# Expected value in coin flipping process

You flip a coin, and if the result is tails, you lose. If the result is heads, you get to play again. What is the expected value of throws before you lose?

My Approach

The expected value is the sum of all the outcomes multiplied by their respective probabilities:$$\sum_{i=1}^{n}V_iP_i$$So for this problem:$$\sum_{i=1}^{\infty}i(\frac{1}{2^i})=0.5+0.5+0.375+…$$I can’t figure out how to find the sum, even though I know it converges.

• This is the expectation of a geometric random variable. The probability of "success" is $p=0.5$, so the expected value is $1/p=2$.
– YJT
Jul 21, 2022 at 16:59
• It seems like your original problem has not much to do with the actual thought experiment popularly known as Schroedinger's cat. I have edited the post to avoid the misleading title (please feel free to modify it again to something more appropriate and which perhaps does not gratuitously mention harming animals...).
– Pedro
Jul 22, 2022 at 14:37
• Yeah, thank you! Jul 22, 2022 at 15:12

In general, starting with a geometric series ($$\sum_{j=0}^{\infty} x^j = 1/(1-x)$$), you can evaluate many related series by taking derivatives or integrals.

• When you need to shift the exponent, multiply or divide by a power of $$x$$;
• When you need to multiply the coefficient by the exponent, take a derivative;
• When you need to divide the coefficient by the exponent, take an integral.

In this case, $$\sum_{j=0}^{\infty} j x^j=x\frac{d}{dx}\sum_{j=0}^{\infty} x^{j}=x\frac{d}{dx}\frac{1}{1-x}=\frac{x}{(1-x)^2}.$$ Substituting $$x=1/2$$ gives $$x/(1-x)^2=(1/2)/(1/2)^2=2.$$

• Note that ending in $j$ steps actually means $j-1$ non-fatal events plus one fatal one, which gives a probability of $x^{j-1}(1-x)$. Jul 22, 2022 at 10:50
• Yes, true. This answer is about summing the series that OP correctly arrived at, not about the original coin-flipping problem. So $x$ here shouldn’t be identified with $p$, the probability of flipping heads (though the same techniques could be used for the $p\neq 1/2$ case). Jul 22, 2022 at 15:10

This is an Arithmetic-Geometric Series. Let $$S=\displaystyle \sum_{i=1}^{N}\frac{i}{2^i}$$. Then, $$S=\displaystyle\frac {1}{2^1}+\frac {2}{2^2}+\frac{3}{2^3}+…+\frac{N}{2^N}.\tag{1}$$ Now, divide the whole equation by $$2$$ and arrange it by shifting the terms by a step: $$\frac{S}{2}=\frac {1}{2^2}+\frac {2}{2^3}+\frac{3}{2^4}+…+\frac{N-1}{2^{N}}+\frac{N}{2^{N+1}}.\tag{2}$$ $$(1)-(2)$$:$$\frac{S}{2}= \frac {1}{2^1}+\frac {1}{2^2}+\frac{1}{2^3}+…+\frac{1}{2^N}+\frac{1}{2^{N+1}}.$$This is a G.P., so sum is $$\frac{S}{2}=\frac 12\left(\frac{1-\left(\frac12\right)^{N+1}}{1-\frac12}\right)$$$$\implies S =\left(\frac{1-\left(\frac12\right)^{N+1}}{1-\frac12}\right)$$ Taking an infinite sum gives us $$S_{\infty}=2$$, which is the required answer.

A different perspective: If the first coin is T then it's lived for 1 minute, and if H, then you're repeating exactly the same process a minute later, so the time you should expect to wait is $$x = \frac{1}{2} \cdot 1 + \frac{1}{2}(x+1) \Rightarrow x=2$$