Ten head isn’t less likely than any other sequence. The probability of all sequences is the same. eg:
HHTTHHTTHH has the same probability as
HHHHHHHHH or
TTTTTTTTTT or
HTHTHTHTHT.
There are 1024 possible sequences, and each of them has a probability of 1/1024
DarrenS you are right when you say that 10 heads in a row is no more likely than any other combination. There is 1 chance 1024 of that happening. There is the same chance for these combinations:
THTHTHTHTH
TTHHTTHHTH
THTHHTHTHT
HHTTHHTHTH
but they are not as obvious. Picking any combination of 10 throws before you start would be exactly as unlikely.
don’t ask and Manduck - you are both right of course - thank you. I mis-explained what it is that I don’t get.
Let me try again. I understand the probability of a H on each trial is .5. What I don’t get is why you have to multiply the probabilities to figure out the cumulative probability of n heads in a row. I know the rule is, for independent events, P = P(a) x P(b) - I just don’t know why that’s the rule.
DarrenS – Try thinking about a more simple example. Consider only two flips. Clearly, the four possible outcomes are HH, HT, TH, TT. So, if we want to figure out the probability of getting HH, first think about the first flip. It has a 50% chance of coming up heads.
If it comes up heads, cool. Now do the second flip (same probability). If, however, the first one comes up tails, the second flip doesn’t matter (the only outcomes are now TH and TT).
So we have a 50% chance for getting heads on the first flip. Then, if it comes up heads, there’s another 50% probability of getting heads on the second flip. Therefore, the total probability is going to be 50%*50%=25% The multiplication makes sense here because it is necessary to get the first outcome correct and then the second outcome correct.
For another analogy, it’s like the one-and-one in basketball. If you miss the first shot, the probability that you’re going to get two points on that trip to the line is zero.
Hope that makes some sense (3:00 in morning means that I’m starting to ramble).
It’s only 12:45 here, so I’ll take a crack at it 
Imagine using a pie chart to represent the probabilities. Say you want to know the probability of the two-flip sequence “H T”
So you know that the chance of a head on the first flip is 50%. So, divide the pie in half. One half represents the probability of a head - label it with an H.
The sequence HT is in the half of the pie in which an H is flipped first. Having flipped that H, the probability that the next flip is a T is 50%, so you divide the H half of the pie in half, making a smaller piece that represents the probability of the HT sequence. The size of that piece is 25% of the original pie, so the probability of HT is 25%.
Does that help?
as a mathematician in a model universe getting HHHHHHHHHH does not increase your chances of a T. AS an emperical scientist I would say that it increases your chances of a H as there is likely something odd with your coin
Maybe the chances of each individual flip coming up heads isn’t exactly 50%. Somebody, somewhere, at some point in history must have got “edge”;).
I agree with you. It doesn’t seem quite right that if you get a long string of heads, the probability for the next flip should favor tails.
One way to look at this is once the flips occur, there is no element of probability or luck. The past is just histrorical facts.
While it is true the next flip’s odds is not affected in any way based upon the past, say it was. Is it rational to predict the future on virtually NO evidence. That is, your 10 or 20 flips, instead of all your flips throughout your life, all the flips of that particular coin, all the flips of every coin in the past?
Also, if we intend to flip a coin 20 times, we can reasonably expect to see at least one head, but our expectation see a head on the 20th is simply .5. Once we have flipped 19 times, the 20th is still .5.
They must run it differently here in Illinois. The Big Game lottery in Illinois (multistate lottery actually) has balls numbered 01-52. You must pick five numbers and a sixth ‘Mega Ball’ number. Each ping-pong ball machine has its OWN set of bals ranging from 01-52 so the same number most certainly can be drawn more than once.
Of course, lotteries can be arranged hundreds of ways and I was just going with what I know. However, in the case of the Illinois Lottery what I wrote is possible.
Let me take a crack at the OP. It’s a very common misunderstanding of probability. People have heard that after a large number of coin flips, the “law of averages” states that heads and tails will be the same. Therefore they assume that if you’ve had more heads than tails, you will have to start getting more tails to satisfy the law of averages (whatever that is).
In reality, the difference in the number of heads and tails you’d expect to get will increase with the number of flips. But it will increase at a slower rate than the number of flips, so that the ratio of heads to flips will still approach 50%.
An example with numbers will help. Let’s say you flip 100 times. How many heads should you expect? You’ll probably see somewhere between 45 and 55, giving a ratio of 45% to 55%. Now do the same thing, but flip 10,000 times. The number of heads you expect will be between 4950 and 5050. So in the first case, it’s reasonable to see 10 more heads than tails. In the second case, it’s reasonable to see 100 more heads than tails. So the absolute number of heads and tails didn’t get closer to the 50% number, they got farther away. On the other hand, the ratio now should be 49.5% to 50.5%, so it has gotten closer to the average.
