coin toss probability

[Fteacher90]

ok so i think i get it, but i wanted to see if someone could help correct me if im wrong.

So lets say you have a coin that is flipped 3 times for a total of 8 outcomes.
how many times can it land 2 HH or 2TT

assuming the 8 outcomes would look something like this:

# of coin flips	1	2	3	4	5	6	7	8
outcomes	HHH	TTT	HHT	TTH	HTH	THT	HTT	THH

since i need exactly 2 HH i would count flips in 3,5,8 out of the total 8 so that would be 3/8 correct ?
the same would apply for exactly 2 TT 3/8

am i doing it right ?

thank you in advance

 

[pka]

So lets say you have a coin that is flipped 3 times for a total of 8 outcomes.
how many times can it land 2 HH or 2TT
assuming the 8 outcomes would look something like this:

# of coin flips	1	2	3	4	5	6	7	8
outcomes	HHH	TTT	HHT	TTH	HTH	THT	HTT	THH

since i need exactly 2 HH i would count flips in 3,5,8 out of the total 8 so that would be 3/8 correct ?
the same would apply for exactly 2 TT 3/8 am i doing it right ?

In these questions precise language is every thing.
I you mean toss a coin three times to get exactly two heads and one tail then $\displaystyle \#3,~\#5,~\&~\#8$ are correct.

 

[Dr.Peterson]

The answer is correct; but the question is wrong!

You can't really ask about "how many times it can ..."; and you can't assume that in any 8 sets of flips you will get exactly the 8 possible outcomes. What you are calculating is not "can" or "will", but probability. That means that if you repeated this experiment not 8 times, but 8000, or 8 million, or 8 quadrillion times, you can expect that about 3/8 of the time you will get exactly two heads.

Your list is not really a list of actual outcomes, but of all possible outcomes; and you are assuming (rationally) that each outcome, in the long run, will be equally likely.

 

[JeffM]

Like the others, I want to stress that being exact in your language is critical in math, and especially in probability, which turns out to be quite un-intuitive. So I shall critique how you posed your question.

So lets say you have a coin that is flipped 3 times for a total of 8 outcomes.

Now this cannot possibly be correct. If you flip a coin three times, you will get a single outcome each time, for a total of three outcomes.

What you mean is that you may get any one of eight possible sequences of outcomes.

how many times can it land 2 HH or 2TT

Again, your language may confuse yourself. 2 HH almost seems like FOUR heads, which is impossible on three flips. What you are really asking is "how many of those sequences contain exactly two H?" One reason phrasing it that way is that it clarifies that you are not interested in the sequence of heads and tails, only the number of heads. If you think back to permutations and combinations, you will remember that whether permutations or combinations is relevant depends on whether order is relevant. You are not interested in order if you are just asking how many.

[QUOTE}assuming the 8 outcomes would look something like this:

# of coin flips	1	2	3	4	5	6	7	8
outcomes	HHH	TTT	HHT	TTH	HTH	THT	HTT	THH

since i need exactly 2 HH i would count flips in 3,5,8 out of the total 8 so that would be 3/8 correct ?
the same would apply for exactly 2 TT 3/8

am i doing it right ?
[/QUOTE]
Your logic is impeccable, but your language may mislead you in the future. Here is how I would reword it.

"The 8 possible sequences are: then your table

Because three of those eight sequences, namely numbers 3, 5, and 8, contain exactly two heads, the probability of two heads is 3/8."

You were right on in your solution. But your language may cause confusion in more complicated situations.

 

[pka]

Here is the outcome space for tossing a coin four times. $\displaystyle 2^4=16$
\(\displaystyle \begin{array}{*{20}{c}}
H&H&H&H\\
H&H&H&T\\
H&H&T&H\\
H&H&T&T\\
H&T&H&H\\
H&T&H&T\\
H&T&T&H\\
H&T&T&T\\
T&H&H&H\\
T&H&H&T\\
T&H&T&H\\
T&H&T&T\\
T&T&H&H\\
T&T&H&T\\
T&T&T&H\\
T&T&T&T
\end{array}\) We see that there are sixteen sets of four each.
But how in the world can one make that orderly listing.
Here is the recipe. We need four & sixteen rows.
In the first column there are eight $\displaystyle H's$ followed by eight $\displaystyle T's$
Next column cut things in half, four $\displaystyle H's$ followed by four $\displaystyle T's$ repeated.
Next column cut things in half again, two $\displaystyle H's$ followed by two $\displaystyle T's$ repeated twice.
The last column the $\displaystyle H's~\&~T's$ alternate. (half of 2 is 1)

This trick can be very useful in many cases. Try it with five tosses.

 

[Fteacher90]

The answer is correct; but the question is wrong!

You can't really ask about "how many times it can ..."; and you can't assume that in any 8 sets of flips you will get exactly the 8 possible outcomes. What you are calculating is not "can" or "will", but probability. That means that if you repeated this experiment not 8 times, but 8000, or 8 million, or 8 quadrillion times, you can expect that about 3/8 of the time you will get exactly two heads.

Your list is not really a list of actual outcomes, but of all possible outcomes; and you are assuming (rationally) that each outcome, in the long run, will be equally likely.

i understand what you are explaining, thank you for all of the responses. I was trying to remember the way the question was worded. i know that the table above is what was being used to ask those questions that i posted. I was trying to understand this concept myself since its still very new to me.

 

[Fteacher90]

Like the others, I want to stress that being exact in your language is critical in math, and especially in probability, which turns out to be quite un-intuitive. So I shall critique how you posed your question.


Now this cannot possibly be correct. If you flip a coin three times, you will get a single outcome each time, for a total of three outcomes.

What you mean is that you may get any one of eight possible sequences of outcomes.



Again, your language may confuse yourself. 2 HH almost seems like FOUR heads, which is impossible on three flips. What you are really asking is "how many of those sequences contain exactly two H?" One reason phrasing it that way is that it clarifies that you are not interested in the sequence of heads and tails, only the number of heads. If you think back to permutations and combinations, you will remember that whether permutations or combinations is relevant depends on whether order is relevant. You are not interested in order if you are just asking how many.

assuming the 8 outcomes would look something like this:

# of coin flips	1	2	3	4	5	6	7	8
outcomes	HHH	TTT	HHT	TTH	HTH	THT	HTT	THH

since i need exactly 2 HH i would count flips in 3,5,8 out of the total 8 so that would be 3/8 correct ?
the same would apply for exactly 2 TT 3/8

am i doing it right ?
Your logic is impeccable, but your language may mislead you in the future. Here is how I would reword it.

"The 8 possible sequences are: then your table

Because three of those eight sequences, namely numbers 3, 5, and 8, contain exactly two heads, the probability of two heads is 3/8."

You were right on in your solution. But your language may cause confusion in more complicated situations.

yes, you were correct in all of the assumptions you posted. I realize that the language is confusing and will try to learn how to properly word it since this is a new concept for me.

So the questions were asking for only 2 HH or 2 TT and so i figured that meant only 2, not 3 in any combination as long as it had 2 heads or 2 tails...again new to the concept so still learning.

i will watch additional videos on permutations and combinations to try to understand better on phrasing and wording sentences.