Coin toss; Not really 50/50 (1 Viewer)

[Grant H]

Before I even read that my immediate thought was to use this info with a woman. I'll make sure I call whatever side is up before the flip should a coin flip ever decide whether or not I get laid!

51 percent? I'll take them odds!

 

[MarkHastings]

Brian, your theory reminds me of some peoples theories where there's a better chance that one particular side of a coin will come up because the opostite side is heavier. Meaning that if one side used more metal to create the raised artwork, that this would make the coin slightly unbalenced and the "heavier" side would land face down more times than the lighter side.

Of course, the weight difference is very minute, but over the course of thousands of flips, they believe it'd have an influence.

 

[andrew markworthy]

Yes, it should. It certainly is true for a die (unless it's one of those v. expensive casino dice, which are calibrated for equally weighted sides). On a die with spots marked by indentations, the '1' side is slightly heavier than the '6' side, so over thousands of throws, the '6' is slightly more probable. The reverse I guess applies to dice with painted spots (i.e. the '6' should be heavier and so the '1' should be more probable).

 

[george kaplan]

Sorry, it doesn't work that way.

Any given outcome has a probability associated with it, but with two outcomes, the probabilities are rarely 50%.

Assuming the probabilities are stable, the chance of something happening twice is less than the chance of it happening once. However, the chance of it happening at least once if you try twice, actually goes up.

The best way to think of this is monetarily. If you really think every event has a 50/50 chance of happening, then you should be more than willing to put up $100 that it will happen to my $1000 that it won't. There'd be a 50% chance you'd win a thousand dollars and an equal chance you'd only lose a hundred.

So, I'll bet you that the sun comes up tomorrow. If it does you owe me a hundred dollars, and if it doesn't I owe you a thousand. If the chance of the sun not coming up is 50/50, then you'd be a sucker for not taking my bet.

 

[Holadem]

2 points made by that article:

1 - If a coin is launched exactly the same way, it lands exactly the same way. The randomness in a coin toss, it appears, is introduced by sloppy humans. Each human-generated flip has a different height and speed, and is caught at a different angle, giving different outcomes.

2 - There is a bias built into the coin.

:rolleyes

Keith said:
EXACTLY!

Would someone tell me what I am missing?

Chance, in many cases, is simply a collection of very physical events that are too minute to observe or replicate without instruments. Same initial conditions will give the same outcome. If the outcomes are different, then all the factors were not accounted for. Why the heck should a coin toss be any different?

--
H

 

[Cees Alons]

The conclusion of the article is wrong. So there seems to be some correlation between the initial position and the final position. So what? The chance that you start the flipping with either heads or tails on top (in your hand) is probably still about 50%. This study only "proves" that the whole manoeuvre following that may be less crucial.

To prove that the outcome may not be 50/50, you must - as far as that study is concerned - also prove that a human tends to favour (knowingly or unknowingly) one of the two initial positions.

Of course, we already knew that a "handy" (dishonest) person can always influence the outcome.

"If a coin is launched exactly the same way, it lands exactly the same way.". Except when that butterfly in Brazil flaps its wings once again.


Cees

 

[Holadem]

...

--
H

 

[BrianW]

The "heavier on one side" theory is bogus when it comes to coins. Since dice are rolled, it definitely matters if the weight is not evenly distributed. But a coin is tossed, not rolled, and will be free to spin about its own center of mass without favoring any heavier or lighter side.

In addition to chaos theory, there's also quantum interactions to consider: interactions with virtual particles, tunnelling of quantum particles within the coin, etc.

I know people who have dropped a contact lens and, after looking for it unsuccessfully, drop the other contact lens in similar fashion, hoping to duplicate the the conditions when the first contact lens was dropped. The idea is that the second contact lens will land very close to the first contact lens. So if you pay attention as you drop the second contact lens, it will lead you to the first contact lens.

This never works.

 

[Cees Alons]

Very true.

And, BTW, the 'landing on a side' argument isn't valid as well. The sum of the chances of all possible outcomes together is 1. It doesn't matter what the chance of landing on a side is, because the other two will be equal and one of those two outcomes is always forced by the procedure.

This is also a correction on the earlier statement that "any situation with only two outcomes has a 50% chance for each outcome". It has already been said before, it's not necessarily dived 0.5/0.5 (50/50 if you talk percentages), so the only valid statement is: the sum of all those mutually exclusive chances must be 1.
In fact, per definition it always is, no matter if we're speaking of a number of possible outcomes of 1, 2, 3, or more.

