Always 50/50 chance?

[Dave1001]

You need independence for these arguments. Take the case where I have a bag with 5 red balls and 5 blue balls. I now draw each ball out of the bag one at a time.

Person A guesses: RBRBRBRBRB
Person B guesses: RRRRRRRRRR
Person C guesses: BBBRBBBRBB

If we do this enough times, person A should be accurate on average 50% of the time. Sometimes he'll be right 7 times, sometimes wrong 7 times, but on average he'll be right 5 out of 10 times. (full range of 0->10 correct)

Person B will be right 5 times out of 10 each and every time. (no range, only 5 correct)

Person C, using the same logic, will definitely have 3 wrong and 3 right. He's only playing with the remaining blue balls. (reduced range of 3->7 correct)

Because the variances are different, the "optimal" guessing strategies can possibly be different.

Yeah, I think this is somewhere close to my initially warped intuition.
My initial warped intuition ranked someone guessing only B or guessing B 50% of the time as likely to be the most correct (50% correct). And someone guessing B 75% (or R 75%) of the time as likely to be the most incorrect. Although I now see why I was off, I'm interested in a model that captures this specific flawed intuition and why it initially appealed to me.

 

[Baron Samedi]

...

Yes, but...

I'm not sure I'd call that cheating or manipulating. It's still just pure guessing. In this case, there's a 50% probability that guessing will result in 10 correct and there's a 50% probability that guessing will result in 0 correct.

Agreed. It's still guessing. And yes, the expected values are the same. It's just that the values are all wonky. Remember, on the "psychic" level, 10 out of 10 is perfect accuracy and a sign from heaven. 0/10 is perfect disharmony, and there must obviously be an evil spirit in the room.

In terms of making strategies better or worse, think of these two betting strategies:

A: Always be exactly 50% accurate
B: Accuracy of 40%, 5 out of 6 games; Accuracy of 100%, 1 out of 6 games

Both strategies have an expected accuracy in the long term of 50%. However, if you play these two games head to head, B loses way too often.

I don't think this is the sort of thing Dave1001 had in mind when he suggested that guessing HHHT... is "worse" than guessing some other sequence. He wasn't very specific, but I figure that if HHHT... had a high probability of being very bad but also had an equally high probability of being very good, it shouldn't be considered worse overall. Nor better, of course. Just different.

Yeah, it's that anti-intuitive thing where people feel that they HAVE to guess at an exact 50% rate to generate random tosses. I asked three people at work yesterday the same question, which would be the most accurate: All Heads; HHHT HHHT ...; or random guessing. Of the three people, all three said Random. Two of these people had stats degrees as well. Go figure.

 

[69dodge]

In terms of making strategies better or worse, think of these two betting strategies:

A: Always be exactly 50% accurate
B: Accuracy of 40%, 5 out of 6 games; Accuracy of 100%, 1 out of 6 games

Both strategies have an expected accuracy in the long term of 50%. However, if you play these two games head to head, B loses way too often.

I'm not sure exactly what situation you're thinking about here. But I agree that if all that matters is whether one wins or loses, so that winning by a lot is no better than winning by a little, then expectation isn't the right number to look at, because it takes into account the size of the win or loss, and not just whether it was a win or a loss.

Google for "nontransitive dice".

Yeah, it's that anti-intuitive thing where people feel that they HAVE to guess at an exact 50% rate to generate random tosses.

That seems like a slightly different issue. But in the general category of statistical misconceptions, to be sure.

I asked three people at work yesterday the same question, which would be the most accurate: All Heads; HHHT HHHT ...; or random guessing. Of the three people, all three said Random. Two of these people had stats degrees as well. Go figure.

Stats degrees? Oh, my.

I'm probably better at math than psychology, but I'll take a guess at the thought process behind that. I suspect it's a matter of getting a bit confused about exactly what question they're trying to answer.

There's a random sequence of coin tosses, and there's some sequence of guesses that they're trying to analyze. The real question deals with the number of correct guesses. To be correct, a guess has to match its corresponding coin toss. Other coin tosses and other guesses don't matter. Either the guess matches its toss or it doesn't. That's the only thing that matters. Of course, the same holds individually for each of the other guesses and each of the other tosses. So, they matter in that sense. But they're all totally separate. For each guess and corresponding toss, you check whether they match; and then, at the end, you count how matches there were.

