline of individual pursuit, the race could end at t=6 minutes when Erics catches Jon, or continue until Erik has finished the designated number of laps, depending on whether only the win counts or the race time must be used either for Record attempt, or to place Eric in the further competition.

I agree with the math, however not with your closing statement as John would be traveling at 4 km per hour assuming a 400mtr track or 2.5kmh and at best Eric would be traveling at 6kmph on the 400mtr track, at these speeds both riders would probably slide off the track on the banking, unless they were riding on the cote d’azur, but for a pursuit race this will normally be blocked, notwithstanding, the crowd would probably get quite angry and start throwing rotten fruit, making the track even more difficult ……

What were we talking about?…..

I got 6 min (to meat, just ober 6 min to overtake) too.

If they start facing opposite directions, they will meet in 72 seconds!

Pat, I love your work but you’re assuming that both riders are impelling their velocipedes in a forward direction. Plus, I think that the use of the word “overtakes” indicates that both riders are travelling in the same direction (albeit, a circular one). Back to The Pedantic School (soon to be renamed The School of Pedantry) for you.

if they started FACING the opposite direction,ie; One faced north, the other south, assuming that both riders ARE impelling their velocipedes in a forward direction, being that they are riding a circular track, then they will be both moving on the same rotational path, (assuming classical physics before anyone chimes in with some Heisenberg’s BS) still 6 or 12 minutes depending on who you believe

Yep, six minutes, not twelve. Careless, Richard!

Perhaps Richard knows less of track pursuit cycling than his readership & meant that Jon & Eric started from left & right sides of the track at the same point. Then the overtake would be at about 12 minutes.

More likely just a cock-up though…

I would say just over six minutes. At six minutes, they would be alongside each other.

To make the twelve minutes answer work, I think that the question should have had them both leaving from the same part of the track, going in the same direction and not from opposite sides of the track.

But “opposite sides of the track” can be defined in different ways, no?

The more central or inner part of the track could be considered to be opposite to the outer part of the track. The riders are on the same radial line(?) and the time for one to overtake the other would be 12 minutes.

Maybe the track is in the shape of a Möbius strip (and the bikes have special wheels that keep them on the track regardless of gravity). That way, “opposite sides of the track” means they’re starting at the same point–one rider right side up, and the other upside down on the other side (or one leaning 90 degrees right and the other leaning 90 degrees left, etc.). Now the answer would indeed be 12 minutes. 🙂

I forgot to add that “round the track” would mean a 360-degree trip around the center, which is only half the distance needed to return to your starting point (same place, same side).

@Ken. Love the Mobius strip idea! That’s more than thinking outside the box; that’s thinking outside the box *factory*.

One problem with your very ingenuous answer, and that is we are told it takes 4 and 6 minutes respectively to go round the track, which surely means getting back to the place you started from. (Which means the same way up of course – you’ve already defined being the other way up at the same point as being on opposite sides of the track).

So I’m afraid it doesn’t work. It’s still 6 minutes

Excellent. I wish I had said that!

FWIW I have a bottle-opener in the shape of a Klein bottle (3D version of the Möbius strip). Well, its 3D projection anyway.

wow, richard posted a perfectly non-controversial puzzle with a straightforward solution on friday, only to ruin it by posting the wrong answer today!

ctj: I think he’s having fun with us. Answers are much more interesting this way.

I’m sure Richard throws the occasional boo-boo just to collect information on how we dispute it. We are incapable of ignoring elephants!

One problem with your very ingenuous answer, and that is we are told it takes 4 and 6 minutes respectively to go round the track, which surely means getting back to the place you started from. (Which means the same way up of course – you’ve already defined being the other way up at the same point as being on opposite sides of the track).

So I’m afraid it doesn’t work. It’s still 6 minutes

oops – comment went on the wrong thread for some reason…

But in the next post I “fixed” the definition of “round the track” to mean a 360-degree trip, which isn’t totally unreasonable. Most people would agree that 360 degrees around a circular track would qualify as “round the track”, no? I think my answer works. 🙂

I posted my reply before I saw your caveat. However, I think you are trying to have things both ways. Round a track surely means a complete lap and not being on the opposite side (using your definition). I think you are trying to have your cake and eat it.

@Steve: Au contraire. I don’t think you’d argue if I claimed that “taking a lap” around a Möbius strip would result in your ending up inverted from your original position. “On the other side” so to speak (even though, technically, there’s only one side). It takes two laps to get back to your exact starting position. 🙂 OK, I admit I’m stretching it a bit. But it was fun to think about.

