Depends. If we pick the final result first, the chances are 50/50, as expected. If we choose it randomly from what remains after you reveal 12 heads, what is probabilty? If you just hapened to pick 12 heads, then the odds are still 50/50 for the next toss. If you knew the order of the tosses (as it sounds like you would) and deliberately revealed 12 heads, the odds are much, much higher that the randomly chosen toss will be tails, since in this case, there was already a finite set of results, and you removed most of the heads from consideration by revealing them (like picking most of the black marbles out of a jar of black and white marbles–mostly white marbles will probably be left).
What I think makes this question interesting is the fact that it’s unanswerable without discussion issues of (or issues which appear to be of) motivation, which aren’t usually part of probability questions. So if I phrase it as “I flip a coin 20 times, then show you that flips 1, 3, 4, 6, 7, 8, 14, 16, 18, and 19 are all heads, what is the probability that 11 is heads” is basically unanswerable without knowing what method I was using to determine which flips to reveal, whether to reveal at all, etc.
This actually depends where he’s playing. American roulette has a Green 0 and a Green 00. My understanding is that European roulette tables have only Green 0. (some single 0 tables can be found in north america, but rarely)
Betting on red in America: 47.37% chance of winning (house edge 5.26%)
in Europe: 48.65% chance of winning. (house edge 2.7%
This coupled with the fact that the payoff doesn’t match the risk means the house always wins. I don’t have the numbers in front of me, but generally a red bet pays 1:1, while the actual risk is 38:36. A single number bet pays off at (I think) 1:35. If you bet $1/spin on 36, and by fluke it came every other number first, you’d be down $3. (38 bets, a single $35 winner)
in chaos,
Pandamonium
–
Ugh. Anybody who says that probability and statistics are easy is nuts. I knew a math guy, a math major, who had taken several math stat & prob theory courses for various requirements and, even though he was an A student in the one I had with him, he still said that this stuff never really makes sense. ::shrug::
Okay, Shijinn, you might benefit from the notion of “conditional probability.” Here’s the introductory joke: A guy is going on an airplane. He remarks to his companion that he is carrying a bomb. In response to his companion’s horror, he says, “Hey! What are the odds of two people carrying bombs on the plane?!” The other introductory joke is from The World According to Garp (or whatever it was called) where the plane crashes into a house for sale and the potential buyer is smitten because the house is “pre-disastered.”
I hope that those jokes are obviously obsurd. A conditional probability is the probability of some event given that some other event has occured. For example, the probabiltiy that a female murder victim was murdered by her husband is (I’m making up a number) 1-in-25,000. Suppose that her husband was a wife-beater, then we have a new probability: The odds that she was murdered by her husband given that he was a wife beater is 1-in-10. This would reflect our experience that wife-beaters tend to murder their wives more than the population of husbands in general.
Another example would be the odds of a batter getting a base hit. His odds are, let’s say, .250, i.e. he’s got a 1-in-4 chance of making a base at bat. But suppose we know that he is facing a left-handed pitcher (sp?), may be significantly worse, say .200, i.e. he’s got a 1-in-5 chance of hitting against a leftie. So, we have a conditional probability: The odds of him getting a base given that he is hitting against a leftie is 1-in-5.
These two examples are significantly different from the two opening jokes because in my examples the odds change based on the “givens”, whereas in the jokes they don’t. The odds of a terrorist carrying a bomb on a plane are in no way affected by whether some random non-terrorist has one as well. These events are independent! A coin is independent of previous flips. This is true even for unfair coins!
Let’s assume that we have a fair coin. You see ten heads in a row. The odds of getting another head given that you’ve seen ten heads are still 1-in-2, and the odds that you see tails given that you’ve seen ten heads is 1-in-2. To say that the conditional probability changes as a result of seeing a certain number of heads is to say that the events aren’t independent and, in this case, is just a rephrasing of the odds pressure joke made by Keilor.
Suppose that instead of flipping a quarter, we’ll spin a penny on a table. Here the odds of getting heads up are significantly smaller than getting tails up—about 30% of the time we’ll see a heads, IIRC, because the heads side is heavier. We can say that the coin isn’t “fair” under these circumstances because it doesn’t give us a 50-50 chance. But the coin will still be independent. The odds of a heads on the first spin is 30%. If we spin and get five heads, the odds of a head given that we’ve seen five heads is still 30%. If we have an a priori assumption that the coin is a ramdomly chosen penny, this will not change no matter how many heads in a row come up.
