This has always bothered me, and no one has ever been able to answer it to my satisfaction. I now turn to the SDMB gurus.
If one were to flip a fair coin 100 times, and each time it came up heads, what are the odds it will come up heads if flipped again?
Now, logic and my experience from Pstat tell me that no matter how many times a coin is flipped, the chances of it coming up heads or tails is 50% each. But what if I were to ask the question a little differently: What are the chances a coin would come up heads consecutively 101 times? I do not know the odds that a coin would come up heads consecutively 100 or 101 times, so let us assume the odds are 1 in a million, and 1 in five million, respectively. The root of my misunderstanding is that the same flip seemingly engenders two different probabilities; namely, the next one flip as a single occurance (probability .5) but ALSO represents the 101st flip, with a probability quite astronomical. Therefore the result would be both 1 in 2 AND 1 in five million. I know it has something to do with “adding up” the .5 probabilities, but I don’t get it.
I asked a question like this a few months ago. Before the surly guys get to you…
The odds of getting heads on a fair coin = 1/2
The odds of getting tails on a fair coin = 1/2
Total 1
While it is possible, it is highly unlikely that you would get 100 heads in a row. However, for the 101st flip the odds of getting heads are 1/2 and tails are 1/2. At a certain point you begin to question the fairness of your coin though.
My dad taught me that odds are determined by “The choices times the choices times the choices…”
Since every flip has only 2 choices, you take 2 to the power of the number of flips.
In your scenario, this would give us 2^100. I make this out to be 1.27 x 10^30. This is a number so big that I’m almost certain there isn’t a name for it. Even a weighted, asymmetric coin would almost certainly defy these odds.
The replies so far have been good, but I want to stress something: The old gambling fallacy of an outcome being ‘due’ is just that, a fallacy. Assuming a fair coin, or roulette wheel, or keno system, or any other random system, you cannot predict the future outcome based on the outcomes that haven’t happened yet. The odds for getting a head after getting 1,000 tails is the same as the odds for getting a head on the first flip.
And all outcomes are equally likely. For example, the odds of getting a thousand tails in a thousand tosses is the same as getting any other thousand outcomes in a thousand tosses. No group of outcomes is special.
(I’m sure snopes has done something on this, but I’ve explained the basics here.)
As others have said, the odds that you would flip a head on the 101th toss is 1/2; and as you said, the chances of getting 101 heads in a row is an astronomical number (1/(2.535 * 10^30)). The reason that the chances of getting a head on the 101th toss is 1/2 and not 1/(2.535*10^30) is because we are calculating an individual outcome. To calculate the overall probability however, you are taking all of the flips into account. What do I mean by that? The chances of getting a head on the first one is 1/2, the chances of getting another head on the second is still 1/2, however, the chances of the two happening as a whole is 1/4 because it’s 1/2 * 1/2. Likewise, you can go all the way up to 100 with this - the probability of getting a head on the 101th toss is 1/2 - the probability of getting 100 heads before that is 1/2^100, and therefore the probability of getting 101 heads in a row is 1/2^100 * 1/2 or 1/2^101.
Looking at it from the opposite side as Ian, there is a theorem in probability that the odds of event A happening given that B happened is equal to the probability of A and B happening divided by the probability of B happening. In other symbols:
P(A|B) = P(A & B) / P(B)
Now if A is the event that we get 101 heads in a row, yes the odds are very small. However, B being the odds that we get 100 heads in a row is nearly as small, so they cancel out. What we’re left with is the probability of the 101st flip coming out heads, given that the first 100 did, is 50%.
Don’t worry about getting brain lock thinking about this problem though. Probability has a way of sometimes seeming unintuitive. For proof, look up in the archives for the Monty Hall Problem (which even fooled Cecil briefly) or see this thread.
The problem in the OP is that you’re conflating the two odds calculations, that of a heads coming up, and that of the 101st heads coming up. The two are entirely unrelated to each other.
You’re correct that the odds of all possible outcomes must add up to 1. In the case of any combination of 101 coin flips, the odds of any one combination coming up are astronomically small, whether it’s 101 heads, or a particular but apparently random order of heads and tails, because there are an astronomical number of possible outcomes, each of which is equally likely. Add them up all up, though, and you get 1.
