On Friday I set this puzzle…….

John wants to toss a coin to make a random decision.

However, he only has a biased coin (that is, a coin that does not have a 50:50 chance of coming up heads or tails). Even worse, John doesn’t know the extent of the bias, and thus has no idea about the likelihood of obtaining a head or tail.

How can John toss the coin to make a 50:50 random decision?

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

Imagine John tosses the coin twice. The chances of getting heads-tails is the same as tails-heads, no matter what the bias of the coin. Thus, all he has to do is toss the coin twice, assigning a ‘yes’ to HT and a ‘no’ to TH. If the coin comes up TT or HH he ignores the trial and repeats the process until he gets TH or HT.

Did you solve it? Any other solutions?

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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If John had no idea about the likelihood of heads or tails, then the choice would have been random itself, so one toss would have sufficed? Seems like the ‘proper’ solution is over complicated.

How would you choose randomly without any random mechanism? Surely you don’t think that, given two choices, a given person has no bias in terms of which one gets assigned to heads.

It’s a good answer Simon, assuming that John did not know if it was heads or tails that was bias. The way I read the question, John knew the coin was bias and therefore which was more likely to come up, he just did not know the extent of the bias. But there is more than one way to interpret the question of course….

Yes, but we are told that John does not know, so the result based on his choice is still random, and 50/50.

Even though i like your solution, as it avoids the two tosses, i don`t think that John’s decision may be unbiased. I would expect that John may be unknowinlgy biased to assign e.g. a favorable alternative to heads and an unfavorable alternative to the number…

Without this “human factor” it would be my most perferred solution!

(I only found the solution with the two tosses…)

Nope, we are told that John doesn’t know the EXTENT of the bias…. he could know that the coin is biased to some degree, and also which side is favoured, just not the extent to which it is likely to come down on that side.

He can’t know which side is biased, as the question doesn’t rule out a negative bias. So one toss is fine.

Even if John knows the bias to be favored to one side (without knowing the extent of the bias), he can still overcome this. Ask someone else who does not know about the bias to assign the choices.

I like Simon’s solution better than the mathematical one.

@One Eyed Jack: Simon’s solution is a mathematical one. There are two mathematical schools of probability: frequentism (which treats probability as long-run frequency) and subjectivism (which treats probability as rational belief in light of available evidence). Richard’s answer is frequentist; Simon’s answer is subjectivist. But both are mathematically correct.

Simon: a 50/50 random decision should be repeatable, with the overall result of many trials approaching 50/50. John wouldn’t be able to achieve this except with the two-toss method of deciding.

So, to elaborate on why this solution doesn’t work: You seem to be assuming that, given two options A and B, you can assign A to heads with probability 1/2. How do you know that you can do this? Do you have a random number generator built into your brain somewhere? Better yet, if you can do this, why bother flipping the coin at all?

The truth is, we all impart some level of bias to processes we carry out. Certain possibilities are more likely to be assigned to heads. (I’m not an expert in Psychology, but I would guess that, for a yes/no question, most people are far more likely to assign the “yes” answer to heads.) With a fair coin, this level of bias doesn’t matter. We could always assign option A to heads, and we’d still get a 50-50 result because we’re getting unbiased randomness from the coin. The coin’s properties negate our bias. Here, we don’t have that assurance, so our bias is actually significant.

The probability theorist E.T. Jaynes once said, “Probabilities do not describe reality — only our information about reality.”

Since you have no info about the bias of the coin, just flip it one time and be done with it!

J: Even if you always choose heads (as I would), it is still random because you have no information to suggest that biased coins are more likely to be biased in favor of heads rather than tails.

Another way to think about it: If we were betting on the outcome of our first flip, is there any reason to think we should assign odds other than 1:1 for heads/tails? Obviously it only works for the first flip, after that we have new information.

I thought the same way as you. The choice itself is randomly. If a coin has 2 tails, but you dont know, it’s still randomly, because it’s 50:50 the answer you will give to tails.

Simon you are confusing ignorance with randomness

I guessed you could just attempt to throw it straight up in the air, If it lands in front of you, it’s heads, if behind it’s tails. There may be some throwing bias though.

Or make a mark on the edge and throw it. If the mark is in a northerly direction it’s heads, otherwise tails.

I agree with Simon.

Another way would be to toss it in the air, snatch it mid-air and slam it on the table.

