Past Events and ProbabilityDate: 04/05/2000 at 06:05:46 From: Peter Subject: Probability (Past affects the future?) Hi, I will break up millions of tosses into groups of tens. I toss a fair coin 10 times and so overall heads and tails will show about 5 times each, or 50%. Runs will occur through these tens, e.g. tttthhhttt, ttttttthth, etc. Is it correct that in a group of 10 throws 8 heads are a more likely result than 10? If this is so, is it reasonable that after the 5 tails (50%) have occurred it is then heads' turn to some degree in this 10-run, overall? For example, hhhhhhhh, after seeing this run, is it more likely we will see tails in the next two tosses? If the answer is no, then how is it more likely that only 8 heads will occur rather than 10? Is the following distribution correct: 0 heads - ?% 1 heads - 10% 6 heads - 40% 2 heads - 20% 7 heads - 30% 3 heads - 30% 8 heads - 20% 4 heads - 40% 9 heads - 10% 5 heads - 50% 10 heads - ?% What is the correct distribution? I'm baffled as to how you can say the past throws will not affect the future throws, because then the odds of 5 tails out of 10 throws would be the same as 10 tails out of 10 throws. Looking forward to your answer. Peter Sundbye Date: 04/05/2000 at 12:01:29 From: Doctor Rick Subject: Re: Probability(Past affects the future?) Hi, Peter, thanks for writing. In probability, we must be very careful to say exactly what we mean. If we look at the same phrase and mean different things, we will not be able to understand each other. So I will go through what you have said, and try to sharpen up the terminology. In at least one place I think you completely misspoke, but I think I know what you meant. >I will break up millions of tosses into groups of tens. I toss a fair >coin 10 times and so overall heads and tails will show about 5 times >each, or 50%. If you were to repeat the set of ten tosses a large number of times, then the average number of heads per set of 10 would be pretty close to 5. In one set of 10 tosses, you could get anywhere from 0 to 10 heads, but the most likely outcome would be 5 heads. >Runs will occur through these tens, e.g. tttthhhttt, ttttttthth, etc. >Is it correct that in a group of 10 throws 8 heads are a more likely >result than 10? There is a greater probability that, in a set of 10 tosses, 8 will be heads and 2 tails, than that all 10 will be heads. See the calculation below. >If this is so, is it reasonable that after the 5 tails (50%) have >occurred it is then heads turn to some degree in this 10-run, >overall? For example, hhhhhhhh, after seeing this run, is it more >likely we will see tails in the next two tosses? Here I am not sure what you are picturing. If the first 8 tosses have all been heads, the next 2 tosses will still each have a probability of 50% of being heads. The same holds true for each of the next 10 tosses. The coin has no "memory" of how many times it has come up heads before; it starts fresh with the same probability each time. However, the probability that the CURRENT set of 10 tosses (of which you've already tossed 8) will have 10 heads is increased. This is not the past affecting the future; it's the past affecting an outcome that includes both past events (the first 8 tosses) and future events (the next 2 tosses). The future events themselves have the same probability as ever, but when combined with the known past events, the overall probability changes. We'll see that next. >If the answer is no, then how is it more likely that only 8 heads >will occur rather than 10? The probability that 8 out of 10 tosses will be heads is greater than the probability that all 10 tosses will be heads (see below). But the problem you are presenting now is different: "What is the probability that 8 (or 10) tosses out of 10 will be heads, GIVEN THAT the first 8 tosses are heads?" We call this a "conditional probability," and we say that we have "a priori" information (when we have partial information about the outcome). This makes it an entirely different problem, and the probabilities are different. This new problem is equivalent to the following: "What is the probability that 0 (or 2) tosses out of 2 are heads?" The answer is: the probability that there are no (further) heads is 1/4, and the probability that both (remaining) tosses are heads is also 1/4. Therefore the two events have equal probability. Going back to the conditional-probability problem, the probability that 8 of 10 tosses are heads, given that the first 8 tosses are heads, is 1/4. The probability that 10 of 10 tosses are heads, given that the first 8 tosses are heads, is likewise 1/4. > Is the following distribution correct: > > 0 heads - ?% > 1 heads - 10% 6 heads - 40% > 2 heads - 20% 7 heads - 30% > 3 heads - 30% 8 heads - 20% > 4 heads - 40% 9 heads - 10% > 5 heads - 50% 10 heads - ?% It's easy to see that your probabilities aren't correct because they don't add up to 100%, the probability that some number (0 to 10) of tosses will be heads. > What is the correct distribution? If you toss a coin 10 times, there are two possible outcomes for each toss. The total number of outcomes is 2^10 = 1024. Of these outcomes, only one has all 10 heads, so the probability of 10 heads is 1/1024. There are 10 ways to get 1 head (it could come on any of the 10 tosses), so the probability of 2 heads is 10/1024. There are 10*9/2 ways to get 2 heads (the number of combinations of 2 out of the 10 events), and so on. The resulting probabilities are: 0 heads: 1/1024 = 0.0009765625 1 head: 10/1024 = 0.009765625 2 heads: 45/1024 = 0.0439453125 3 heads: 120/1024 = 0.1171875 4 heads: 210/1024 = 0.205078125 5 heads: 252/1024 = 0.24609375 6 heads: 210/1024 = 0.205078125 7 heads: 120/1024 = 0.1171875 8 heads: 45/1024 = 0.0439453125 9 heads: 10/1024 = 0.009765625 10 heads: 1/1024 = 0.0009765625 ------------ 1.0000000000 >I'm baffled as to how you can say the past throws will not affect the >future throws, because then the odds of 5 tails out of 10 throws >would be the same as 10 tails out of 10 throws. I don't understand how you get this conclusion. If my comments above haven't cleared up your bafflement, could you please explain this? I hope my discussion above clears up some confusion, but I doubt that it will answer all your questions. I'd like to hear from you again so we can work through this. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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