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Suppose you have a fair six-sided die, one fair coin, and one biased coin. The biased coin has probability 0.25 of showing Heads. First you roll the die. If the number on the die is less than three, you then toss the fair coin until you see a Head; otherwise, you toss the biased coin until you see a Head. Compute the expectation of the number of tosses, giving your answer correct to two decimal places.

I have no clue how to approach this.any hints would be helpful

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    $\begingroup$ Conditional expectation on the result of the die. $\endgroup$
    – YJT
    Commented Dec 23, 2020 at 17:34
  • $\begingroup$ Are you familiar with conditional probabilities? $\endgroup$ Commented Dec 23, 2020 at 17:34
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    $\begingroup$ Please tell us what part of the problem you do understand. Can you compute the expected number of tosses of the biassed coin until heads shows, for example? $\endgroup$
    – saulspatz
    Commented Dec 23, 2020 at 17:35
  • $\begingroup$ Not really sure on how to start this- i don’t understand how to combine the events to get expectations- would they be a partition of the sample space(IE rolling less than 3 and rolling more than 3?) $\endgroup$
    – user849459
    Commented Dec 23, 2020 at 17:35
  • $\begingroup$ Ignore everything else, can you solve the same question if you just toss the fair coin? $\endgroup$
    – YJT
    Commented Dec 23, 2020 at 17:37

2 Answers 2

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Based on your comments, I would suggest revising the concepts of expectation values before attempting this question. Nevertheless, I have provided a solution below with some references and definitions to guide you towards the correct path. I haven't provided the explanation for everything so that you spend some time understanding the concepts and finally figure it out on your own(if you need some more explanation or are stuck somewhere, please reach out via the comment section). Hope it helps.

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  • $\begingroup$ How did you identify this was a geometric distribution ? $\endgroup$
    – user849459
    Commented Dec 23, 2020 at 18:49
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    $\begingroup$ It is the most popular interpretation of geometric distribution. It is actually defined in a similar way(search wiki). I just used this from my set of tools to do things faster. Ideally, you should be able to calculate the expectation irrespective of this knowledge(which you should try out once to understand geom. rvs.) $\endgroup$ Commented Dec 23, 2020 at 19:03
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If the probability of showing a Head (H) in a flip of a coin is $p$ then expected number of flips to see the first H is $\displaystyle \frac{1}{p}$.

Using that, find expected number of flips $\small E(F)$ for the fair coin and expected number of flips for the biased coin $\small E(B)$.

Now probability of die showing less than $3, \small P(X \lt 3)= \displaystyle \frac{2}{6}$ (die shows either $1$ or $2$)

So expected number of flips to see first H $ \small = P(X \lt 3) \times E(F) + (1-P(X \lt3)) \times E(B)$

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