A different, but related, question has to do with stastical inference: How do we decide whether a coin is indeed “fair.” Suppose we know nothing about a coin, how do we determine whether it is a fair coin? Well, that answer will be handled better by others, and it deserves another thread, IMO.
So if we leave aside the question of statistical inference regarding the fairness of the coin, the odds on a coin flip are completely unaffected by any pattern of outcomes that preced the current flip. To say otherwise is to say that the universe somehow makes fine adjustments to the physics of a coin flip to ensure that the odds come out “fairly.” You simply can’t wait for some pattern to appear and act on it, if you have reason to assume that the coin is fair. And even if the coin is unfair, you just change the probability of a given flip, but the independence doesn’t go away. So the magician who can flip heads 90% of the time through sleight of hand is bound by the same rules of probability as the gambler in a casino. The chance that the magician will flip a head is 90% no matter how many heads he has flipped previously.
(If you want to infer whether you are dealing with a magician’s trick or a regular person flipping a coin, then you’re into stastical inference. That is something I am not going to address.)
I wanted to note that in high school we engaged in casually gambling based on flipping quarters. If I could get away with it I could flip a quarter in the air and see the quarter into my hand. Even though it was spinning rapidly I could tell how it would land. Then when I transferred the coin from my right hand to reveal it on the back of my left hand I could flip it with my palm to achieve the desired result.
I can’t say I played with the sharpest tools in the shed, but I am sure some of them still believe that some people are just lucky.
If you want to win a quick coin flip against someone and can catch them off guard, the old “Heads I win, Tails you lose” is always fun.
obviously i’m the only one who don’t get this. bangs head on wall
well i’m here to learn right? i must thank everyone for your patience, do bear with me while i try to frame this question again. my question is sort of different from the OP’s *, i’m looking at the beginning of that 1000 head instead of the end.
- first i already understand the premise that varying your bets does nothing to change the expectation or the odds. the house will statistically get their cut in the long run.
- second i also understand that every single game is independent of each other. getting a sequence of heads does nothing to affect the next throw. that is, every throw is 50/50 whatever happened before.
- third i see there is a betting limit. let’s just ignore that for this question.
here’s my assumption - that 11 bad flips will only happen about 1 in 2048 games. i walk in dumping $1024 with the idea that i have 2047 games to win >$1024 ($1 at a time…) by jumping in after 1 bad flip everytime.
i know that every single flip is independent of each other, i know that every single throw is 50/50. but…, i also know that a streak of 11 comes 1 in 2048 games. why can’t i start on the second game where i can look forward to 2047 games thinking that a streak of 11 can only happen about once? why is the first bad flip so important that the probability is halved with only one bad flip out of eleven?
why does perspective matter? for a person trying to flip 11 heads in a row, he’d think he would need to flip at least 2048 times for it to happen. to an observer who sees the first head appear, it would take only 1024 times instead. surely it does not take 1024 attempts just to get your very first head?
where is that wall?
- World Eater, i just realised this is a long hijack, hope you don’t mind…
Mr 
These two statements are inconsistent. If you really understand that each game is independant, you must know that once the first game is played, its outcome can have no effect on the rest of the games.
There’s your error. A streak of 11 does not come 1 in 2048 games. (Game is very vague BTW. I assume you mean one flip of the coin is what you are calling a game.)
A streak of 11 heads will occur, on average, over the very long haul, about 1 in 2,048 games. That’s a very different statement than the one you made. Ponder that for a while, see if you can feel the difference.
In the Monty Hall puzzle, the results may be counterintuitive, but they are easily derivable and unarguable, as long as the question is stated unambiguously. The probelm is that the question is usually stated verbally in a way that allows for several mathematical interpretations, which lead to different results. The arguments almost always add up to various parties saying, “No no, the question was asking this, not that!”
If I understand your point, the correlation between them is a failure to understand when conditional probabilities apply. In Monty Hall, the guesser doesn’t understand that their choice affects what Monty does, and therefore needs to be taken into account. In the gambler’s fallacy, the gambler falsely believes that past events affect future ones. In one case, conditional probabilities are not applied when they should be, in the other case, they are applied when they should not be.