My solution: Toss and guess if the head image on the coin will be mostly upright or mostly upside down (retoss if it comes down tails)

I had 4 solutions:

1) the original solution from Richard. Figured that it does not work in case of a 100% bias. It requires infinite tossing.

2) davebakedepotato’s coin-pointing solution: make an arrow on both sides of the coin, throw it on a flat floor, and the person to which the arrow points (closest) wins

3) Simon Taylor’s solution (after that one had been spoiled friday). But it will only work once or a couple of times. The bias will show up soon

4) Spin it on a table, smack it, and it may give an unbiased heads or tails.

Richard did say that only one decision was required!

It’s not wrong, I just explain the disadvantage.

I went for throw it on a chess board….

Nice

My solution: ask someone else to call wads or tails. Does need an extra person though.

My answer was exactly the same as the “official” answer given by Richard. I am therefore feeling smug because it seems that lots of other people had various “wrong” answers, and because the “correct” answer is not just being repeated by everybody as if it’s obvious. :p

He could draw a line and toss the coin vertically over the line and allocate H & T status to whether the coin land

s left or right of the line.

that was my solution

This is very likely to be non-random. I dare say (for example) a left handed thrower would end up with a H/T bias that was opposite to a right handed thrower. If you start adding new things in the room you might as well ignore the biased-coin and use a non-biased one that you find on the table!

In any case, you would need to do a statistically significant amount of test tosses, to confirm the randomness of this method: So you might instead just use those throws to measure the bias of the coin. At any rate, you’ll need much more than just a few tosses.

The official solution is incorrect.

If the coin is biased, so is the first transition, in other words, if P(H) > P(T), then P(H,T) > P(T,H), and vice versa.

To illustrate: imagine a large bias such as 1 T coming out only for every 100 H – that means H is likely to come up first, and hence the transition H to T is more likely.

i have been trying to get my head around the official soln and micheals comment is the only one so far that makes sense to me. if the two top spin results are head then head and head then tail richards solution must still be biased.

thank you micheal for putting in Mathematics what my intuition tells me. i still stand by my suggestion in the original puzzle thread. mr wiseman is a social scientist not a real scientist so he is prone to such mistakes as this.

“not a real scientist”! How rude!

Dharma – you’re welcome, but don’t be so harsh on Richard. Probabilities are hard for everyone, and even “real” scientists (let’s keep the quotes firmly on here) are easy to mislead — just do a search for “Monty Hall problem”.

Ok: 99% of the throws is H, 1% is T. Then the chance for getting H and then T = 0.99 * 0.01 = 0.0099. The chance for getting T and then H = 0.01 * 0.99 = 0.0099. That is exactly the same, so no bias!

“To illustrate: imagine a large bias such as 1 T coming out only for every 100 H – that means H is likely to come up first, and hence the transition H to T is more likely.”

Micheal: H is equally more likely to come up SECOND. There’s your error!

no, because (in your example) you still have a vastly higher probability to come up with HH after the first H, than a HT.

HH = 98.01%

HT = 00.99%

TH = 00.99%

TT = 00.01%

(Sorry, posted this below too)

In other words: when you’ve thrown an H, there is a 99% chance that the next will also be an H, so you’ll have to throw again. However, if you’ve thrown a T, there is a 99% change that the next will be an H, so you’ll have a desicion. So HT is evenly unlikely as TH, because in both scenario’s you’ve had the 1 out of a 100 events. TT is very unlikely as that happens 1 out of 10.000 times.

Completely agree with M, it’s fair. The important note here is that is done in trials of 2 coin throws at a time, completely ignoring any previous results. HT is exactly as likely as TH.

Nice explanation -M-

thank you micheal for your kind words. i did look up the monty hall problem and it is very revealing. i think time will show that the majority answerers here are making a similar error.

-M-:

In a continuous string of trials the probability of either transition is indeed equal, but the transition you likely

encounter firstis from the high probability toss to the low probability toss.However, one can rescue the situation by considering strictly

pairsof tosses, and discarding those with equal outcomes. In this case, yes, the ordered pairs HT and TH are, of course, equally likely. A “trial” must always consist of two tosses, not one.If you, however, toss continuously until you simply get the first transition, that probability is exactly equal to the one of the first toss.

That was the actual solution Richard wrote. Why would anyone keep the first H until they get a T? You gotta start over each time.

