# Probability with biased coin problem

Jules César gives Astérix a biased coin which produces heads 70% of the time, and asks him to play one of the following games:

Game A : Toss the biased coin 99 times. If there are more than 49 heads, he will be sent to feed crocodiles.

Game B : Toss the biased coin 100 times. If there are more than 50 heads, he will be sent to feed crocodiles. If there are exactly 50 heads, he is granted a fair coin. If the fair coin produces a head, he will be sent to feed crocodiles. Which game will Astérix choose to play?

My hunch is that it will be game A, but I want to calculate the probability. For Game A, I wanted to compute the probability that he would get more than 49 heads. Using the binomial distribution, I found:

$$\sum_{i=49}^{99}{99 \choose i} (0.70)^i (0.30)^{ 99-i}$$

I tried calculating this through Wolfram Alpha but got 0.999999 which doesn't seem right. Is there an easier way to compute this sum, or probability for that matter?

• Calculation in principle will do it, but you are only asked for comparison. Jul 13, 2016 at 4:13
• Here is a hint: math.stackexchange.com/questions/657886/… Jul 13, 2016 at 4:23

The only case where there may be a difference between the outcomes of the two games is if there are $49$ or $50$ heads after $99$ tosses. Call the probability of these two situations $p(49)$ and $p(50)$.

If the $99$ toss game were played with a fair coin, the probability of getting $50$ heads and $49$ tails would be the same as getting $49$ heads and $50$ tails, because the binomial coefficients match.

With the unfair coin, the coefficients still match, but single-throw probabilities increase the chance of getting exactly $50$ heads. Taking the coefficient as $k$, we have $p(49) = k0.3^{50}0.7^{49}$ whereas $p(50) = k0.3^{49}0.7^{50}$, Thus $p(49) = \frac{0.3}{0.7} p(50)$.

Set $a:=p(49)+p(50)$. Then $p(49) = 0.3a$ and $p(50) = 0.7a$. In game $1$, these directly become good and bad outcomes. In game $2$, the first additional flip takes us to $0.09a$ at $49$ heads, $0.42a$ at $50$ heads and $0.49a$ at $51$ heads. So, after the additional fair coin flip is taken, we have a good outcome in this case of $0.09a+0.21a = 0.3a$ and a bad outcome of $0.21a+0.49a = 0.7a$ - exactly the same as for Game $1$.

Thanks for nothing, César.

Indeed, even if it had made a difference, it wouldn't make a practical difference, because we're working with a very low probability case in the first place. Astérix would do well to get busy with his potion.

Suppose Asterix tosses the biased coin 99 times in whichever game he plays. He can get

• More than 50 heads. In which case it doesn't matter which game he plays, the crocodiles get a meal.

• Exactly 50 heads. In which case playing the second game gives him a chance to escape from the crocodiles. (How much?)

• Exactly 49 heads. In which case playing the second game gives him a chance to be fed to the crocodiles. (How much?)

• Less than 49 heads. In which case the poor crocodiles go without whichever game he plays.

So the difference in the games lies in the middle two scenarios.