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So this is more of a problem with getting an intuitive understanding. I'm sure something like this has been asked before but I couldn't find it because I didn't have a clue what my problem was to start with. Here goes:

Question:

You have flipped a fair coin 9 times and it has landed on tails all 9 times in a row. What is the probability that the next flip will be tails?

My understanding:

If you've flipped 9 heads in a row, and are asked what is the probability that the next flip will be a head, that's not the same as asking what's the probability of flipping 10 heads in a row.

I do understand that each coin flip is completely independent, and so will always be a 50-50 chance of heads or tails.

The problem:

Yet, I'm not quite sure why the probability of flipping 10 heads in a row is different from flipping a 10th head. Is it because we're only being asked to calculate the probability of that one event happening rather than the entire set of events?

I think I may have inadvertently solved my own problem by asking this question because I had to think so much to ask it haha!

I'm going to post it anyway just to ask if you can perhaps give me an example that will help me better grasp this?

EDIT:

I found another discussion here that asked a better, more illustrative question, and the Gambler's Fallacy, specifically the part on coin tosses, best explains the logical problem in my question.

I'm placing this here in the hopes that it helps one of you future readers:

Wikipedia: Gambler's Fallacy

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  • $\begingroup$ everything you have said sounds perfectly correct to me but your question title is a bit misleading $\endgroup$
    – M A Pelto
    Commented Sep 11, 2018 at 4:25
  • $\begingroup$ Yeah, I just edited that because I had a different example in there earlier so I've changed that. Thanks for pointing it out! @MattAPelto $\endgroup$ Commented Sep 11, 2018 at 4:27

2 Answers 2

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As Matt said, you have correctly answered your own question.

This is what is called the "memorylessness" or "memoryless property". Which states that the probability of an event happening does not depend on past trials.

To describe the difference intuitively, I would say, it is like taking a multiple-choice exam and randomly guessing the answers (a,b,c, or d). Here, if you are talking about a specific question, then the probability you have correctly answered it is 1/4. However, the probability of answering the entire exam correctly is quite minute.

There are multiple ways to see this, for example, you could also consider the event space. For each of those questions, you have different types of elements, and different sizes of the event spaces.

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  • $\begingroup$ Thank you! That example you gave really helped me get to the root of the intuitive problem I was having. Great explanation :) $\endgroup$ Commented Sep 11, 2018 at 5:49
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You have the correct understanding, regardless of the previous events, the probability of flipping either H or T is 50-50.

The difference that you note is the correct one! That the probability of a given set of events occurring is different from the probability of the last event occuring given the ones before it. Consider these questions: What are the odds that if you flip a coin three times, you get exactly H,T,H, in that order? What are the odds that you'll get 2H 1T in any order? What are the odds that, given you previously had flipped twice and got H both times, that you will get T the third time? If you can answer these three questions, then you've got the distinction you want. It should also be fairly simple to see how this extends to 9,10, or $n$ flips.

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  • $\begingroup$ Thank you so much! The examples you gave were really helpful! $\endgroup$ Commented Sep 11, 2018 at 5:49

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