# Flip a coin 6 times. What is probability of at least 4 heads?

I can figure out the much simpler case of the probability of getting at least 2 heads in 3 coin flips: There are 8 (2^3) ways to flip a coin 3 times: HHH, HHT, TTT, TTH, HTH, HTT, THT, THH. 4 of these contain 2 or more heads. Therefor the probability of at least 2 heads in 3 coin flips is 4/8. How could I have done this without writing out all the possibilities and counting the ones with 2 or 3 H's?

Today the Indiana Pacers won the first game in the 4 out of 7 series with the Miami Heat. Assuming that either team has a 50% chance of winning each of the remaining games, what is the probability of the Heat winning 4 of the remaining games?

• Have you heard of Binomial coefficients? (or nCr combinations?) May 19, 2014 at 1:08
• Yes, I've heard of nCr combinations. 6C4 = 15. May 19, 2014 at 1:15
• Sorry - I just realised the 'at least' modifier. You have to also add 6C5, and 6C6. May 19, 2014 at 1:18
• So the probability is then simply number of ways you get what you want, divided by number of possible outcomes. May 19, 2014 at 1:20
• Please clarify: If Heat loses four of the first four games, are the remaining games played? Answers using the naked binomial distribution may not be correct if the series terminates when one team achieves four losses. Jan 24, 2018 at 1:13

Today the Indiana Pacers won the first game in the 4 out of 7 series with the Miami Heat. Assuming that either team has a 50% chance of winning each of the remaining games, what is the probability of the Heat winning 4 of the remaining games?

There are $6$ remaining games. The desired criteria is that Heat wins at least $4$, when given that Heat lost the first 1. This is a binomial distribution; so named because of the use of the binomial coefficient to count number of permutations of outcomes that match the desired criteria.

The probability of exactly $k$ successes in $n$ trials with probability $p$ of success in any trial, is: $${n\choose k}p^k(1-p)^{n-k} \;=\; \,^n\mathrm{\large C}_k\;p^k(1-p)^{n-k} \;=\; \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$

So: $\mathbb{\large P}(\text{win at least }4\text{ more of }6) = {6\choose 4}\left(\frac 12\right)^4\left(\frac 12\right)^2+{6\choose 5}\left(\frac 12\right)^5\left(\frac 12\right)^1+{6\choose 6}\left(\frac 12\right)^6\left(\frac 12\right)^0$. $$\therefore \mathbb{\large P}(\text{win at least }5\text{ more of }6) = \frac 1{2^6}\left(\frac{6!}{4!2!}+\frac{6!}{5!1!}+\frac{6!}{6!0!}\right)$$

• = 42/64, or .65625 (Actually, the Pacers have won the first game, not the Heat. So the probability of the Heat winning the series is 1 - .65625 = .34375) May 19, 2014 at 21:45
• @NotSuper actually, $(1+6+15)/2^6=22/64=11/32\simeq 0.34$
– glS
May 29, 2019 at 9:16

You rightly suspect that there may be some simpler, formulaic method to find these combinations.

It's given by:

$$\dfrac{n!}{k!(n-k)!}$$

Where $$!$$ denotes the factorial ($$n!=n\cdot(n-1)\cdot(n-2)\cdot ...\cdot 2\cdot1$$), $$n$$ is the 'number of things', and $$k$$ is 'how many things are chosen'.

So in this specific case, $$n=6, k=4$$. That gives us the number of ways we can get 4 heads from 6 flips.

You can probably see how to extend this to your sports example, and lose the restriction that their odds are only as good as a coin-flip.

• Thanks very much. I get 11/32 as the answer to my coin flipping question. However, I don't understand why I should "lose the restriction that the odds are only as good as a coin-flip" for these NBA games. Or if I did put the probability of the Heat winning each remaining game at .55, how would I calculate the probability of the Heat winning the series? May 19, 2014 at 1:47
• Okay, so when you do $P(Win)=\frac{N_{WINS}}{N_{TOTAL}}$, you're implicitly doing $P(Win)=\frac{N_{WINS}}{N_{TOTAL}}\cdot\frac{P(WinGame)}{P(LoseGame)}$. Of course, $P(LoseGame)=1-P(WinGame)$. If you set the odds at 50/50, then you're just multiplying by 1, so you left it out. May 19, 2014 at 1:54
• Just to make sure that I understand you: P(Win) is the probability of the Heat winning the series, and P(Win) = (22/64)x(.55/.45) = .420. Is this correct? May 19, 2014 at 2:09
• Yep. Though I'm not liable for any gambling losses ;) May 19, 2014 at 2:17