Sorry, I hadn’t gotten down this far yet. Max, I agree with you… since the methods/motivations weren’t revealed in the question, it is impossible to specify the appropriate mathematical model.
You took 10 paragraphs to say what I just said, and gave me an “ugh”? 
My feeling is, probability and statistics are easy. It’s trying to verbally describe what they say that’s hard.
Nope, I gave probability an “ugh” because it is hard. You may find it easy, and that is your good fortune. John Allen Paulos can’t grasp even the most basic ideas in economics, if his critique of More Guns, Less Crime is any indication; David Suzuki can’t grasp the notion of an externality; and I went to school with an (ABD) mathematician who often had a terrible time with some pretty basic economic concepts; yet, all that is more-or-less obvious to me. I can’t explain why I get it and others don’t; but to blow off other people’s struggle as an inability to grasp something “easy” is both dishonest and unfair. Ditto for probability & statistics.
Suppose it were true that since 11-head streaks occured once in 2,048 games, one could wait for an 11-head streak to occur and then bet on it not occuring again for another 2,048 (or so) games. The it would also be true that if there were a streak of 1 head, we could wait for that 1-head streak and bet against heads on the next flip.
There is no reason why an 11-head streak is more special than a 10-head streak, so we could apply that strategy to a 10-head streak. And to a 9-head streak, and to an 8-head streak, and so on. When we get to a 1-head streak, we would find, if the strategy worked, that for a given flip there is a 50-50 chance of heads, but on the next flip the chance of heads would be considerably smaller. If the 11-head strategy you describe works, then the odds on coin flipping would flip-flop with every toss—heads on the first flip implies 90% chance of tails on the next flip, which implies (if you get a tails on flip 2) a 90% chance of heads on the third flip, and so on. But then independence is ruined.
I don’t mind at all Shijinn, my question has been answered, and I finding the the discussion to most interesting.
If anyone feels like explaining that Martingale stuff, feel free, just remember to use real small words and to talk very…slowly. 
I’d be interested in hearing the basic idea behind martingales, myself. I ran screaming after the first semester of graduate statistics, seeing only a rewording of real analysis and measure theory. I know martingales were to be covered in the second semester.
On the other hand, is this more a topic for a new thread?
just the basic you say? the idea behind martingales is to bet a large sum of money against a streak of results. the appeal of say, a 10-step martingale is that a casual gamer do not often see streaks of 10, thus the assumption is that if a streak of 10 do not occur, the gamer will be ‘guaranteed’ to win. the catch is that streaks of 10 do happen, and especially so with the large number of games needed to win enough just to finance the capital needed for such a system.
i don’t see the difference, in the very long haul the idea of stepping in later would still net a profit so long as the probability remains close to 1/2048. well it doesn’t matter anyway, i don’t see any practical way to repeatedly step in halfway using the martingale. with the same view, i still don’t see why a single step in the sequence will half the required games to get a streak of 11…
not so. if i flipped a coin 1000 times i will very well expect to get about 500 heads. this does not ruin independence, just what we would expect to happen. (even if it doesn’t)
thanks everyone, especially js_africanus. thinking about that 1 head streak made me realise my mistake is to interpret 1/2048 as ‘probably happen about 1 time in 2048 trials’ instead of ‘odds of event happening is 1 out of 2048 combination’ or just ‘probability of 5%’. it seems so simple now i’m not sure why i got it wrong in the first place. must have been greed.
Okay, if that’s all the detail you can give with “just the basics”, try an explanation that would work on someone who’s got half a year of graduate-level probability theory under his belt and isn’t afraid to use it.
that would be someone else’s job then, or perhaps wikipedia can help?
There is no required games to get to a streak of 11, that’s your fallacy.
You continue to believe that there are 2,047 flips in a row (stop using the term games, that is meaningless) without a streak of 11 heads, and then suddently on slip 2,048, the heads start rolling. That just isn’t true. The sequence of 11 heads can happen at any time during the 2,048 flips. It can happen on flip 1. Or not at all, or twice, or 18 times. You seem to accept this but then you make statments like the above which just don’t make any sense. You don’t know when the streak may come, so there’s no way to jump into some point of the sequence and increase your odds.