That you’re connecting the two is exactly the Gambler’s Fallacy.
Hey, any hustler will tell you that a coin that flipped heads 100 times in a row is more than likely not a “fair coin”. If I were betting, I am betting heads, and I’m flipping the coin.
This is an application of Bayes’ theorem. When interpreting a long-shot probability, you have to calculate it as a ratio against other long-shot probabilities.
For instance, let’s say that it’s 1/100,000 that a coin should come up heads every time (let’s say that we’re talking about 16 flips or so). Let’s also say that the odds are only about 1/10,000 that the game is rigged. Therefore, the true odds that the next flip will come up heads is about 10:1, and you should bet heads.
Precisely, Stone. In a real situation like this, the information that the coin is fair is not (can never be) an assumed axiom-it’s a piece of empirical information in its own right, and you have to consider the possibility that it’s wrong. Even with a high degree of certainty *a priori[/a], after 100 flips of heads the confidence level in the ‘fair coin’ hypothesis will fall to virtually zero. Bet heads, bet heads …
I believe Rosencrantz and Gildenstern covered this the best… :after 89 consecutive coin flips with heads as the outcome: “List of possible explanations. One: I’m willing it. Inside where nothing shows, I am the essence of a man spinning double-headed coins, and betting against himself in private atonement for an unremembered past. Two: time has stopped dead, and the single experience of one coin being spun once has been repeated ninety-times. On the whole, doubtful. Three: divine intervention. Four: a spectacular vindication of the principle that each individual coin spun individually is as likely to come down heads as tails and therefore should cause no surprise each individual time it does.”
Thought I’d throw in a brief tidbit my statistics prof mentioned a couple years ago, though it’s to do with lottery numbers (6-49), not coins. He asked why no one ever chooses the consecutive numbers 1,2,3,4,5,6 on their tickets - that combination has the exact same probability of coming up as any other six number combo. Kinda makes you feel silly purposely trying to choose the “most likely” numbers on the lottery.
Perhaps this belongs elsewhere, but this probability thread just happened to come up.
Actually, there are quite a few people that pick 1-2-3-4-5-6 for their lottery line (according to an unnamed official), knowing that it’s just as likely to be a winner as 3-7-10-22-28-30. So it’s not a good idea to pick that, not because it won’t win, but because you’ll definately have to split it with a handful of math nerds. :D:D
since this has been covered quite well there’s little left to say but I will try and cover new ground and make it a little more accessible.
let’s say that the odds for heads are exactly 0.5 and the odds for tails are exactly 0.5 Then the odds for 10 heads in a row is 1/1024. These presets are just to eliminate the negligible presence of alternate possibilities.
Then the odds for 11 in a row are 1/2048. The error made is assuming that those are still the odds when you reach 10. What happens is with the first flip the probability that a head would come up went from 0.5 to either 1 if it is heads came up or 0 if tails comes up.
So if you were to flip a coin 10 times in a row and each time a head comes up your odds equation would look like this
0.5 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 1/1024
1 * 0.5 *0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 1/512
1 * 1 *0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 1/256
etc etc until you get to
1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 = 1
and after your 10 flips you’d logically get a 100% chance that the 10 flips you just finished were heads. And when you factor in the odds for the next flip 1 * 0.5 is 0.5 again
sorry if I seemed a little condescending but most of the math I do is aimed at kids.
or how about an even wierder possibility, the coin does not come down at all. a la Newsradio. But this could also occur with some of the most extremely wierd and unlikely workings in our universe. One that I can think of is that, by some super improbablility that would make all our calculators explode in frustration, all the protons in the coin happen to decay at the same time. I think I recall a guesstimate that the half-life of a proton was 1 x 10 ^ 31 s. So multiply that by the number of protons in a dime and that should be the odds of that coin well. I’m not sure exactly what would happen to it, but you would never see it again. I may have calculated the probability of decay from the half-life wrongly, it’s been awhile since I took a 400 level physics course.
Thought these answers do make sense, I wish I would have read them earlier. I was in Vegas last year and saw a roulette<Sp? table that had landed on black over a dozen times in a row…I watched as it hit black three more times. I was so sure that this HAD to end soon, I drpped forty bucks on red…Needless to say, I lost my A** and it kept right on hitting black. I guess thats why it’s a gambling.