Suppose I flip a coin 4 times and get n heads, and I flip the coin another 5 times and get m heads. What is P( n < m )? Can I just add another coin flip to the first and think I get .5 more heads?

if the coin is fair, the answer is 0.5 (and it's true for any N flips vs. N+1 flips).the easiest way is to also consider the dual problem with tails, i.e., Pr(4-n<5-m). by symmetry of a fair coin, this has to be same as Pr(n<m). the trick then is to realize Pr(4-n<5-m)=Pr(n+1>m)=Pr(n>=m) and Pr(n<m)+Pr(n>=m)=1

An alternative approach might be the following. Let A and B be the respective coins. Define 3 events as follows:E1: A generated more heads than B in first 4 tossesE2: A and B generated equal number of heads after 4 tossesE3: A generated less heads than B in first 4 tossesLet Prob(E1)=p, then Prob(E2)=p by symmetry. Since Prob(E1)+Prob(E2)+Prob(E3)=1, then Prob(E3)=1-2pProb(n<m)=Prob(E1)*1+Prob(E2)*1/2+Prob(E3)*0=p*1+(1-2p)*(1/2)=1/2where m is the number of heads coin A produces, n is the number of heads coin B produces.The logic is that if E1 occured in first 4 tosses, then m will be greater than n with probability 1. If E2 occured in first 4 tosses, then m will be greater than n with probability 1/2 (fifth toss of coin A should be heads in order to have m>n). If E3 occured in first 4 tosses, then m will be greater than n with probability 0.

NumOfCoins = NumOfHeads + NumOfTails 4 = h + t 5 = H + T (H-h) + (T-t) = 1 the random variables (H-h) and (T-t) by symmetry have equal probability of being >0, So Probability(H>h)=0.5