7

Where we do NOT require that the heads or tails be consecutive (though they may be!)

Obviously, this expectation, $E[T]$, is bound as follows: $N < E[T] < 2N - 1$

And obviously $E[T] = \sum_{i=N}^{n=2N-1}i*P[Game \ Ends \ On \ i^{th} \ Round]$, where $\sum_{i=N}^{n=2N-1}P[Game \ Ends \ On \ i^{th} \ Round] = 1$

But how would one find such a probability for an arbitrary $i \in \{N, N+1, ..., 2N-1\}$?

TheMathBoi
  • 415
  • You might be asking a question related to negative binomial distribution. check this link https://en.wikipedia.org/wiki/Negative_binomial_distribution – Vamshi Kumar Kurva Aug 09 '19 at 09:41
  • 1
    Well this is Banach matchbox problem... :-) https://en.wikipedia.org/wiki/Banach%27s_matchbox_problem – Olivier Aug 09 '19 at 10:45
  • And, according to the same reference, your expectation will satisfy $2n - \mathbb E[T_n] \sim (2/ \sqrt{\pi}) \sqrt{n}$ - I have not done the calculation again... – Olivier Aug 09 '19 at 10:48
  • I don't think this problem is quite the same as the Banach Matchbox Problem, but it is certainly similar! I'll go and play around with it for a bit. – TheMathBoi Aug 09 '19 at 13:02
  • 1
    @Olivier: according to my answer, $2n-\mathbb{E}[T_n]=2n\frac{\binom{2n}{n}}{4^n}\sim\frac{2n}{\sqrt{\pi n}}$, which agrees. – robjohn Aug 13 '19 at 13:32
  • @TeenPhilosopher : let's bet ? ;-) More seriously, which difference do you see ? – Olivier Aug 13 '19 at 14:58
  • @Olivier: In the matchbox problem, the game goes until a matchbox is found empty upon attempting to get a match rather than when taking the last match. This game ends when the last match is taken from a box. Given that game, it could go on for $2N+1$ turns. – robjohn Aug 13 '19 at 17:36
  • Yep, time to reach the N-th 0 or 1 (whichever comes first) here rather than the N+1-th in the matchbox (time the match is found empty). The parameter has to be adjusted. "Au temps pour moi" :-) – Olivier Aug 14 '19 at 06:53

2 Answers2

2

The probability that the game ends at round $i$ is:$$\binom{i-1}{N-1}2^{1-i}$$If e.g. the game ends with a tail at $i$-th round then among the $i-1$ results in the former rounds must be $N-1$ tails. There are $\binom{i-1}{N-1}$ configurations for that and each of them has probability $2^{1-i}$ to occur. Further we must multiply with the probability that a tail will appear at $i$-th trial which is $\frac12$. Same outcome if the game ends with a head at $i$-th round so a multiplication with $2$ is also needed.

So to be found is:

$$\sum_{i=N}^{2N-1}i\binom{i-1}{N-1}2^{1-i}=2N\sum_{i=N}^{2N-1}\binom{i}{N}2^{-i}$$

drhab
  • 153,781
  • Absolutely brilliant! Thank you so much! I would like to think that I gave this problem an honest and fair try, but it seems that my combinatorics abilities still has quite a ways to go! – TheMathBoi Aug 09 '19 at 13:01
  • You are welcome. Just keep on developing these abilities. Your mathematical maturity will grow then. – drhab Aug 09 '19 at 14:01
1

Preliminary Formula $$ \begin{align} a_n &=\sum_{m=n}^{2n}\frac1{2^m}\binom{m}{n}\\ &=\sum_{m=n}^{2n}\frac1{2^m}\left[\binom{m-1}{n}+\binom{m-1}{n-1}\right]\\ &=\frac12\sum_{m=n-1}^{2n-1}\frac1{2^m}\left[\binom{m}{n}+\binom{m}{n-1}\right]\\ &=\frac12\left[a_n-\frac1{2^{2n}}\binom{2n}{n}+a_{n-1}+\frac1{2^{2n-1}}\binom{2n-1}{n-1}\right]\\[3pt] &=\frac12(a_n+a_{n-1})\\[9pt] &=a_{n-1}\tag1 \end{align} $$ and since $a_0=1$, we have $$ \bbox[5px,border:2px solid #C0A000]{\sum_{m=n}^{2n}\frac1{2^m}\binom{m}{n}=1}\tag2 $$

The Answers

The number of ways to get to $n$ heads on flip $m$ and not on flip $m-1$ is $$ \binom{m}{n}-\binom{m-1}{n}=\binom{m-1}{n-1}\tag3 $$ This is also the number of ways to get to $n$ tails on flip $m$ and not on flip $m-1$. Thus, the probability of getting to $n$ heads or $n$ tails on flip $m$ and not on flip $m-1$ is $$ \frac2{2^m}\binom{m-1}{n-1}=\bbox[5px,border:2px solid #C0A000]{\frac1{2^{m-1}}\binom{m-1}{n-1}}\tag4 $$ Note that $(2)$ shows that $$ \sum_{m=n}^{2n-1}\frac1{2^{m-1}}\binom{m-1}{n-1}=1\tag5 $$ That is, the probability of getting $n$ heads or $n$ tails by flip $2n-1$ is $1$. The expected duration is then $$ \begin{align} \sum_{m=n}^{2n-1}\frac{m}{2^{m-1}}\binom{m-1}{n-1} &=2n\sum_{m=n}^{2n-1}\frac1{2^m}\binom{m}{n}\\ &=\bbox[5px,border:2px solid #C0A000]{2n\left(1-\frac1{4^n}\binom{2n}{n}\right)}\tag6 \end{align} $$

robjohn
  • 353,833