I bet that when thinking about the HHHT... guess sequence, people tend to think to themselves, ok, let's see here, I know that a random sequence of coin tosses will have about half heads and half tails, while this sequence of guesses has three-quarters heads and only one-quarter tails. That can't be good. This must be a bad way to guess.

But the question isn't, which sequence of guesses has the same ratio of heads to tails, overall, as the sequence of coin tosses. All that matters is whether each guess matches its toss. If every single guess is wrong, for example, the head-to-tail ratio in the sequence of guesses will be about the same as in the sequence of coin tosses! (Namely, both near 50:50.) The two questions are not the same at all.

 

[lenny]

Can you give me an example of a single event that does not have a 50/50 chance of happening?

has anyone asked what "probability" means in the case of a single event?

 

[Dave1001]

has anyone asked what "probability" means in the case of a single event?

Doesn't bayesian analysis attempt to answer that? I'm also thinking meta-probabilities. I think one would still have to reference the past to attempt a forward looking assessment of the probabilities of a given unique event. Such as the probabilities a particular thing would go wrong on Y2K. Or the probabilities the WTC towers would fall.

 

[Dave1001]

I'm not sure exactly what situation you're thinking about here. But I agree that if all that matters is whether one wins or loses, so that winning by a lot is no better than winning by a little, then expectation isn't the right number to look at, because it takes into account the size of the win or loss, and not just whether it was a win or a loss.

Google for "nontransitive dice".





That seems like a slightly different issue. But in the general category of statistical misconceptions, to be sure.





Stats degrees? Oh, my.

I'm probably better at math than psychology, but I'll take a guess at the thought process behind that. I suspect it's a matter of getting a bit confused about exactly what question they're trying to answer.

There's a random sequence of coin tosses, and there's some sequence of guesses that they're trying to analyze. The real question deals with the number of correct guesses. To be correct, a guess has to match its corresponding coin toss. Other coin tosses and other guesses don't matter. Either the guess matches its toss or it doesn't. That's the only thing that matters. Of course, the same holds individually for each of the other guesses and each of the other tosses. So, they matter in that sense. But they're all totally separate. For each guess and corresponding toss, you check whether they match; and then, at the end, you count how matches there were.

I bet that when thinking about the HHHT... guess sequence, people tend to think to themselves, ok, let's see here, I know that a random sequence of coin tosses will have about half heads and half tails, while this sequence of guesses has three-quarters heads and only one-quarter tails. That can't be good. This must be a bad way to guess.

But the question isn't, which sequence of guesses has the same ratio of heads to tails, overall, as the sequence of coin tosses. All that matters is whether each guess matches its toss. If every single guess is wrong, for example, the head-to-tail ratio in the sequence of guesses will be about the same as in the sequence of coin tosses! (Namely, both near 50:50.) The two questions are not the same at all.

If the question was "which sequence of guesses has the same ratio of heats to tails, overall, as the one randomly generated?" Would a 50:50 sequence be the best guess? Or would once again, it not matter if we picked a 50:50 sequence, all heads, or HHHT?

 

[lenny]

If the question was "which sequence of guesses has the same ratio of heats to tails, overall, as the one randomly generated?" Would a 50:50 sequence be the best guess? Or would once again, it not matter if we picked a 50:50 sequence, all heads, or HHHT?

there are two very different questions here: are you trying to predict "the ratio of heads to tails overall" (in which case you need never consider any particular sequence) or are you trying to pick "the" sequence which has the higherst number of correct individual calls (in which case, it is unlikely to matter which sequence you take, given that the trials are IID and each outcome equally likely on each trial.)

 

[lenny]

has anyone asked what "probability" means in the case of a single event?