Cees

 

[george kaplan]

Although if you want to get really, really technical, there are certain events that are not Borel measurable, which can do all kinds of whacko things to probability. Also, infinity can screw things up, and you can have a collection of events, all of which have zero probability of happening, but still have a non-zero probability that at least one of them will happen. A good example is picking a (real) number from 0 to 1. If you pick, say, 0.0345283503802, then the probability that that number comes up is 0, since there are an infinite number of outcomes. On the other hand, the probability that the number will be between 0 and 0.5 is 50%.

However, in the real world, what Cees said is still basically all true.

 

[Cees Alons]

George,

And what do you think is the sum of the infinite number of zero chances in that collection?

Cees

 

[RobertR]

Is that similar to asking what an infinite number of infinitely small rectangles adds up to (as in calculus)?

 

[BrianW]

Robert, I think it is, but I can't prove it.

--------------------------------

A coin toss is 50/50, right? But if you toss a coin only three times, you will never get a 50/50 distribution of heads vs. tails. In any series of three consecutive tosses, you will end up with two-thirds of one (either heads or tails) and on third of the other, or all three tosses the same.

Of course, mandating an odd number of tosses is what eliminates the possibility of having a perfectly even distribution of heads vs. tails. It is also well known that the more tosses you mandate, the more the results will approach 50/50. So it seems to me that a series of, say, ten coin tosses is a good, quick way to allow for the possibility of having a 50/50 distribution of results (by having an even number of tosses) while at the same time having enough tosses to effectively, though not always, “smooth out” the occasional streak.

So allow me to propose a very generous wager: You and I will toss a coin ten consecutive times and note the results. If the ratio of one outcome vs. the other is 5/5 (i.e., five of one and five of the other), then I will give you $100. But if the ratio is 6/4 (i.e., six of one and four of the other), then you must give me $100. And in order to prevent the windfall of the occasional streak of seven or more identical tosses, all other outcome distributions are ignored for the purposes of the wager.

There’s no trick, and it's an even-money wager. I’m allowing you to bet on the 50/50 outcome of a series of ten coin tosses. What could be a more sure thing? I’m betting only on the 60/40 outcome – admittedly the next best thing – and no other outcome is allowed to benefit either of us.

So, is it a bet?

 

[RobertR]

Hmm, so if the coin is tossed 10 times, there are 2^10, or 1024 possible outcomes, correct? So the question is, how many of those outcomes are 60/40 vs. 50/50...

 

[andrew markworthy]

Brian, you've got yourself a bet, just as soon as the Loch Ness Monster makes a personal appearance on Elvis's Comeback Special.

I can't be bothered to work out the maths (sorry, I mean 'math'), but the odds are just a teeny weeny bit in your favour, aren't they?

Let's suppose you toss a coin ten times and note now many times heads comes up. You then repeat the exercise several thousand times, each time noting how many heads come up in a series of 10 tosses. What you'll find is that when you plot the results, you'll get something resembling a bell-shaped distribution (aka normal distribution or Gaussian curve). An outcome of 5 heads will be the most frequent occurrence, but outcomes of clustered around this (i.e. of 4 or 6 heads) will be nearly as frequent and the total outcome of 4 *or* 6 will be greater than 5 *only*. So accordingly, the odds are in your favour.

 

[Keith Mickunas]

Brian, I'll take your bet if you win only on 6 heads or on 6 tails, but not if either come up. In other words you have to declare before the tosses which will have the majority, heads or tails.

 

[BrianW]

Keith, you seem like a good sport with an eye for a fair deal. I'll tell you what: You and I can bet on a single coin toss, and I'll declare right here, in advance, and in writing, no less, what I'm betting on before the toss. No two-to-the-power-of-ten stuff, no bell curves - just a single, uncomplicated, sporting coin toss which will result in only one out of two possible outcomes. I'll even put up $110 to your $100, just to give you something for your trouble. How can you refuse a deal like that?

Here's my declaration: Heads = I win, Tails = You lose.

Is it a deal?

 

[Cees Alons]

(bolding by me - C.)

Side track, or is there some resemblance to Brian's bet proposal?

Cees

 

[MarkHastings]

LOL Brian. That reminds me of the old bet "I bet you I have more money in my pocket than you do!?"

I'd win every time. Even if you had $100 in your pocket and I had 1 penny in mine. The trick is, we're not talking about YOUR pocket, you don't have ANY money in MY pocket. All of the money in MY pocket belongs to ME.

 

[george kaplan]

The sum is 1, not 0. It's another oddity of infinity that if you add up enough things with zero probability, you get a non-zero sum. Of course, all of this stems from the fact that while you can talk about randomly selecting 1 thing from an infinite collection, you could never actually do so.

All of this is part of something called Measure Theory. I'm rusty though, my coursework was back in the 80s, and I haven't taught any courses that use it since the 90s.