The joys of understanding!

http://math.stackexchange.com/questions/146605/improving-von-neumanns-unfair-coin-solution

“There’s a fair bit of literature on this problem.”

Indeed.

Michael, P(H,T) is not greater than P(T,H), it is exactly equal to it. -M-‘s maths are spot on. However, in your example P(H,H) is much greater than P(T,T).

Dharmaruci, intuition does not always give the correct solution. In fact much of Richard Wiseman’s line of work demonstrates that our senses can be easily fooled, and our intuition can often lead us astray.

I don’t think that Richard will be happy with your “not a real scientist” remark, and I think his understanding of probability theory is just fine.

Making a continuous sequence does not make sense. You need strict pairs. Otherwise one toss is enough, because you know what to toss next to get a winner. It would be like HHHHHHHHHHHHT <– Winner = HT. It has the exact bias from the original "one toss" case, and you would know who is going to win this at the first toss.

Using strict pairs is fair:

HH – retry

HH – retry

HH – retry

HH – retry

HH – retry

HH – retry

T…

Big chance for TH.

Each coin toss is an independent event the outcome of the first throw doesn’t effect the outcome of the second.

Therefore

P(H,T) = P(H)*P(T) = P(T)*P(H) = P(T,H)

Monty Hall concerns events that aren’t independent.

This seemed unikely and I had started to test it by brute force in Excel when I realised it wasnt’ needed. Instead of deciding based on which event is first, give the coin to Alice and Bob. If both gets heads or tails, repeat. If Alice gets heads and Bob gets tails, then the answer is “yes”. If Alice gets tails and Bob gets heads, then it’s “no”. It’s clearly true that neither Alice nor Bob is more likely to get one answer than the other, regardless of who goes first. When you realise this, you realise you don’t need to give the coin to anyone else but you can do it as in the original problem. Richard’s solution is correct.

HH = 98.01%

HT = 00.99%

TH = 00.99%

TT = 00.01%

I had another solution, which is similar in principle. Just toss it an even number of times, and every other toss make whatever comes up the opposite.

I like this one – better than my idea of throwing the coin away and doing rock, scissors, paper.

I immediately took the heads/tails issue out of the test, just like Graeme King above. My only difference was to draw a line on the ground, perpendicular to a stony wall. Toss the coin at the wall above the line and see if the coin falls left or right of the line…

An even more minimal solution would be to toss the coin and see if the design on the obverse (whether heads or tails) is facing towards you or away from you.

This is not working for a 100% biased coin.

got it!

now please explain the bear joke :(

Line on the floor strikes me as the simplest solution for ordinary people, and is the one I first thought of.

Once someone had hinted at the one-toss solution, I presumed that had to be the “real” answer. I was hoping it would be, because it is a lovely lesson in a priori probabilities, which are often misunderstood. In effect, if you don’t know which is the advantaged side, it is your initial choice of H/T that gives the 50/50 probability. (Though if someone has told you which way the coin is biased, but not by how much, then this solution is no longer available.) Disappointed Richard did not present it.

Richard’s actual solution will have you tossing coins for ages if the coin is very biased. Correct procedure is essential – if you toss HH you have to start again. I confess I did consider various options of double tosses for HT, on the grounds of symmetry, but I was looking for something that could always be achieved in precisely 2 tosses and failed to find Richard’s suggestion, obvious as it is once you know it.

I had so many solutions for this I tossed a coin to see which I one I should use today :- 1000 tosses to determine the bias then start flipping again for decision.

I guessed the ht-th or single toss would be the Monday solution but I’d actually use the line on the floor (I’d throw it to the dog an see if he sniffed it)

toss the coin while its flat and give it horizontal spin, if it lands with the head side facing up, if the head is pointing to the left or right should be random :)

almost the same

But mine was

HT, TH for yes

HH, TT for no

In case of a 99% H / 1% T coin:

HH = 98.01%

HT = 00.99%

TH = 00.99%

TT = 00.01%

Which makes: HH+TT = 98.02 and HT+TH = 1.98%.

I’ll join your game. And I win if it’s HH or TT :) Wanna bet?

Good to read all the ingenious solutions that substitute another mechanical process for the simple task of tossing the coin and observing Heads or Tails. Yes, drawing a line on the floor or dropping the coin on a chessboard will work — so long as you avoid introducing some bias into the process. That’s thinking outside the box!