I’m afraid I would argue with the idea that taking a lap round a mobius trip means being inverted at the same point in time. Like any other circuit, completing a lap means ending up at the point you commenced it. Often that’s marked with a line (although a virtual one will do if we define the start of a lap arbitrarily – it makes no difference). You will pass that line only once per lap, and it cannot be on both sides of the strip. There’s no sens in which a line drawn on the surface of a mobius strip appears on the other.

The usual definition of completing a lap involves returning to your starting point…

Yes, and Ken has already defined the corresponding “inverted” point as being on the opposite side of the track, so I don’t see how something can simultaneously be “on the opposite side” and “at the same point” of the lap.

Ken Haley’s answer was great. Unfortunately, it seems flawed. A Mobius strip is one-sided. There is no other side.

Technically true. It was an intuitive reference to the point on the part of the strip that would be “the other side” if the strip wasn’t twisted to make it a Möbius strip. But that’s only a semantic problem. Doesn’t affect the proposed solution.

Eric over takes John at the very beginning of the race. His time puts him faster than John, so Eric is always ahead. Eric comes even with John on the track at 6 minutes. For Eric to lap John takes another 6 minutes, by this time Eric has gone 3 laps while John has traveled only 2 laps. I guess the meaning of overtakes is dependent on whether you are looking at time or distance. You win a race by overtaking your opponent in time not distance, so I say Eric overtakes John from the start.

Ok, I got -6 minutes, something must have gone wrong.

I think we should kidnap Richard Wiseman and squirt custard into his ears until he confesses.

I already tried that. He said that he didn’t know what I was shouting about and that someone in a bear costume has been posting under his name. I probably would have got more information out of him but I got distracted by how tasty his ears were.

I did the same too, together with handfulls of jelly, whipped cream and spongecake. Richard said that he could not hear me because he was a trifle deaf.

I expect you’ll get your just desserts for using that pun.

You’ll get taken into custardy.

That’s pudding it lightly.

No more gorillas – they’ve been replaced by meringue utans

Here’s what I hope is a simple if slightly long-winded way of assessing the problem.

We don’t need to know the length of the circuit or even use any algebraic substitutions for its distance.

All we need to know is a) where Jon and Eric are relative to each other at the start and b) that Jon takes one and a half times as long to complete a full circuit as Eric (four minutes versus six minutes).

The latter means that Eric runs one and a half times as quickly as Jon.

If you imagine Jon running but his position in space being fixed, with the circuit rotating beneath him in the opposite direction and at the same speed as him, Eric’s speed is then the equivalent of his own speed minus Jon’s, i.e. one and a half times Jon’s speed minus Jon’s speed, which equals half of Jon’s speed.

This means that if Eric had to complete a full circuit relative to a fixed point, it would take him twice the time Jon would take if the track weren’t rotating, i.e. 12 minutes.

But because Jon starts on the other side of the track, Eric only has to cover half the distance, so it takes him six minutes.

For the puzzle to be anything other than completely unsolvable, we have to assume that “overtake” means “draw level with” and we have to assume that they’re about the same size as each other, so their centres of gravity or any other set points we choose – shall we say their noses? – coincide. Any other interpretation introduces pointless unknowns and associated distractions.

I am new here and don’t want to upset anyone by posting the wrong thing in the wrong place at the wrong time but I have a simple answer that (I think) makes sense of Richard’s 12 minutes.

Do I post here?

Please ignore.

Over 100 posts. Is this a record?

Not yet… richardwiseman.wordpress.com/2014/07/21/answer-to-the-friday-puzzle-267

but when it is, how far will the needle move?

I don’t know about the rest of you, but I miss Richard’s normal Tuesday, Wednesday, Thursday blog-posts. Have those stopped for good? Hope not…

Just type ‘optical illision’ into google images if you’re missing the non-puzzle related blogposts.

A reminder that Richard is doing hos sleep and dream workshop tonight (full details at the top of this page).

It will help many of the regular contributers to this website to know when they are awake and when they are dreaming. Distinguishing the two are clearly hard for some who post here.

That is evident in the way they cling to outmoded ways of problem solving while dismissive of better and more intuitive ways.

Astrology does not predict the future. Yet the principles behind it easily allow any one of us to approach the puzzles of this website with confidence. No less too the puzzles of Life itself.

