# A biased coin flipping game strategy.

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?

• Have you tried computing the expected value of each strategy? Apr 21, 2021 at 18:31
• 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.
– lulu
Apr 21, 2021 at 18:33
• @lulu 'What would be the best strategy?' to play the first or the second game, why it is unclear?
– MaPy
Apr 21, 2021 at 18:36
• Since you are sure to win either game, what difference can it make?
– lulu
Apr 21, 2021 at 18:37
• 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).
– lulu
Apr 21, 2021 at 19:22

## 1 Answer

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 $$\mathbb{E}[X_1]>\mathbb{E}[X_2]$$

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