gambling probability problem

We are given a fair coin. We start out with 5 dollars. We keep tossing the coin. If the outcome is different than the previous one, we are awarded another 5 dollars. However, we do not get anything if the outcome is the same as the previous one. Let's say we toss the coin X times in the long run. How much do we expect to have in the end?

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Note that for a fair coin this is the essentially the same as ignoring the first toss and winning with heads / losing with tails. –  Hagen von Eitzen Jan 18 '13 at 16:52
And if it is a biased coin that comes up heads with some probability x then what bias would we really seek in this case?! –  user58749 Jan 18 '13 at 16:56
Well, the case of biased coin would be very different. –  Hagen von Eitzen Jan 18 '13 at 17:22

Given the current state: time $n\geq 0$ and the amount of money $5x$ dollars, the probability that we have $5(x+1)$ at the next step is the same as the probability that we stay with $5x$, and of course both are $\frac12$. So as Hagen has mentioned, your expected profit is $$5+\mathsf E\sum_{i=0}^X \xi_i = 5\left(1+\frac X2\right).$$ since $\xi_0 = 0$ and $\xi_i$ takes values $\{0,5\}$ equiprobable.
The odds for each toss is $50\%$ for a win and $50\%$ for a "loss" where a loss does not actually constitute any loss other than time. So given that you begin with $\$5$you should expect to gain$\$(5x \cdot 0.50)$ more, i.e., you should expect to gain about $\$5$for every other flip. - By$x$, do you mean the number of tosses? This site uses MathJaX, which is based on LaTeX. The dollar sign is used to begin and end mathematical formulas in LaTeX, so you have to be careful about using it. See this tutorial. – N. F. Taussig Jul 11 at 8:21 Yes. @N.F.Taussig Also looking back my answer seems too sure of itself. I was simply stating my line of reasoning. – Peter Jul 11 at 8:23 This problem may be solved using generating functions which is admittedly more powerful than strictly necessary in this case, but it does illustrate the method, which is worth knowing. Let$p_A(n)$be a polynomial in$u$such that$[u^k] p_A(n)$is the probability of having a capital of$k$dollars and the last coin came up heads and similarly$p_B(n)$for tails. Then we have the following system of equations $$p_A(n) = \frac{1}{2} p_A(n-1) + \frac{1}{2} u^5 p_B(n-1) \\ p_B(n) = \frac{1}{2} u^5 p_A(n-1) + \frac{1}{2} p_B(n-1)$$ with initial conditions $$p_A(0) = \frac{1}{2} u^5 \quad \text{and} \quad p_B(0) = \frac{1}{2} u^5.$$ Now introduce the generating functions $$P_A(z) = \sum_{n\ge 0} p_A(n) z^n \quad \text{and} \quad P_B(z) = \sum_{n\ge 0} p_B(n) z^n.$$ Multiply the recurrences by$z^n:$$$p_A(n) z^n = z \frac{1}{2} z^{n-1} p_A(n-1) + z \frac{1}{2} u^5 z^{n-1} p_B(n-1) \\ p_B(n) z^n = z \frac{1}{2} u^5 z^{n-1} p_A(n-1) + z \frac{1}{2} z^{n-1} p_B(n-1)$$ Sum over$n$to obtain the following system of equations $$P_A(z) - \frac{1}{2} u^5 = \frac{1}{2}z P_A(z) + \frac{1}{2}u^5 z P_B(z) \\ P_B(z) - \frac{1}{2} u^5 = \frac{1}{2}u^5 z P_A(z) + \frac{1}{2}z P_B(z).$$ The solution to this system is $$P_A(z)=-{\frac {{u}^{5}}{{u}^{5}z-2+z}}\\ P_B(z) =-{\frac {{u}^{5}}{{u}^{5}z-2+z}}.$$ The generating function for the expected capital after$n$steps is then given by $$f(z) = \left( \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} = -10\, \left( 2\,z-2 \right) ^{-1}+10\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}.$$ Now we have $$[z^n] f(z) = 5+5/2\,n,$$ which is precisely what was obtained above. We can also compute the variance, which is where the generating functions start to shine. First compute the expected factorial moment$Q(Q-1)$where$Q\$ is the capital. This has generating function $$g(z) = \left( \frac{d}{du} \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} = -40\, \left( 2\,z-2 \right) ^{-1}+140\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}-100\,{\frac {{z}^{2}}{\left( 2\,z-2 \right) ^{3}}} .$$ This gives $$[z^n] g(z) = 20+{\frac {115}{4}}\,n+{\frac {25}{4}}\,{n}^{2}.$$ The variance can now be obtained from $$Var[Q] = E[Q^2] - E[Q]^2 = E[Q(Q-1)] + E[Q] - E[Q]^2,$$ which gives $$Var[Q] = {\frac {25}{4}}\,n.$$ We could in fact compute all factorial moments if desired.