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?
 A: 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.
A: 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.
