1
$\begingroup$

I had an assignment where I had to calculate the probability that B brings more tails if B tosses a fair coin n+1 times than A who tosses the coin n times. That probability was $\frac{1}{2}$.

Our professor suggested that the other way round question (that is: A tosses n, B tosses n+1, what's the probability A brings more tails than B?) is a more difficult one and I became curious. I tried using the probability I've already found by following the thought process below:

Assume A and B both have the same probability to bring more tails than the other (by symmetry). Let E be the situation where A gets more tails than B, so:

$\mathbb{P}[A_T > B_T] = 1 - \mathbb{P}[A_T \leq B_T]$

However, I know that $\ \{ A_T \leq B_T \} = \{ A_T < B_T \} \cup \{ A_T = B_T \} $.

I know the probability $\mathbb{P}[A_T < B_T]$ from the assignment question, so I went on to look for $\mathbb{P}[A_T = B_T]$. Both $A_T$ and $B_T$ follow a binomial distribution, so I went ahead and calculated that, but the result is attrocious.

I went on here and found this question, which is the same: What is the probability to get more heads in n-1 tosses than in n tosses ? (fair coin) , but the answer provided was a bit difficult for me, as I don't know what $C_n$ is. Our professor also suggested that the correct answer is again $\frac{1}{2}$, so I'm not sure.

(Sorry in advance if uploading the same question is invalid, but I couldn't understand the explanation provided and seeing as it's from 2018 I'm unsure whether I'll get an answer. :/)

$\endgroup$
8
  • $\begingroup$ Assuming, that the original assertion is accurate, re A:$(n)~$ tosses, B: $(n+1)~$ tosses, probability equals $(1/2)$ that B gets more tails than A, you have a shortcut to the reverse problem of A getting more tails than B. That is, focus on the probability that A:$(n)~$ tosses, B:$(n+1)~$ tosses results in a tie, re the number of tails tossed. $\endgroup$ Nov 10, 2021 at 20:25
  • $\begingroup$ Re previous comment, the tie could result from each of A, B getting exactly $k$ tails, where $~k \in \{0,1,2,\cdots,n\}.$ See also Binomial Distribution, specifically $\displaystyle \binom{n}{k}p^kq^{(n-k)}.$ $\endgroup$ Nov 10, 2021 at 20:28
  • $\begingroup$ I tried doing that by saying that $\mathbb{P}[A_T=B_T]$ is basically the probability that they have a tie at $n$ throws and then B gets heads on his $(n+1)^th$ throw plus the probability that A has $k$ tails by his $n^th$ throw, B has $k-1$ tails and the latter brings tails in his $(n+1)^th$ throw. Thus: $$ \mathbb{P}[A_Tn=B_Tn] \cdot \mathbb{P}[B_Tn+1=H]+\mathbb{P}[A_Tn=k and B_Tn=k-1]$$ I know that $\mathbb{P}[A_Tn=B_Tn]\mathbb{P}[B_Tn+1=H]$ is $(1-2p) \frac{1}{2}$ and then I tried calculating the rest using the binomial distribution, but my result was very weird and didn't look like 1/2 :( $\endgroup$
    – Tita
    Nov 10, 2021 at 20:30
  • $\begingroup$ That is not the approach that I am suggesting. If B tosses the coin $6$ times, his chance of exactly $3$ tails is $\displaystyle \binom{6}{3} \times 2^{-6}$. Correspondingly, if A tosses the coin $5$ times, his chance of exactly $3$ tails is $\displaystyle \binom{5}{3} \times 2^{-5}$. So, the chance of A and B both getting $3$ tails is $\displaystyle 2^{-11} \times \binom{6}{3} \times \binom{5}{2}$. $\endgroup$ Nov 10, 2021 at 20:33
  • 1
    $\begingroup$ Assume that A tosses $n$ times, B tosses $(n+1)$ times. Let $R$ denote the chance that $B$ tosses more tails, $S$ denote the chance that B tosses the same number of tails as A, and $T$ denote the chance that A tosses more tails. Presumably, $R$ is known, you want to compute $T$, and you know that $R + S + T = 1$. Also, $$S = \sum_{k=0}^n \left[2^{-(2n+1)} \times \binom{n}{k} \times \binom{n+1}{k}\right].$$ $\endgroup$ Nov 10, 2021 at 20:56

0

You must log in to answer this question.

Browse other questions tagged .