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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?

NotSuper
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2 Answers2

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

Graham Kemp
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  • = 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) – NotSuper May 19 '14 at 21:45
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    @NotSuper actually, $(1+6+15)/2^6=22/64=11/32\simeq 0.34$ – glS May 29 '19 at 09:16
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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.

OJFord
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  • 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? – NotSuper May 19 '14 at 01:47
  • 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. – OJFord May 19 '14 at 01:54
  • 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? – NotSuper May 19 '14 at 02:09
  • Yep. Though I'm not liable for any gambling losses ;) – OJFord May 19 '14 at 02:17