I am hopeful that someone will raise this with Mr Wiseman himself at the talk tonight. I would ask myself were I to be there. I am sure a fascinating discussion would ensue unless the likes of @simon and @steve heckled down the dialogue.

why isn’t there a new friday puzzle? me sad…… 😦

He’s busy this morning at the local velodrome

Or still trying to work out what’s wrong with his answer to last week’s puzzle.

Or maybe busy deciphering one of Barry Goddard’s comments.

I followed a slightly different route to everyone else. I created a spreadsheet, with the lap times for both riders and the length of the track as constants, then worked out the speed of each rider.

I then created a list with whole minutes in column A, Jon in column B (starting at 0) and Eric in column B (starting at [track length]/2). To account for the circular track, if ((distance travelled) > (track length), (distance) – (track length), (distance)).

Regardless of what track length I chose, they both met at the 6 minute mark, so technically Eric would pull ahead by 6 minutes 1 second. If you wanted to be extra pedantic, you could take into account the average length of a bicycle and calculate how long it would take Eric to pull a bike’s length ahead of Jon…

The length of the bike won’t matter as both riders should have started with the front tyre just behind their respective start lines. As a rider is (usually) deemed to have passed another when the front tyre of one is ahead of the other (that’s how races are decided), the time difference is fractional.

I suppose the one doing the passing will be just traveling fractionally further at the point where he overtake if he does it on the outside.

Well, in the absence of a question from Richard today, here is one from me

Someone picks, at their will, two cards from a deck of cards. The cards have

different numbers, so one is higher than the other. (In other words, the person

picks two distinct numbers in the inclusive range 1 through 13.) The cards are

placed face down on a table in front of you. You get to choose one of the cards

and turn it face up. Now, you will select one of the two cards (one of whose

face you can see, the other one you can’t). If you select the highest card, you

win. Design a card-selection strategy for which your chance of winning is

strictly greater than 50%.

Hey, good on you for showing some initiative!

Hmm, instinctively I would keep any turn-up higher than a 7 and swap if the turn-up is less than 7. However, this assumes that the person who initially picks the two cards does so randomly. Once the initial picker detects that I’m using 7 as my swapping point he/she would just choose two cards less than 7 (or greater) each time and my odds of winning would be no better than chance (50%).

Obviously I should always keep a King and swap an Ace but other than that, all that I know for sure, is that I have a 50% chance of turning up the higher card. What I need to do is to randomise the swapping point that I use so that the initial picker is not able to develop a strategy to thwart me.

If I randomly choose a number from 1 to 12 (say, with a dodecahedral die) then, if the card which I turned up has a higher value than my random number, I keep the card, otherwise (if it is less than OR equal) I swap it. If both of the cards are less than/equal the random number, or both are a greater, then my chances of selecting the higher card are still only 50% – nothing we can do about that.

However, if one card is less than/equal and one is greater (ie the swapping point is “between” them), I will always win keeping a card which is higher in value and swapping everything else. I think that this will give me a slight edge in the game. Someone else can work out the odds.

The odds are 22/7.

This is a very easy puzzle.

If I reveal the ace I turn over the other and win.If I reveal the king I keep it and win.

Thus I have 2 in 13 chances of winning automatically “out of the box” so to speak.

In the other 11 choices it does not matter. I have a 50% chance whatever I do.

Therefore I win 6.5 + 2 = 8.5 out of 13 times no matter what the opponent does.

Obviously I could labour to “prove” that with 13 separate similtaneous equations. Yet why bother? The solution is intuitive and obvious and a good example of how to approach puzzles

Er, nope. All I do is make sure that I never give you an A or a K and you’re back to pure chance. Like whether you get a puzzle right or wrong. Simon’s solution looks like it will do the trick.

Very good puzzle. I have no clue …

I await confidently for Richard to give the answer. Then the doubters will be silenced.

I love the initiative but am less keen on the puzzle. Sorry to be blunt. Here’s one I gave to Richard last year. He was polite enough to thank me for it but he didn’t publish it so here you go.

What are the next three letters in this sequence?

OUR

FAT

HER

WHO

ART

I’m speechless – well nearly so

The next three letter word in that sequence is

AVE

INH its the Lords Prayer. Poor puzzle

@Anonymous. INH isn’t a word. It’s not as straightforward as that.

Alison, the question asks for “the next three letters”. It does not have to be a word

September

Good answer, but you have not worked out the year, which is 1965 BC

Hi Richard & Co. Awhile back I posted a magic comment, but there is a spelling mistake, how can I correct it? thanks Fonda Zenofon

It’s fixed now.

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