# (Coin Toss) Probability of Getting 4 Heads

I am stuck on a problem for my stats class. I have not done any work but I will explain why. Here is the problem;

• A coin is flipped until you get a tails. What is the probability of getting at least 4 heads?

I have done probability with coins before, but this question stumped me. How? Because we only have ONE coin, and we don't know how many times the coin is tossed. I know that with one coin, the probability of getting a head is 1. And the number of outcomes is 2. However, I don't know the next step after this, especially when I'm not given information on how many times the coin should be tossed.

Any help would be great. Or maybe just a tip on looking at this problem from a different perspective? Thank you!

• Can you work out the probability that you’ll get tails in four tosses or fewer? – amd Apr 27 '17 at 20:23
• But there is information: the coin is flipped until you get tails – drhab Apr 27 '17 at 20:25
• The coin will be tossed until you get a tails, then the number of times you got a heads will be counted. What is the chance that you got 4 heads? you can get one heads then another then another, you will only stop when you get tails or get 4 heads (when you get 4 heads you don't care about what happens if you keep flipping the coin). That means the probability of getting at least 4 heads is the probability of getting 4 heads consecutively from the beginning, that is 1/16. – Donat Pants Apr 27 '17 at 20:26

I've also started statistics as well.

How I look into this question is the other way around:

Rather than looking for 4 in a row, I look at the probability of not having 4 heads in a row (having the compliment of the probability).

Let say P(A) = Having at least 4 heads before first tail

$P(A') = P(T)+P(H\cap T)+P(H\cap H \cap T)+P(H \cap H \cap H \cap T)$

$P(A')+P(A) = 1 \rightarrow P(A) = 1-P(A')$

$1-P(T)+P(H\cap T)+P(H\cap H \cap T)+P(H \cap H \cap H \cap T)$

$1-[1/2 + 1/2^2 + 1/2^3 +1/2^4] = 1-15/16 = 1/16$

You want the probability of flipping at least four heads before obtaining the first tail.

Thus you want the probability that at least the first four tosses are heads.

This is $1/2^4$.

The probability of getting a heads first is 1/2. The probability of getting 2 heads in a row is 1/2 of that, or 1/4. The probability of getting 3 heads in a row is 1/2 of that, or 1/8. The probability of getting 4 heads in a row is 1/2 of that, or 1/16.

After that... it doesn't matter... you have at least 4 heads.

• This works for $a=1$, but not for other values. – Trurl Apr 27 '17 at 21:06
• Good point... I took the phrase "a tails" to mean one tails. Pretty sure that was the intention of the OP, but your version is more interesting. – Jed Apr 27 '17 at 21:23

There are several possible approaches.

Getting 4 heads to start has probability $(1/2)^4 = 1/16$ as in the comment by @DonatPants.

More formally, outcomes that satisfy your condition are HHHHT, HHHHHT, HHHHHHT, etc. So the total probability is the geometric series with probability $A = (1/2)^5 + (1/2)^6 + (1/2)^7 + \dots .$

There is a formula for summing this series. If you don't know it, you can note that $(1/2)A = (1/2)^6 + 1/2)^7 + \dots,$ so that $A - (1/2)A = = (1/2)A = (1/2)^5$ and $A = (1/2)^4,$ which is the same as the previous answer.

I do not know why @MarkusStuhr has withdrawn his Answer. The explanation by @Joel (+1) as does the answer by @manmood (+1) that appeared while was typing this. I hope one of these explanations is clear to you. The key points throughout is that we're assuming the coin is fair [$P(H) = 1/2$] and that the tosses are independent.

Also, if your book includes the geometric distribution, you should look at that because it is related to this problem. I don't want to discuss the geometric distribution in this Answer because there are at least two versions of it, and discussing the wrong one might be confusing.

Well, h=heads, t=tails Sample space={(hhhh),(hhht),(hhth),(hhtt),(hthh),(htht),(htth),(httt), (thhh),(thht),(thth),(thtt),(tthh),(ttht),(ttth),(tttt)} There are 16 total outcomes, and only 1 of these outcomes results in 4 heads. This means the probability of landing all 4 heads in 4 tosses is 1 out of the 16. So the answer is 1/16. As for the "before flipping a tail", it doesn't seem to matter because no matter the outcome there is still only 1 in 16 chances to get 4 heads flipped.