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So I'm stuck on a problem. It's related for some work I'm doing but I'll keep it simple here.

Here's the scenario: I flip a coin 11 times and I record the results without telling you what the results are. Now you have to try and predict what I tossed.

What are the odds that you will guess 8 out of those 11 tosses correctly?

Parker Ellertson
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  • Before solving this we need a bit more information. Do we assume the coin is fair? And do you care about the order of the tosses, that is if you tossed HTH and I guess HHT is that still considered correct? – Maarten Punt Jun 21 '17 at 07:31
  • @MaartenPunt yes the coin is fair, and the order matters. – Parker Ellertson Jun 21 '17 at 07:45
  • (1) Do the 8 correct guesses need to be in sequence, or are any 8 out of the 11 positions fine? (2) Are you interested in exactly 8 correct guesses, or at least 8 correct guesses (i.e., would 9 correct guesses count or not)? – Stephan Kolassa Jun 21 '17 at 08:03
  • @StephanKolassa 1: the 8 can be in any sequence 2: I'm interested in exactly eight correct guesses. – Parker Ellertson Jun 21 '17 at 08:10
  • https://en.wikipedia.org/wiki/Binomial_distribution – Glen_b Jun 21 '17 at 08:53

1 Answers1

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If your coin is fair and you are not interested in which 8 out of your 11 guesses are correct, then you are interested in

the number of successes in a sequence of $n$ independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome

which is modeled by a binomial distribution, in your case with parameters $p=0.5$, $n=11$ and $k=8$. The probability you are interested in is given by the probability mass function:

$$ {n\choose k}p^k(1-p)^{n-k}$$

You can calculate this, or use built-in tables or probability density functions from your favorite statistical package. Finally, because I'm never sure I haven't made an error, you can simulate. If all three approaches give the same result, things are looking good. In R:

> choose(11,8)*0.5^8*(1-0.5)^3
[1] 0.08056641
> 
> dbinom(8,11,0.5)
[1] 0.08056641
> 
> nn <- 1e6
> set.seed(1)
> sims <- matrix(runif(nn*11)<0.5,nrow=nn)
> sum(rowSums(sims)==8)/nn
[1] 0.080538
Stephan Kolassa
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  • Given that the OP is asking for odds rather than probability shouldn't it be 0.08/0.92, that is roughly 9:100? – Maarten Punt Jun 21 '17 at 09:26
  • @MaartenPunt: good point, thanks. Considering that the OP accepted the answer, I assume (s)he either used "odds" in a non-technical sense, or did intend the technical sense and was able to work out the odds from the probability himself. – Stephan Kolassa Jun 21 '17 at 09:44