# I Sequential vs simultaneous probability

1. Sep 15, 2016

### e2m2a

If I toss 1000 coins, one at a time, the laws of probability predict about half will end up heads and the other half tails. But what does probability say if I toss all 1000 coins at the same time? I assume the probability remains the same: half will end up heads, the other half tails. So, my real question is are the laws of probability not dependent on if events are sequential or simultaneous? Assume the events are not dependent on each other in any way.

2. Sep 15, 2016

### Simon Bridge

The coin tosses are called "independent" because the outcome of any individual coin does not depend on the outcome of others.
But not all events are like that. When the probability of an outcome depends on the past outcomes, then those are dependent events and the probability is said to be conditional.

You can see if the pile you get from throwing coins one at a time is the same sort of shape as when you toss all the coins at once.
Maybe look at what happens to magnets if they are dumped together in one go vs having one magnet added at a time?
If you put magnets around the rim of spinners, then they will have different odds when spun together to what they'd read when spun separately or one at a time. ie - if the spinners stick to the table once they stop spinning.

I take it this is the sort of thing you mean, rather than something like sampling without replacement changes the odds?

3. Sep 16, 2016

### e2m2a

Ok, you pretty much answered my question. Thanks. If the events are independent, then it is irrevelant if the events occur one at a time or simultenously. Where I am going with this is if one was investigating a phenomenon and it consisted of lets say 10000 independent events that had a possible 3 equal outcomes. It would take a tediuous amount of time to run 10000 seperate tests sequentially to record the outcomes. Alll one would need to do is have all 10000 events occur simultaneously, as long as the events do not affect any other event. Then the outcome would be the same in either case as far as the distribution of the outcomes. Doing the test simulatenously would save a lot of time.

4. Sep 16, 2016

### Simon Bridge

Yes that's right: and it wasn't what I was thinking you meant... so it's simpler.

If you were to pull a card from a std 52 card deck, the probability of drawing the ace of spaces (or any one specific card) would be 1/52
If you repeated the draw 52 times, replacing the card and shuffling each time, then the outcomes of previous draws does not affect the outcome of the next one.
You would get the same maths if one trial is repeated 52 times with one deck or 52 times on 52 decks or twice each on 26 decks etc.

But if you were to draw cards without replacing them, then the probability changes depending on the outcome of the previous draw ...
ie. on the second draw, if the first was the ace, then the probability of getting the ace is zero ... if the first draw was not the ace, then the probability is now 1/51.
If you drew 52 times, then the probability of getting the ace on one of those draws is 1.

Interestingly, tossing 1000 coins all at once, as a bunch, would make the outcomes fairer since the interaction of the coins bouncing off each other while they fall further randomizes them ... even if the coins were not quite fair to begin with. It also reduces experimenter bias - where someone finesses the physics of tossing a coin to guarantee an outcome.

Many identically prepared systems that are independent is called an "ensemble", and is a useful tool in statistical physics.

5. Sep 16, 2016

### PeroK

Here's a different way to answer your question. If you toss the coins at the same time, there are essentially only three cases:

a) Heads and tails are equally likely.

b) Heads are more likely than tails.

c) Tails are more likely than heads.

So, if it is not a), then which of b) or c) would it be?

6. Sep 16, 2016

### e2m2a

Not sure what you mean. Case a could never be true for a sufficiently large sample (law of large numbers). So, case a would be contrary to the laws of probability. It could never happen. So, I don't see a meaningful answer for case b or case c, since case b or case c could never occur for a sufficiently large sample.

7. Sep 16, 2016

### PeroK

Isn;t it obvious that if heads and tails are equally likely for a single coin toss, then they are equally likely for every coin, regardless of whether other coins are tossed before it, after it or at the same time is it?

How would a coin know that other coins are being tossed at the same time and behave differently?

8. Sep 16, 2016

### e2m2a

Yes. I see your point. You are right. Inanimate objects cannot "know" their environment. Mmmm, but then how do electrons "know" they are being observed, and thus, give a different diffraction pattern in the double slit experiment?

9. Sep 16, 2016

### EnumaElish

You could formulate your original post as, throw the same single coin 1000 times, versus throw 1000 coins simultaneously. If you assume independence, then you predict probabilistic equivalence between the two. Is your assumption of independence tested and verified? It is usually a good enough assumption, but suppose you decide to test it. You find out that for some reason, after you obtain heads, you are somewhat more likely to obtain heads if you throw the same coin again. You conclude independence assumption can be rejected for sequential tosses. At that point the probabilistic equivalence between the two experiments breaks down. If electrons behave as if they can sense other electrons in a simultaneous experiment but not in a sequential experiment then these two experiments (with electrons, not coins) would not be equivalent. That does not imply anything about coin tosses per se.

10. Sep 16, 2016

### e2m2a

Sorry, I got off on a tangent. The double slit experiment is a completely different problem. With coins, assuming no interaction between the coins, sequential tosses or one large simultaneous toss should yield the same probalistic outcome. I see that now.

11. Sep 16, 2016

### EnumaElish

I now see that my statement of the problem makes it a little more complicated than yours. In my version of the problem another assumption is needed for probabilistic equivalence: that the single coin tossed 1000 times over is "representative" (e. g. the "average") of the 1000 separate coins in the simultaneous experiment. To make the point, if the 1000 coins thrown simultaneously are ordinary coins but if the one thrown sequentially says heads on both sides then the two experiments would clearly not be equivalent.