Why an Extra Coin Gives Bob a 50% Chance?

[Kakashi]

Homework Statement

Alice and Bob have 2n+1 coins, eahc coin with probability of heads equal to 1/2.Bob tosses n+1 coins, while Alice tosses the remaining n coins. Assuming independent coin tosses, show that the probability after all coins have been tosses, Bob will have gotten more heads than Alice is 1/2.

Relevant Equations

N/A

Whether Bob has more heads depends on both Bob and Alice outcomes. For Bob to win Alice must get fewer heads. If bob gets x heads then Alice must get 0,1,2,..,x-1 heads. The number of sequences with k heads is $$ {n} \choose {k} $$. So for a fixed x the number of Alice sequences with less than x heads is $$ \sum_{k=0}^{x-1} {{n} \choose {k}} $$. The number of bob sequences with exactly x heads is $$ {n+1} \choose {x} $$.

 

[PeroK]

I'm not sure whether you've posted an attempt or just some initial calculations.

You need to be clever. Here's a big hint. If they toss $n$ coins each, then there is a probability $p$ that Alice has more Heads; the same probability $p$ that Bob has more Heads; and, a probability $q$ that they have an equal number of Heads.

Take it from there.

 

[Kakashi]

Wow I dont know why I was overcomplicating things.

The probability that Bob wins= p+q*1/2=p+(1-2p)*1/2=1/2

My counting attempt was My attempt was that every sequence with x heads for Bob can be paired with every Alice sequence having fewer than x heads. So the total number of pairs for this fixed x is $$ {{n+1} \choose {x}} \sum_{k=0}^{x-1} {{n} \choose {k}} $$
Summing over all x
$$ \sum_{x=1}^{n+1} {{n+1} \choose {x}}*\sum_{k=0}^{x-1} {{n} \choose {k}} $$
Then $$ P(\text{Bob wins})=\frac{\sum_{x=1}^{n+1} {{n+1} \choose {x}}*\sum_{k=0}^{x-1} {{n} \choose {k}}}{2^{n+1}} $$. Since there are 2^{n+1} Bob sequences and 2^{n} Alice sequences.