So the law of averages applies to the average, not to the absolute number. In general, if you double the number of trials, the absolute number difference will increase by the square root of two (about 1.4). But the ratio, or average, will get smaller by a factor of 1.4 / 2, which is about 0.7.
Maybe I should be more clear about that. Instead of one machine with balls numbered 01-52 they have six separate machines with balls numbered 01-52 in them. Each one spits out a single ball.
Now I think on it I’m not 100% sure I’m right about that game but I do remember seeing one of the Illinois Lotteries using multiple machines which would allow for same number combinations. I checked the Illinois Lottery web page but it isn’t very clear about any of this.
A wee bit off topic, but this has always been a favored conversation topic of mine when I want to start an argument about statistics.
Remember “Let’s Make A Deal”? Monty Hall would ask you to pick from 3 curtains. Behind one curtain was a terrific prize, behind the other 2 boobie prizes.
One thing Monty would do from time to time was allow you to change your mind. Contestant initially picks curtain 1, Monty says, “well, let me show you what’s behind curtain 2”, and it’s a boobie prize. Monty now asks “Would you like to change your mind and choose curtain 3?”
Statistically, you should always change your mind. In the above situation, chances are 1 in 3 that curtain #1 is correct. Chances are 50/50 that it’s behind curtain 3.
Reason is, when you initially chose curtain 1, chances were 1 in 3 that you were correct. However, now that you’re choosing between 1 and 3 (since 2 has been eliminated), the chance of 3 being correct are 50/50.
Counter-intuitive, but correct. If you don’t believe me, get a piece of paper, have someone play “Monty” with you, and keep track of your “winnings” when you’ve changed your mind.
NOTE: The KEY here is that Monty always shows you a boobie prize curtain before allowing you to change your mind.
You may have the right notion, but your presentation is hard for me to understand. You are saying that you’re more likely to win if you switch, right?
Interestingly, this assumes that either you or the person placing the prize chooses doors with equal probability. If you both have a preference, all bets are off.
Actually, in the Monty Hall problem (search the site, it’s been covered many times including by Cecil) you always switch and you have a 2/3 chance of winning, not 50/50. MH always knows where an empty door is, he’s giving you essentially a choice of two doors against your one door. Always switch, try writing a small computer program to do the simulation.
The easiest way to understand the Let’s Make a Deal trick, is to clone Monty and the studio so there are three.
Have each contestant choose a different door. Switching means one will go from winning to losing, but the other two go from losing to winning.
That particular problem has always been a favorite of mine, too. You are right that you are more likely to win if you switch, but not for the reason given. In fact, the chances that the prize is behind door number 3 are 2/3, rather than 50/50.
The reasoning behind this is that the chance that you picked the correct door on your first try is 1/3. Knowing what is behind door number 2 doesn’t change this at all. Since the probability of the prize being behind the second door is 0, and since the probability must sum to 1, the chances of it being behind door number 3 is 2/3, not 1/2 as most people answer.
To make the logic more clear, consider 1,000,000,000 doors. then the chance that you will get it right on your first try is only one in a billion. Now, the host opens 999,999,998 of the other doors without the prize behind them, and offers you the chance to switch. Is the probability of it being behind each door now 50/50? No. If you stay with your original choice you’d only win once for every billion tries. I’d rather go with the other door, which will be correct 999,999,999 times out of every billion.
Needless to say, I’d hate to be the person that actually has to open all those doors. 
Actually, that’s not quite stated correctly. You seem to be saying that Monty let’s you change your mind only some of the time, but in those times that he does let you change your mind, he shows you a boobie prize. You really need to state that Monty always allows you to change your mind, and always shows a boobie prize before doing this.
Actually, with the problem set up as I described above, there’s a 2/3 chance it’s behind curtain 3.
I have got to start getting up earlier in the morning. CurtC explained beautifully the source of the misconception of “things should average outin the log run, shouldn’t they?” I was all set to post, but couldn’t do a better job of it myself. Read it again, if it didn’t jibe the first time.
Here is a way to look at the tenth flip in a visceral way that can let you understand the fifty fifty chance on a gut level.
Getting ten identical flips in a row is an incredible coincidence.
You just flipped nine straight heads.
Now the fact is that you are in the process of your ten flip test. Getting nine straight heads is an incredible coincidence that has already happened! What are the chances now that the last nine times you flipped, you got heads? Pretty good chance, huh? OK, now you got to flip again. The chance is fifty fifty. If it comes up tails, you have to wait to try the tenth in a row flip until the first nine come up again. Ten in a row is real unlikely. The tenth in a row is fifty fifty, after you have already done nine.
Tris
Oh, and by the way, the easiest way to do the Monte Hall Problem is this:
Dealer lays three cards face down. Contestant picks a card. Dealer discards the lower of the two remaining cards (use suit ranking to differentiate identical cards). Contestant chooses one of the two remaining.
Play for ten hands each strategy, switch or not switch. It becomes obvious, very quickly.
I got this one wrong at first. But then, I had never paid attention to the show. I thought Monte just threw away one at random.
Tris