Since the different sides of a penny have different amounts of metal for the patterns, the sides of a penny are asymmetrical.

Because of this, is there a very minute advantage to either side landing face up?

You can probably make an argument that theoretically there might be an advantage to tails landing up (it looks to me like Lincoln uses more metal the the treasury, or whatever that building is). But the effect would be so small it would be lost in the noise of other variables, such as air currents, wear, manufacturing differences. That is, it makes for interesting discussion but don’t seize on it as a way to build your retirement nest egg.

Here’s one previous discussion:

http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=89668

Another question to consider. Say you tossed it 1000 times in a row. What are the chances that you will end up perfectly 500 heads and 500 tails?

50%. Easy to understand if you do it for smaller numbers.

I believ all options are equally likely: there are 1000 possible endings (all heads to all tails) so 1/1000?

All options are equally likely, but half of those options include a 50/50 split between heads an tails.

BTW there are not 1000 possible endings but 2^1000 different combinations. You can’t just count the number of heads & tails, you have to count how many ways each number can occur. Work this out with 4 tosses instead of 1000 to see how it works.

For a reasonably large number N, an even split(or as even as possible if N is odd) will be about .8 divided by the square root of N.

Here N=1000, therefore the probability is about .025.

No. Only one option results in “perfectly 500 heads and 500 tails.” Out of 1000 flips, the number of flips that result in heads ranges from 0 to 1000 (and the number that result in tails ranges from 1000 to 0). Exactly 500 flips resulting in heads (or tails) is only one out of 1000 possible results.

A coin can land on other than heads or tails. I was flipping a coin at a local DQ a while back and it landed on its side next to the wall on the table. It rolled a little bit but stayed directly on its side. So, you won’t always get a heads/tails option.

Good find, although I could find only one post (**gr8guy**’s that directly addressed the OP in this thread, and he made a claim without providing a cite.

No. No, no, no. You are assuming that flipping 1 head and 999 tails is the same probability as flipping 500 of each. It is not. Let me start with a lower number.

Out of 4 flips, there are 16 possible outcomes, not 4.

HHHH

HHHT

HHTH

HHTT

HTHH

HTHT

HTTH

HTTT

THHH

THHT

THTH

THTT

TTHH

TTHT

TTTH

TTTT

8 of those outcomes include 2 heads and 2 tails. The chance of flipping perfectly 2 heads and 2 tails is 50%.

The same mathematics scales up to any number of flips.

Damn, I counted wrong but the point is that it’s not 1/1000. I’ll look for an actual formula.

brianmendelez: Sort of. There are 1000 possible values for the result ``number of heads thrown’’, and 500 is one of those values. But all possible values are not equally likely, so the probability of seeing 500 is not 1 in a thousand. As CookingWithGas said, there are not 1000, but 2^1000 possible sequences for the coin tosses, some of which (but I don’t think exactly 50% of which) will give 500 heads. We can expand this for a few tosses:

2 tosses:

HH

HT

TH

TT

Here there are 4 possible toss results (2^2) and three outcomes for #heads; the probability of 50% heads is 50%

4 tosses:

HHHH

HHHT

HHTH

HHTT

HTHH

HTHT

HTTH

HTTT

THHH

THHT

THTH

THTT

TTHH

TTHT

TTTH

TTTT

yielding 16 outcomes (2^4), 5 values for #heads, and 6/16 chance of exactly 50%heads. I don’t have my stats textbooks at hand, and don’t really feel like deriving the distributions for large N from scratch (it *can* be done ), but I’d be inclined to believe that aahala did look up the tables to come up with his answer…

The confusion apparently seems to be on getting *exactly* 500 versus getting *at least* 500. The probability of getting *at least* 500 will be 50%. 500 is also going to be the mean (the most likely outcome) but it will have a small overall chance.

If you make a few assumptions (that the outcome is only heads or tails[sup]*[/sup], and the probability of heads or tails is a fixed value each time), the value is easily found.

The formula for this (the Binomial Distribution) is

P(k successes in N trials) = C(N,k)* p[sup]k[/sup] q[sup]N-k[/sup]

Where p is the probability of success, and q = 1-p (probability of failure). C(a,b) is the choose function, or combinations (see the link for more).

So if N=1000, p=0.5, the answer is about 0.025.

As N gets bigger, the distribution approaches the Normal Disribution (the Gaussian or ‘bell’ curve).

You can recalculate with a different p (representing a weighted coin). For instance if p= 0.51, P(500) = 0.021.

[sup]*[/sup] There is a version with more than two outcomes, called the ‘multinomial’ distribution, which is about as simple. If you assign a low probability to getting a ‘sider’, you can calculate the odds again.

OK, CookingWithGas’s last (2) posts were not there, even when I previewed

I am making no such assumption. I was responding to your statement that “all options are equally likely, but half of those options include a 50/50 split between heads an tails.” You yourself just pointed out that it is not true that “all options are equally likely.” It is also untrue that half the possible results “include a 50/50 split between heads an[d] tails.” The *most likely* results may include an even heads-tails split, but the “most likely” results is not the same thing as “all possible” results. Exactly 500 flips resulting in heads (or tails) is only one out of 1000 possible results.

Forgot to point out that **aahala** had the right answer first; using the approximation when N is large that it approaches the normal distribution.

And I meant **CookingWithGas**’s possible confusion, hopefully anyone who thought it was only 1/1000 sees that there’s more than one way to get exactly 500.

I don’t know how that approximation formula was determined. I presume it had something to do with the binomial formula. I believe the approximation becomes increasingly close to the exact answer as N increases.