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Two gamblers are playing coin toss game: Gambler A has (n+1) coins and B has n coins. What is the probability that A will have more heads than B if both flip all their coins.

Not sure how to go about solving it. Any ideas? What if A has (n+2) coins.

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    $\begingroup$ Possible duplicate of A probability theory question about independent coin tosses by two players $\endgroup$ – lulu May 29 '16 at 0:32
  • $\begingroup$ Hint: After $A$ has tossed $n$ of her coins, and $B$ has tossed all of hers, either (i) $A$ is leading or (ii) $B$ is leading or (iii) they are tied. Note that (i) and (ii) are equally likely. $\endgroup$ – André Nicolas May 29 '16 at 0:39
  • $\begingroup$ @lulu: This one has the additional $A$ has $n+2$ coins question. $\endgroup$ – André Nicolas May 29 '16 at 0:44
  • $\begingroup$ @AndréNicolas Ah, missed that. Thanks. Doubt that case is easy... $\endgroup$ – lulu May 29 '16 at 0:45
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Assuming a fair coin.

Let $X$ be the count of successes in the first $n$ flips by A, $Y$ the indicator that the last flip by A is also a success, and $Z$ the count of successes in $n$ flips by B.

Because $Y$ is independent of $X$ and $Z$, then:

$$\begin{align}\mathsf P(X+Y>Z) ~=&~ \mathsf P(Y=1)\,\mathsf P(X\geq Z)+\mathsf P(Y=0)\,\mathsf P(X>Z)\\ = &~ \tfrac 12(\mathsf P(X=Z)+2~\mathsf P(X>Z)) \end{align}$$

Note by symmetry that $\mathsf P(X>Z) = \mathsf P(Z>X)$ and so ...

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Suppose there are $n+1$ coins for Gambler A and $n$ coins for Gambler B.

The possibilities for Gambler B are $2^n$ and the possibilities for Gambler A are $2^{n+1}$ or $2^n \cdot 2$. There are twice as many possibilities and half of them in which there is an additional heads. Simply put Gambler A has a $50\%$ chance of having strictly more head flips.

But the real insights come when you generalize the problem to say that Gambler B has $n$ coins and flips $k$ heads. What are the chances that Gambler A flips more than $k$ heads given they have $n + m$ coins.

Well the chance, $p$, that someone flips a given number of heads or more than a given number of heads is given by the binomial distribution which can be used to show that the probability of Gambler A achieving at least more heads than Gambler B is given by the following equation:

$\begin{align*} p =\sum_{k=0}^{n} \left({n \choose k} \cdot 0.5^k \cdot 0.5^{n-k} \cdot \sum_{j=k+1}^{n+m} \left( {n+m \choose j} \cdot 0.5^j \cdot 0.5^{n+m-j}\right)\right) \end{align*}$

I'll add more when I can. But in the mean time you may be able to see that $p$ remains $0.5$ no matter how large $n$ becomes as long as $m$ is $1$.

However if $m$ is $2$ the situation becomes exceedingly more complicated because the answer is not constant. Here are the first few cases:

  • $n$ is $1$ then $p$ is $0.6875$.
  • $n$ is $2$ then $p$ is $0.65625$.
  • $n$ is $3$ then $p$ is about $0.63671$.
  • $n$ is $1000$ then $p$ is about $0.50891$.

Gambler A loses his comparative advantage the more coins that are tossed for both Gamblers. But Gambler A will always maintain the advantage and has quite a heavy advantage when few coins are tossed.

When Gambler A has one extra coin he may as well just toss one coin. When Gambler A has more than one extra coins, I don't see a way to simplify things here (which doesn't mean that it doesn't exist).

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  • $\begingroup$ how about the case when m>1, say m=2 $\endgroup$ – Gerry May 29 '16 at 12:37
  • $\begingroup$ @Gerry When $m=2$ the highest $p$ will be is when $n=1$, which is $0.6875$. As $n$ increases the relative advantage of Gambler A decreases and $p$ approaches $0.5$. $\endgroup$ – Marcus Helmer May 29 '16 at 17:18
  • $\begingroup$ Thanks. Can you please elaborate on how you arrived at expression for p in your solution. $\endgroup$ – Gerry May 29 '16 at 20:26
  • $\begingroup$ @Gerry The largest summation, that is $\sum_{k=0}^{n}$ covers all cases of Gambler B flipping $k$ heads from 0 heads to $n$ heads. The second part comes from the binomial distribution and is the probability that Gambler B flips exactly $k$ heads. That is multiplied by the chance that Gambler A flips $k$ or more heads ($k+1$ to $n+m$ heads). So it is a summation of binomial distributions so to speak. The summation essentially adds the chance of Gambler A flipping more than $k$ heads for each possible number of heads that can be flipped for Gambler B. Does that make it clear? $\endgroup$ – Marcus Helmer May 29 '16 at 20:49

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