Doesn't bayesian analysis attempt to answer that?

attempt, yes: but most Bayesian sects do so by redefining "probability" relative to one's beliefs, thereby distancing it from empirical applications, and limiting is apparent relevance to the sciences (other than psychology)

 

[Jekyll]

attempt, yes: but most Bayesian sects do so by redefining "probability" relative to one's beliefs, thereby distancing it from empirical applications, and limiting is apparent relevance to the sciences (other than psychology)

Personally, I like to think of the Bayesian approach as abstracting away the t parameter of the frequential approach. Instead of the probability being the ratio of the set of experimental outcomes as the parameter t tends to infinity, it's the ratio of outcomes in the set of possible events conforming to experimental set up.

The subjectivity of the Bayesian approach is often optional and I find it a lot easier to spot compared with the frequential methodology where it is often buried in your choice of hypothesis.

 

[strathmeyer]

Then closed at #137 by a pompous twat?

No, apparently they're still going at it.

 

[69dodge]

If the question was "which sequence of guesses has the same ratio of heats to tails, overall, as the one randomly generated?" Would a 50:50 sequence be the best guess? Or would once again, it not matter if we picked a 50:50 sequence, all heads, or HHHT?

A 50:50 sequence of guesses would be best, because the sequence of random coin tosses probably will be near 50:50 and is more likely to be exactly 50:50 than to be any other ratio. (Not more likely than all other ratios combined, but more likely than any other particular ratio.)

 

[Foolmewunz]

@Hammer of Thor

Welcome, also, to your first thread-jacking. If you think the coin tossing got 'em going, try mentioning "Monty Hall". Go on, I dare you!

@ Strathmeyer
Awwww Shaddup! It's the math and science boards. They're funnier than humor, at least! And they make more sense than Politics or Conspiracy Theories. I come here to relax and remind myself that there's always at least one person in the world who knows more than me on a certain topic, and in the case of Math/Science/Computers, there are obviously hundreds, if not thousands. (I usually just read but every now and then, I've actually come to the lads and lasses for a little help - and they've always been cool about it, putting up with my complete inability to add, subtract, multiply or divide.)

 

[69dodge]

attempt, yes: but most Bayesian sects do so by redefining "probability" relative to one's beliefs, thereby distancing it from empirical applications, and limiting is apparent relevance to the sciences (other than psychology)

Redefining?

What else could it mean?

Empirical applications never involve infinitely many trials, so defining it that way won't really work. We can talk about what we imagine would happen if the trial were to be repeated infinitely often, but then we're back to belief again.

"A tossed coin will land heads with probability 1/2." If this means, "in a long sequence of tosses, it will probably land heads about half the time", then what is the meaning of that "probably"? Shall we repeat the long sequence itself many times? That won't help; there will still be a "probably" in there.

I don't see any other way out of the infinite regress, besides invoking belief.

 

[Baron Samedi]

No, apparently they're still going at it.

Dude, I think I see why you're confused. The "cool" forums are down the hall and to the left three doors, where they're talking about what Britney and Lindsay really did in rehab. We're still stuck on coin tosses here.

@Hammer of Thor

Welcome, also, to your first thread-jacking. If you think the coin tossing got 'em going, try mentioning "Monty Hall". Go on, I dare you!

...

Now, what is the optimal solution to Month Hall -- choosing all Heads, HHHT, or a random selection of Heads and Tails -- when factoring in the Baysian priors of you being selected and somewhat sober enough to not toss your lunch on Monty's cheesy jackets?

 

[JonnyFive]

This thread was awesome until after post #18. Then it was ruined by the slow witted. Then it was ruined by the nerds.

You're just jealous of our sexy brains.

when factoring in the Baysian priors of you being selected and somewhat sober enough to not toss your lunch on Monty's cheesy jackets?

That's an irrelevant factor. Monty's so wasted from drinking the night before that he doesn't even care. Just choose a door and take your goat, dammit!

 

[Baron Samedi]

You're just jealous of our sexy brains.



That's an irrelevant factor. Monty's so wasted from drinking the night before that he doesn't even care. Just choose a door and take your goat, dammit!

Ha! You had to mention the goat. It always bothered me whenever somebody mentioned this problem. What if the guy picking between Door #1, Door #2, or what's in the box really, reaaaaaaaally, wanted the sexy goat? Now, he knows the goat CAN'T be in the box, because as we all know, there's nothing in the box and you'd be stupid to pick it, so what should the crazy pervert do to maximize his chances of winning the goat and not the lame-ass car?