I thought within the box and came up with Richard’s solution. Having nothing much to do Sunday, I wrote an excruciatingly detailed explanation for my blog aharmlessdrudge.wordpress.com. (Hm. How do I make that a link?). I posted that a few minutes ago at about 04:45 Pacific Daylight Time.

I’ll post the generalization (how does John base a choice among n options using only the biased coin?) when I get some time.

I got yellow

Couldn’t you just spin the coin on a table?

I’m with Simon Taylor and Co. on this. If I have no idea which way the bias goes, then even if the coin is a 100% bias, whichever I choose will be random?

Exactly what I thought.

I’d love to know why it’s wrong.

davebakedpotato’s solution is used in badminton for choosing the first server – throw the shuttle straight up into the air. It will almost always land on its side. Whichever side of the court the head points to is the winner of the toss.

The explanation given by Richard and by -M- is correct. The argument made by Michael is wrong, and I don’t understand it at all. It is clearly stated that the coin is tossed in two-at-a-time, not in a continuous sequence.

This is how I imagined the question:

p=probability of H, so (1-p)=probability of T.

p(HH) = yellow

p(HT) = blue

p(TH) = green

p(TT) = red

By observation, blue = green

http://i45.tinypic.com/2j10x7a.jpg

How would this solution be changed to deal with the case when (to contrive an example) the lightbulb in the room flickers randomly; staying on for a random duration, then staying off for a random duration. If this flickering was biased (e.g. more on than off) how would you avoid the problem of being more likely to see on->off than off->on when you looked at the bulb?

You could toss a coin I guess

my solution was a bit different.

– you toss the coin until the you get a different side than the initial one

– depending on the number of times you had to toss, you make you decision (odd: option A, even: option B)

Very clever, cozdas!! This not only works, but given any sequence of coin tosses, you arrive at a decision no later than Richard’s solution, and often much sooner!

My vote goes to this as the best answer, unless someone can show that it’s not 50-50. I haven’t proven it, but I’m nearly certain that it is..

It is not 50-50. For a coin that is nearly fair, you will flip it twice about 50% of the time. Since you will sometimes flip it 4, 6, 8, etc. times, an even number of flips will be more likely. In fact, you will flip an even number of times about twice as often as an odd number of times for a coin that is nearly fair.

It’s not 50-50. The probability of getting an odd number of tosses comes out to -(p^2 – p + 1)/(p^2 – p – 2), where p is the probability of flipping heads; this is only equal to 1/2 when p = 0 or p = 1. If you want an easier argument, notice that, with a fair coin, you stop after 2 flips half of the time.

J: you are right, my solution is invalid and actually it seems that you can never get 50/50 chance. My simulation showed that your formula is off by a factor of 2, so the chances of getting odd number of flips changes between 2/3 and 1.0 depending on the bias of the coin.

Fail ! :o)

Melt the coin in a furnace and forge a new, unbiased one.

My solution was basically the same:

The first toss of the coin would decide whether he should choose heads or tails. If it comes up heads, choose tails, and if it comes up tails, choose heads.

I’m disappointed that the answer isn’t by removing the bias from the equation and getting the result in one toss.

Toss it upwards to a magnet.

Simple.

My first thought was a non-mathematical solution I’ve yet to see here.

Pick up the coin, close your eyes, flip it around in your hands until you are confident you don’t know which way up the coin is facing. (Make sure to do this without letting go of the coin at any point, thereby *hopefully* negating the flipping bias.) Place it on the table. Job done.

Thoughts?

Flick it up spinning in the air with your thumb, catch it in the air and slam it down. Done.

Here a different solution:

He places the coin on a table and makes it spin on its edge. Than he slams with his flat hand on the spinning coin.

Through rotating the coin is extremely stable. Thereby the bias that is a problem when tossing the coin is irrelevant. So far it is simple physics.

The only factors effecting the result now, are how exactly and when the hand hits the coin.

This is totally random.

And so there is a 50:50 random decision.

He has to hit the coin a little carefully though, for it might hurt if he dose not. :)

The puzzie says “How can John toss the coin …” which implies one toss!

The correct answer is to catch the coin in mid toss in which the bias will make no difference because you will catch it randomly in the spin.

It doesn’t imply one toss.

And yet the answer does not start with “Imagine John toss the coin twice” but with

“Imagine John tosses the coin twice”.

