Math Help - n coin flips

1. n coin flips

The probability of an unbiased coin landing on heads or tails is 50/50. What are the chances of flipping the coin 2 times and having them both land on heads. Calculate the probability of flipping the coin 100 times and it landing on heads 100 times in a row. Calculate the probability of it landing nth times in a row with nth flips.

Despite each individual flip having a 50/50 chance to land on heads or tails, explain why as the number of flips increase, the probability of flips landing on heads consecutively drops dramatically

I obviously understand that the coin has a 50% chance to land on heads. For the next flip it also has 50% chance to land on heads but to work this out, do I take the percentage chance and divide it by the number of flips?

$\frac{0.5}{2}\rightarrow \frac{0.25}{3}\rightarrow \frac{0.08333}{4}\rightarrow \frac{0.0208325}{5}=?$

not sure if I'm doing this correct, god this is tough lol

What am I doing wrong? And why does the chance of getting another head in a row get less an less probable despite having a 50/50 chance?

**EDIT** I just googled to see what the answer is and for 10 heads with 10 flips the probability is $\frac{1}{1024}$ but I can't figure out the formula.

*EDIT** also just googled for 100 heads with 100 flips and the insane answer is $7.88644\times 10^{-31}\%$ or $\frac{1}{126,799,924,934,444,438,921,525,105,176,6 10}$ lol that is insane!

2. Re: n coin flips

The probability of getting a heads on any particular toss is indeed 0.5, but when we are asked what is the probability of getting $n$ heads in a row, we must consider the is asking what is the probability of getting heads the first time AND getting heads the second time AND getting heads the tirhds time all the way yup to $n$ times. So, we apply the special multiplication rule:

$P(\text{heads }n\text{ times})=\left(\frac{1}{2} \right)^n=\frac{1}{2^n}$

This is actually a special case of the binomial probability formula, where the number of favorable outcomes is equal to the number of trials:

$P(X)={n \choose n}p^n(1-p)^{n-n}=p^n$

3. Re: n coin flips

Hi,
I hope the 2 attachments shed some light on the problem.

4. Re: n coin flips

Originally Posted by MarkFL
The probability of getting a heads on any particular toss is indeed 0.5, but when we are asked what is the probability of getting $n$ heads in a row, we must consider the is asking what is the probability of getting heads the first time AND getting heads the second time AND getting heads the tirhds time all the way yup to $n$ times. So, we apply the special multiplication rule:

$P(\text{heads }n\text{ times})=\left(\frac{1}{2} \right)^n=\frac{1}{2^n}$

This is actually a special case of the binomial probability formula, where the number of favorable outcomes is equal to the number of trials:

$P(X)={n \choose n}p^n(1-p)^{n-n}=p^n$
Just out of curiosity, how does this differ to winning the lottery, the odds of winning are 1 in 14million. How does this compare to the odds I posted of 100 flips with 100 heads. How many more times are you likely to win the lottery than you are to get 100 heads with 100 flips

5. Re: n coin flips

You are about $9.05464714448735\times10^{22}$ times as likely to win the lottery as to flip a coin 100 times and get a predetermined outcome for all 100 flips, but you are about $6.36280810788257\times10^{37}$ times as likely to flip a coin 100 times and get a predetermined outcome for all 100 flips as you are to shuffle a standard deck of 52 playing cards and get the cards in a particular order. The number of ways to arrange a deck of playing cards is about 84 billion times as great as the number of atoms in our solar system.

6. Re: n coin flips

Wow... that just goes to show how much greater your chances are to win the lottery compared to 100 predetermined coin flips, and then how much greater your chances are to get 100 predetermined coin flips compared to a predetermined card deck shuffle