SO, I drop a piece of bread and jam repeatedly. It lands either jam face-up or jam face-down and I know that jam side down probability is $P(Down)=p$

I continue to drop the bread until it falls jam side up for the first time. What is the expression for the expected number of drops? By considering the binomial expansion of $(1-p)^{-2}$, can you that the expected number is $1/(1-p)$?

Also, please include a rough sketch of the probability distribution function for the number of times it falls jam side down for N drops, where N is large

My Progress

The probability that jam side is down for n drops: $p^n$

The probability that jam side is up for the first time at the $n^{th}$ drop is: $p^{n-1}(1-p)$

How do I continue from there?

Thank you very much! :)

  • 2
    $\begingroup$ Please edit your question to include a description of what you've tried and where you're stuck. This will help people give useful answers at an appropriate level. In general it's not good practice to simply post homework problems, especially when they contain imperative demands to "show" or "give." $\endgroup$ Dec 10, 2014 at 17:07

2 Answers 2


Average number of throws to get first up results = $\Sigma N Q_{N}$ where $Q_{N}$ is the probability of getting first $N-1$ throws down, then an up.

$Q_{N} = p^{N-1} (1 - p)$ So $\Sigma N Q_{N} = (1 - p) \Sigma Np^{N-1} = \frac{1 - p}{(1 - p)^2}= \frac{1}{(1 - p)}$

The probability distribution function for getting N consecutive ups from the start is $p^N$. For large $N$ this drops swiftly with p.


Let $X$ be the random variable for the number of drops until the bread lands jam-side up. You found that $P(X=n)=p^{n-1}(1-p)$. The expected value is therefore


Now the problem mentions the binomial expansion of $(1-p)^{-2}$, so I assume you know that


Can you put all this together?


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