Toss it once and don’t look.

Guess whether its heads or tails.

If he’s right, pick choice “A”, if he’s wrong, choice”B”.

I’d toss coin once and decide what to do if it is “H”. But I wouldn’t do what I decided if it is an “H” and I would do what I decided if it is a “T”.

I can show you why the solution from Simon does not work (i.e. you just make up something and toss)… Everybody who said that this will work forgets that you probably still have a bias choosing and you don’t know how much bias you have… But anyway couple of examples to show that it does not provide a 50:50 chance:

so You choose heads 60% and tails 40% of the time, also the coin is biased heads 80% and tails 20%, as your choosing and the tossing are independent events: the following probabilities are there:

HH : 60% * 80% = 48%

HT : 60% * 20% = 12%

TH : 40% * 80% = 32%

TT : 40% * 20% = 8%

which of course nicely sums up to 100%, now in all of these cases you guessed the flip right at HH and TT which gives 48+8 = 56%

so in this case you had a 56:44 chance of guessing right, which as far as i know is not a fair chance…

(you can try it with different numbers yourself…)

so this is clearly not the right solution… What Richard in the post described is the solution, although of course that one won’t work for the coin that always comes up on one side, but is a good solution for real life situations…

If you *know* the bias is like that, then you’re right. However, if you don’t know the bias is like that, there’s also some chance that the coin is biased exactly the other way: which you also have to take into consideration in the probability calculation.

Since the question explicitly stated that you *don’t* know the bias, I think a one toss solution is fine. (assuming, that the coin is equally likely to be biased towards heads as tails)

to answer Ronald: That is the point, if I can show you an example where your solution does not work, than your solution does not work, end of story… I could show many others… actually your solution only works if either the coin is unbiased or John is unbiased, for the first the puzzle states the opposite, and the second is improbable…

Now for the “official solution”, it works all the time when there is at least some chance that both sides are possible…

This thread really shows, that probability is a hard concept to understand for most people…

This thread is a veritable cornucopia of wrong headed assertions this week – many of them from people who seem unable to read the question. Come back Hugh Janus – where are you when we need you?

I assumed he could simply assign one option to heads, one to tails and toss the coin. This works if he doesn’t know which side is more likely to ‘win’. Compare, for instance, a case where someone tosses the coin, catches and covers it, and then has someone ‘call’. It’s now 100% certain whether heads is up or not, but their ignorance suffices to make the process fair.

Some people point out that the question only says that John is ignorant of the EXTENT of the bias, but not its direction. I agree my method wouldn’t work if, for instance, he knew that heads was more likely, just not by how much. However what the question said is “John doesn’t know the extent of the bias, and thus has no idea about the likelihood of obtaining a head or tail.”

The second part of that sentence says he has *no idea* about the likelihood of obtaining a head or tail. This rules out the possibility that he knows heads to be more likely, just not how much more likely. And if he really has no idea, then he only needs to toss the coin once.

again, the question is how can he make a 50:50 random decision. what a lot of people, including you say in this thread, that the coin does not matter, he can just call head or tails randomly. Which is of course not true, maybe he really has a true random process in his head and then yes he can do a random decision. But! We don’t know that, it is more probable that John is biased in one direction… so this solution does not work… and the solution given by Richard works….

@Istvan.

You’re not understanding the claim. Yes, it does require that he can call Heads or Tails – the process could never get started otherwise – but it doesn’t require that he can do so 50/50 randomly.

Suppose that John ALWAYS calls Heads. It still works: he calls Heads, what are the odds of Heads? For all we know (and all he knows), Heads is no likelier than Tails. We know that objectively one is more likely, but we have no idea which.

Perhaps the coin is biased 80% Heads (to use your example from above). In that case, of course, the combination of him calling Heads and a bias in favour of Heads means he’s more likely than not to ‘win’. But the coin might (equally, for all we know) have been 80% biased in favour of Tails. We have no more reason to expect Heads than Tails, or vice versa.

As I said in my original post, often someone will toss a coin, catch it in one hand, and flip it (covered) onto their other forearm, before then asking someone to call. At this point, either Heads is uppermost or it isn’t – we can consider this the ultimate (100%) biased coin. Nonetheless, so long as the caller has no idea whether it’s Heads or Tails that is uppermost, it doesn’t matter.

