How to split a pot of $100 when a game of flipping coins until first person gets 10 heads is interrupted.(A variation of the points problem)

Question

Question :

Andy and Beth are playing a game worth $100. They take turns flipping a penny. The first person to get 10 heads will win.

But they just realized that they have to be in math class right away and are forced to stop the game.

Andy had four heads and Beth had seven heads. How should they divide the pot?

This is a question from the book Probability: With Applications and R by Robert P. Dobrow .Now the given answer is :

P(Andy wins) = 0.1445. Andy gets $14.45 \;and\; Beth\; gets\; $85.55

But I am not getting this answer . So can someone check my solution and tell me where I am wrong , or else provide a better solution . Thanks in advance.

Proposed Solution : Let Andy be A and Beth B: A needs 6 heads to win, and B needs 3 heads to win.

Let $X_a$ be the number of times A needs to flip the coin to get 6 wins , and $X_b$ be the number of flips for B to get 3 wins . Both $X_a$ and $X_b$ are Negative Binomial Variables with the $X_a$ having parameters 6 and $p=\frac{1}{2}$ and $X_b$ having parameters 3 and $p=\frac{1}{2}$.

Now if A needs $k$ flips to win. Then the required probability is :

$P(A \;wins) = P(X_a = k)*P(X_b > k)$(i.e A wins in $k$ moves and B gets 3 heads after $k$ flips)

$P(X_a = k)=(^{k-1}C_5)*(\frac{1}{2})^6(\frac{1}{2})^{k-6}=(^{k-1}C_5)*(\frac{1}{2})^k$(Negative Binomial Distribution formula)

$P(X_b > k) =$ The 3rd head should come after $k$ trails , therefore until $k$ trails only $0\;or\;1\;or\;2$ heads can occur.

Therefore : $P(X_b>k) = \sum_{i=0}^2{^{k}C_i}*(\frac{1}{2})^i(\frac{1}{2})^{k-i}=\sum_{i=0}^2{^{k}C_i}*(\frac{1}{2})^k$

So $P(A\;wins) = (^{k-1}C_5*(\frac{1}{2})^k) * (\sum_{i=0}^2{^{k}C_i}*(\frac{1}{2})^k) $ . Now minimum 6 tries are required for A to win, so $k:6 \to \infty$

$\begin{equation} P(A\;wins) = \sum_{k=6}^{\infty}\biggl( (^{k-1}C_5*(\frac{1}{2})^k) * (\sum_{i=0}^2{^{k}C_i}*(\frac{1}{2})^k) \biggr) \\ = \sum_{k=6}^{\infty}\biggl((\frac{1}{2})^{2k}*(^{k-1}C_5)*(^kC_0+^kC_1+^kC_2)) \biggr) \\ =\sum_{k=6}^{\infty}\biggl(\frac{1}{2^{2k+1}}*(^{k-1}C_5)*(k^2+k+2)\biggr) \end{equation}$

Now I calculated the sum on my calculator for $k : 6 \to 150$ and it is converging to 0.052. Which is very far off from the given answer of $P(Andy\; wins) = 0.1445$. So is my method wrong ? Or is the sum convergence slow ?

Can someone solve this question , employing an entirely different method if necessary.

Answer 1

No, your answer is right. I got the same thing by this formula, in which the number of failures is represented by the summation indices.

$$\sum_{k=0}^\infty{5+k\choose k}.5^{6+k}\left(1-\sum_{i=0}^{k+3}{2+i\choose i}.5^{3+i}\right)$$

Code is not shown here, but it gives the answer after 1000 iterations: 0.05258345

A simulation shows that this is about correct.

x=rnbinom(100000, 6, .5)
y=rnbinom(100000, 3, .5)
mean((x+6)-(y+3)<0)
0.05303

Answer 2

Another approach would be considering the rest of the game as a Markov chain.

The states of the game are tuples $(a,b)$, with $a = 4 \dots 10$, $b = 7 \dots 10$, and the states $(10,\cdot)$ and $(\cdot ,10)$ are the state in which Andy or Beth won the game respectively.

From each state $(a,b)$ we can go to the states $(a,b)$, $(a+1,b)$, $(a,b+1)$ and $(a+1,b+1)$ with equal probability $1/4$ if we allow both of them to toss the coin. One exception is state $(9,9)$, from which we can only go to $(9,9)$ with probability $1/4$, and states $(9,10)$ and $(10,9)$ with probability $3/8$. The possible outcomes of both tossing the coin if both have $9$ heads is either

• tails,tails in which case they stay in $(9,9)$ independent of whose turn it is when they leave for class,
• tails,heads in which case they go to $(\cdot,10)$ or $(10,\cdot)$ depending on whose turn it is when they leave for class,
• heads,tails in which case they go to $(\cdot,10)$ or $(10,\cdot)$ depending on whose turn it is when they leave for class,
• heads,heads in which case they go to $(\cdot,10)$ or $(10,\cdot)$ depending on whose turn it is when they leave for class.

If we considers $(\cdot,10)$ as a single state, and $(10,\cdot)$ also as a single state, we have the following transition matrix \begin{align*}\begin{pmatrix}1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&1/4&0&0&1/4&0&0&0&0&0&0&0&0&0&0&0&0&1/2&0\\ 0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&1/4&0&0&1/4&0&0&0&0&0&0&0&0&0&1/2&0\\ 0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&1/4&0&0&1/4&0&0&0&0&0&0&1/2&0\\ 0&0&0&0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&1/4&0&0&1/4&0&0&0&1/2&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&1/4&1/4&0&1/4&1/4&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&1/4&0&0&1/4&1/2&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1/4&1/4&0&0&1/2\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1/4&1/4&0&1/2\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1/4&3/8&3/8\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1 \end{pmatrix} \end{align*} where we order the states as $(4,7)$, $(4,8)$, $(4,9)$, $(5,7)$, $\dots$, $(\cdot,10)$, $(10,\cdot)$.

In Mathematica this transition matrix seems to converge after $30$ iterations to a probability of $0.9321$ for Beth to win, and $0.0679$ for Andy to win.