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What's the probability of flipping an even number of heads after tossing 99 fair coins and one weighted coin. I don't know if this matters, but let's say the probability of the weighted coin landing heads is p.

Is the following correct:

P(even in 99 flips)*P(tails on last flip) + P(odd in 99 flips)P(heads in last flip) (1/2)[(99/100)(1/2) + (1/100)(1-p)] + (1/2)*[(1/2)(99/100) + (1/100)(p)]

This gets 1/2, but this doesn't logically make sense to me, because what is the last coin is p=0. Shouldn't it be less/more likely?

Thanks!

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If we flip $99$ fair coins, the probability of getting an even number of heads is $\frac12$. To see this, observe that for any $k$ from $0$ through $99$, the probability of getting $k$ heads and $99-k$ tails is the same as the probability of getting $k$ tails and $99-k$ heads. The probability of getting an even number of heads is

$$\sum_{k=0}^{49}P(2k\text{ heads})=\sum_{k=0}^{49}P(2k\text{ tails})=\sum_{k=0}^{49}P(99-2k\text{ heads})\;,$$

which is the probability of getting an odd number of heads, since $99-2k$ runs through the odd numbers from $99$ down through $1$. Since the two are equal, both are $\frac12$.

Thus, it doesn’t matter how the weighted coin lands. If it lands tails, the probability that the total number of heads is even is the probability that the $99$ fair coins produced an even number of heads, which is $\frac12$. If it lands heads, the probability that the total number of heads is even is the probability that the $99$ fair coins produced an odd number of heads, which is also $\frac12$.

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Let $p_{n}$ be the probabilty of getting an even number of head with $n$ fair and one biased coin. To get an even number of heads with $n+1$ fair and one biased coin means to either get an even number with the first $n$ fair plus one unfair coins and tails with the last fair coin, or an odd number with the first and heads with the last coin. So $$ p_{n+1}=\frac12p_n+\frac12(1-p_n)=\frac12.$$

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