Ignorance is as good as genuine randomness here (indeed, if the universe is determinate – above the quantum level – it’s all we ever have). There’s no need to assume that he has a randomness function in his head.

Here are a few tips for you guys:-

1. The bias in the coin is physical. Any physical solution e.g. throw it against a wall, eat it and see if it comes out heads or tails on exit etc, will not overcome the bias. Unless you come up with a physical way of eliminating the bias, the bias remains. “Good luck with that” (eliminating the physical bias) as they say.

.2. The Simon Taylor solution – i.e. as you do not know what the bias is, it is random, does not work either. Ignorance of the bias does not eliminate it. If the bias is such that it always comes up with one side 99% of the time, under the Simon Taylor method once you have chosen heads or tails there is still a 99% to 1% bias once the coin is tossed. The ST “solution” just perpetuates the bias. It does not eliminate it or bring it to 50/50.

3. The “If the coin has two heads it is biased” theory. Biased towards what? There is only one possible outcome – heads. “Bias” implies one (or more) outcomes, with one (or more) of them favoured. This is not the case with a double headed coin.

What I find interesting about this thread is that no one has come up with an answer other than that which you will get by Googling the problem. There is at least one other, more elegant, solution than RW’s, albeit a variant on RW’S but differently presented, but no one has stated it.

@Anonymous.

Read the example in my last post. Have you never seen a coin toss conducted like that? Or do you think they’re all defective?

Yes, once you call you are now more likely to win if you called ‘correctly’ (the side with the bias in its favour), but with no idea of the bias you’re no more likely to call correctly than wrongly. To take the simple (100%) example: if you call Heads and it is Heads then you definitely win, but if it is Tails then you definitely lose. Similarly, if you call Tails and it is Tails you definitely win, but if it’s Heads you definitely lose.

Now suppose the coin has a 60% chance of coming up Heads. Then if you call Heads you get a 60% chance to win whereas if you called Tails you only get a 40% chance to win. But again this possibility has to be balanced against one where the coin has a 60% chance of coming up Tails – so the same call would go the other way. If you consider all the possibilities in this way, then you’ll see that your expectation of winning (with no knowledge of the bias) remains 50%.

Having called, yours odds of winning are no longer 50%, but that’s not the point that matters. The point is that when you call, with no knowledge of the bias, neither Heads nor Tails is any better.

“Having called, yours odds of winning are no longer 50%, but that’s not the point that matters. The point is that when you call, with no knowledge of the bias, neither Heads nor Tails is any better.”

Where we disagree is that my view is that not knowing what the bias is does not make the call a 50/50 AKA random one.

Once John has chosen your methodology he has chosen a non-random, non 50/50 mechanism before he has even tossed the coin. John leaves 50/50 behind once he choses your system, not on the point of call.

I fear that you may be confusing ignorance with randomness.

It’s not *confusing* ignorance with randomness if the claim is that ignorance amounts to epistemic randomness.

Your theory requires the assumption that there is genuine indeterminacy in the world, which is far from clear, and cannot explain what we often do even in clearer cases (such as the one I described where the coin is tossed before the call). In both cases, ignorance is all we have, but it’s enough.

Suppose I tell you I’ll toss a biased coin and you can call Heads or Tails, though you have no idea which is more likely. Which will you call then?

“Having called, yours odds of winning are no longer 50%, but that’s not the point that matters. The point is that when you call, with no knowledge of the bias, neither Heads nor Tails is any better.”

See? That is exactly the point, you are saying it, having called it your odds of winning are not 50%!

BUT the puzzle wanted a solution where you had a chance of winning of 50% no matter what you called!!!

Which you can get with the described method. In your method when you call something you don’t have an even chance… With the method of the real solution, even after you called it (does not matter what you called), you have a 50% chance…

Why is that so hard to understand, I cannot see… Grrrrrr…

“the puzzle wanted a solution where you had a chance of winning of 50% no matter what you called!”

So where does it say that? All it says is “make a 50:50 random decision”, but obviously it ceases to be 50/50 at some point, if only once the coin has been tossed and the result revealed. (For the record, I’m not denying that the method Richard gave ALSO works – but it is more complicated than necessary to meet the requirements of the question.)

“Why is that so hard to understand, I cannot see.”

I understand what you’re saying, I just don’t agree. You keep imposing an additional requirement on the lottery that is neither specified in the question nor perhaps actually able to be met.

