Need Help finding probability of events happening consecutively given a number of attempts

Billiam

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Jul 20, 2021
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I know that flipping a coin and getting heads 3 times in a row is .5*.5*.5
however what is the formula for flipping the coin 10 what is the probability that i will have 3 heads in a row within those attempts
 

JeffM

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I presume that “flipping the coin 10” means flipping the coin ten times in a row.

Let’s think about a similar problem. What is the probability that if you flip four times in a row, you will get three heads in a row?
 

Billiam

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I presume that “flipping the coin 10” means flipping the coin ten times in a row.

Let’s think about a similar problem. What is the probability that if you flip four times in a row, you will get three heads in a row?
yeah! i forgot to add "times"

would it me around 25% cause 1 extra chance right? with 1 more toss
 

Dr.Peterson

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yeah! i forgot to add "times"

would it me around 25% cause 1 extra chance right? with 1 more toss
No, the calculation isn't that simple. You'll need to be more precise in your thinking.

How do you calculate a probability? One way, for the simpler problem with 4 tosses, would be just to list outcomes. That can help you think more clearly about the larger problem as well.

You also need to decide whether you mean "exactly 3 heads in a row" or "at least 3 heads in a row". You may also have to take into account that some sets of 10 tosses can include more than one set of 3 in a row. JeffM's simpler problem lets you ignore at least that last issue for the moment.
 

JeffM

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yeah! i forgot to add "times"

would it me around 25% cause 1 extra chance right? with 1 more toss
The concepts involved in probability theory are quite simple; applying them is complex because it is so easy to forget something relevant.

First, please read Dr. Peterson’s wonderful answer.

Second, although it is frequently the hardest method computationally, I find that what is most intuitive for me is to think conceptually using the method that follows. If A and B are mutually exclusive, then the probability of A or B is just the sum of the probability of A and the probability of B. Note that this method will not work unless A and B are mutually exclusive, which means that the probability of A and B is zero, just plain impossible.

Third, you are of course correct that the probability of flipping a fair coin three times and getting three heads is

[math]\left (\dfrac{1}{2} \right )^3 = \dfrac{1}{8} = 12.5\%[/math].

So far, so good. However, it is unclear what you mean thereafter. Why do you add 25%, which would give a total of 37.5%? Or are you adding 12.5%, which would give a total of 25%? Let’s not worry too much about answering those questions because both answers are wrong.

Let’s try my method of thinking (while remembering it may not be the only way of thinking and may not be computationally efficient).

In what ways can I get at least three heads in a row with four flips?

3 heads then a tail.
A tail then 3 heads.
4 heads.

Are they mutually exclusive?

What are their respective probabilities?

So what is their sum?

Fourth, with a problem this small, you can CHECK your answer by listing the sixteen possibilities and counting. Obviously that is not going to work with 10 flips. And as Dr. Peterson indicated, there is a complication with 10 flips that is rather trivial with 4 flips. So here is a hint.

Solve the problem for 5 flips. Use that answer to get the answer for 10 flips.

EDIT: I did not read Dr. Peterson’s answer carefully enough. I assumed that the question was “at least three heads in a row.” There is a reason why our guidelines say to give the problem completely and exactly.
 
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