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I have been thinking of the following game and the best strategy to follow.

Let us assume I have a biased coin which gives head with probability $0.8$ and tails with probability $0.2$. I have been offered two options:

  • Toss the coin $10$ times. If the number of heads is greater than $8$, I win; if it is less than $8$, I lose; if it is $8$, we play again.
  • Toss the coin $20$ times. If the number of heads is greater than $16$, I win; if it is less than $16$, I lose; if it is $16$, we play again.

What would be the best strategy? Also, what would be the optimal number of tosses should I select have I given the option?

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  • $\begingroup$ Have you tried computing the expected value of each strategy? $\endgroup$
    – Alex R.
    Apr 21, 2021 at 18:31
  • $\begingroup$ It's not clear what you are asking. In both cases you appear to be able to play infinitely often, so the probability that you win (eventually) is $1$. We have no information as to what you win so I can't see why one branch should be preferred to the other. $\endgroup$
    – lulu
    Apr 21, 2021 at 18:33
  • $\begingroup$ @lulu 'What would be the best strategy?' to play the first or the second game, why it is unclear? $\endgroup$
    – MaPy
    Apr 21, 2021 at 18:36
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    $\begingroup$ Since you are sure to win either game, what difference can it make? $\endgroup$
    – lulu
    Apr 21, 2021 at 18:37
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    $\begingroup$ No, there isn't. You keep playing until the tie is broken. That's why we can subtract $p_8$ from the probabilities in the first game (and $P_{16}$ in the second). $\endgroup$
    – lulu
    Apr 21, 2021 at 19:22

1 Answer 1

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Let's assume you want to maximize your expected value. The first strategy is a random variable $$X_1= \begin{cases} 1 & \textrm{if } k > 8,n=10 \\ 0 & \textrm{if } k < 8,n=10 \\ Y & \textrm{if } k = 8,n=10 \\ \end{cases} $$ where $X_1 \sim Y_1$. So $$\mathbb{E}[X_1]=(p_{9,10}+p_{10,10})+p_{8,10}\mathbb{E}[Y_1]$$ $$\mathbb{E}[X_1]=\frac{p_{9,10}+p_{10,10}}{1-p_{8,10}}$$ Similarly the second strategy has expected value $$\mathbb{E}[X_2]=\frac{p_{17,20}+p_{18,20}+p_{19,20}+p_{20,20}}{1-p_{16,20}}$$ Those $p_{k,n}$ are all binomial probabilities. Turns out that $\mathbb{E}[X_1]>\mathbb{E}[X_2]$

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  • $\begingroup$ Could you word this answer in a way that more directly relates it to the question? THX $\endgroup$
    – PiGuy314
    Sep 7, 2021 at 13:07

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