You’re probably right that it’s fruitless to discuss this further, but if you would like to be enlightened then I’d recommend:

George Sher ‘What Makes a Lottery Fair?’ Nous (1980)

and

Peter Stone ‘On Fair Lotteries’ Social Theory and Practice (2008) (or his more recent book, The Luck of the Draw).

“Suppose I tell you I’ll toss a biased coin and you can call Heads or Tails, though you have no idea which is more likely. Which will you call then?”

I will probably take my own coin (which is probably biased) and do the algorithm from the solution and call the result. That way I made sure that my chances of winning are fair. :)

(you could say that i could have just called whatever, but you forget that priming people is very easy – just ask a magician – so the chance that i choose the wrong side could be great, if the other person knows the bias, and also know how to prime my subconscious…)

also the best solution here is to refuse to play :)

Thanks Istvan

I’m leaving the thread now – we have made the point several times, but Dr B seems unable to take it in.

So many contributors to these discussions seem unable to Read the question.

I refer you to my last reply to Istvan – I see your point, I just disagree, and you don’t seem to have taken in what I’m saying. What part of the question have I not read? (The only bit Istvan was able to cite doesn’t actually appear to be there…)

Hi Anon, Istvan, Ben – interesting comments, thanks. Don’t get discouraged from discussing this – it’s a live issue in philosophy of probability, and a subtle one, and whatever the answer is it’s not obvious. Although the majority of professional philosophers agree with Ben on this, I don’t think things are quite as clear-cut as he’s making out.

The probabilities got by the two methods are different in kind. The HT/TH method gives an objective statistical-mechanical probability of 50:50, Simon’s method gives a 50:50 probability which is less fundamental and ‘more epistemic’ – it’s more about our ignorance and less about the world. Ben would argue that these kinds of probabilities are equivalent for all practical purposes. But that’s not necessarily true (though it may be true in this particular case as specified.)

One way they could come apart is if a ‘bookie’ running the experiment could control which way the coin was biased and knew which way the guesser is biased in their guessing. Then they could ensure that the guesser guessed wrong more often than right. The HT/TH method can’t be exploited in this way.

Another way they can come apart is via observation selection effects, as in anthropic reasoning, the Doomsday argument, and the Sleeping Beauty problem. There’s a big literature on all this.

Conclusion: it depends what kind of 50:50 probability is demanded by the rules of the problem, i.e. how objective the ’50:50 random’-ness is required to be.

One very last comment on this. I guess i am coming from a mathematical angle, and that is why the philosophical answer is unsatisfying for me. But as you said the HT/TH method works every time independent from any philosophical assumptions. And the “epistemic” solution presupposes that we have chosen a coin randomly from all the possible biased coins (where the distribution is uniform), in which case it also works…

Ok :). I read a bit about epistemic probabilities and randomness, and I see where you are coming from. So I guess we both were right in our ways, we just solved different problems. But I maintain that in the real world, you can only say that epistemic randomness is enough, if you really sure, that you have no way of knowing which way the coin is biased (i.e. nobody is trying to manipulate or prime you one way or another, and not even you subconscious knows where the bias lies).

But thank you Ben for pressing the issue, as I learned something today.

Glad we made some progress (thanks to Anon JULY 18, 2012 AT 9:53 AM).

As I said above, I’m happy to admit that Richard’s answer (as you defended Istvan) works. My only claim was that it was unnecessarily complicated as a solution to the question stated.

My alternative epistemic solution (as also defended by some others above) wouldn’t work if either a) John needed to make a series of decisions or b) he knew which way the coin was biased, but the question explicitly said that he wants to make a decision (singular) and has no idea of the bias. Given those assumptions, I actually think my solution slightly better – first because it works even with a 100% bias (as in the coin on arm case) and second because it only requires one toss rather than a very long series.

As to which mechanism is better ‘in the real world’, that’s a much larger question. I accept that epistemic randomness may be worrying – if someone else has the coin and subtly tries to prime you to call one way. (But note that – absent priming – this is ok, and probably explains why it’s usual for one to toss and the other call.) On the other hand, while objective chance would be nice, I don’t see how we can know that we have it, as opposed to mere ignorance. Any coin you use might be biased without your knowledge.

But I don’t think we need to settle these issues to answer Richard’s question: either of our answers works for that.