 

[JonnyFive]

Ha! You had to mention the goat. It always bothered me whenever somebody mentioned this problem. What if the guy picking between Door #1, Door #2, or what's in the box really, reaaaaaaaally, wanted the sexy goat? Now, he knows the goat CAN'T be in the box, because as we all know, there's nothing in the box and you'd be stupid to pick it, so what should the crazy pervert do to maximize his chances of winning the goat and not the lame-ass car?

Well, assuming the standard Monty Hall setup, he would be better off sticking with his original choice.

Assuming the setup where Monty is so drunk he can't remember which door is which and picks at random... well, let's see.

You have 18 possible intial scenarios. In 6 of them, success is guaranteed (Monty picks the car, so you automatically get a goat). Thus, those scenarios can't count either for or against the probability of a switch being beneficial, as either would result in a positive outcome.

You're left with a wash, where there are 6 outcomes resulting in a switch being sucessful, and 6 resulting in a switch not being successful.

And there's a 50/50 chance you'll get your precious, precious goat. Assuming Monty is drunk enough not to know what he's doing.

I think that answers the OP's question, anyway.

 

[Baron Samedi]

Well, assuming the standard Monty Hall setup, he would be better off sticking with his original choice.

Assuming the setup where Monty is so drunk he can't remember which door is which and picks at random... well, let's see.

You have 18 possible intial scenarios. In 6 of them, success is guaranteed (Monty picks the car, so you automatically get a goat). Thus, those scenarios can't count either for or against the probability of a switch being beneficial, as either would result in a positive outcome.

You're left with a wash, where there are 6 outcomes resulting in a switch being sucessful, and 6 resulting in a switch not being successful.

And there's a 50/50 chance you'll get your precious, precious goat. Assuming Monty is drunk enough not to know what he's doing.

I think that answers the OP's question, anyway.

Damn. So everything is a 50/50 chance after all? Life would be so much simpler if a drunken Monty Hall ruled the world.

 

[rtalman]

Since I feel like being flamed today, I will again state that in the Monty Hall scenario, you have a 2/3 chance of getting the prize instead of the Zonk if you switch when Monty opens a non-prize holding door.

A) Prize behind door #1
I) You pick door #1, Monty shows you door #2 or door #3. Switch = lose. Don't switch = win.
II) You pick door #2, Monty shows you door #3. Switch = win. Don't switch = lose.
III) You pick door #3, Monty shows you door #2. Switch = win. Don't switch = lose.

B) Prize behind door #2
I) You pick door #1, Monty shows you door #3. Switch = win. Don't switch = lose.
II) You pick door #2, Monty shows you door #1 or door #3. Switch = lose. Don't switch = win.
III) You pick door #3, Monty shows you door #1. Switch = win. Don't switch = lose.

C) Prize behind door #3
I) You pick door #1, Monty shows you door #2. Switch = win. Don't switch = lose.
II) You pick door #2, Monty shows you door #1. Switch = win. Don't switch = lose.
III) You pick door #3, Monty shows you door #1 or door #2. Switch = lose. Don't switch = win.

Score:
Switch 6, Don't switch 3.

 

[JonnyFive]

Since I feel like being flamed today, I will again state that in the Monty Hall scenario, you have a 2/3 chance of getting the prize instead of the Zonk if you switch when Monty opens a non-prize holding door.

But if, and only if, Monty is aware of the contents of the doors and always chooses a non-prize door. If he doesn't know and chooses at random, you have the same probability of winning whether you stay or switch.

This is because, in that case, 1/3 of the time Monty chooses the car and the game ends at that point. In the 2/3 of the cases where the game actually progresses, you have a 1/2 chance of getting a car either way.

I say this only because not every statement of the problem is explicit on this matter, but it's very, very important to the setup.

Of course, if Monty is so wasted that he sometimes chooses the door you chose (I'm not sure what consequences of this would be in the game... automatic loss? Forced to pick another door? Monty arrested for drug use?), that just messes everything up, and you might as well go home.

I'm such a nerd!