Probability question: A fair coin is to be tossed repeatedly. For integers r and s, not both zero...?

Question

A fair coin is to be tossed repeatedly. For integers r and s, not both zero, let P(r,s) be the probability that the total of r heads are tossed before a total od s tails are tossed so that P(0,1)=1 and P(1,0)=0.
Explain why, for r and s greater than or equal to one that: P (r,s) = 1/2P(r-1,s) + 1/2P(r,s-1)

Answer 1

Directly before the (r+s)th toss, we already have (r+s-1) tosses.

If among these tosses, the first r tosses are exactly heads, and the remain s-1 are exactly tails, that is, (r, s-1), then now there's a probability of 1/2 that we get another head, to realize (r, s).

If among these tosses, the first r-1 tosses are exactly heads and the remaing are r tails, the next toss, be it head or tail, won't realize (r, s). So I think P(r, s)=1/2P(r, s-1).

I'm not sure if I understood your problem correctly though.

Answer 2

Let H be a Head, T be a Tail

Consider P(r,s)
ie there have been r+s tosses, the last one being a T
Before this there were r+s-1 tosses, r being H, s-1 being T
The probability of this ie P(r,s)
is given by:
P(r,s) = (r+s-1) C r (1/2)^r . (1/2)^ (s-1) . (1/2)
ie P(r,s)= (1/2)^ (r+s) . (r+s-1) C r (1/2)^r

Now consider P(r-1,s)
ie there have been r+s-1 tosses, the last one being a T
Before this there were r+s-2 tosses, r-1 being H, s-1 being T
The probability of this ie P(r-1,s)
is given by:
P(r-1,s) = (r+s-2) C (r-1) (1/2)^(r-1) . (1/2)^ (s-1) . (1/2)
ie P(r-1,s)= (1/2)^ (r+s-1) . (r+s-2) C (r -1)

Now consider P(r,s-1)
ie there have been r+s-1 tosses, the last one being a T
Before this there were r+s-2 tosses, r being H, s-2 being T
The probability of this ie P(r,s-1)
is given by:
P(r,s-1) = (r+s-2) C (r) (1/2)^(r) . (1/2)^ (s-2) . (1/2)
ie P(r,s-1)= (1/2)^ (r+s-1) . (r+s-2) C (r)

1/2P(r-1,s) + 1/2P(r,s-1)
= (1/2) [ (1/2)^ (r+s-1) .( (r+s-2) C (r -1) + (r+s-2) C (r))
=(1/2)^(r+s) . { (r+s-2)! / [ (r-1)!(s-1)!] + (r+s-2)! / [ r!(s-2)!]
=(1/2)^(r+s) { [r(s+r-2)! +(s-1)(s+r-2)!] / (r!(s-1)!)
=(1/2)^(r+s) { (s+r-2)!(r+s-1)/[r!(s-1)!]
=(1/2)^(r+s) { (s+r-1)! / [ r! ( (r+s-1 -r)!]
=(1/2)^(r+s) . (s+r-1) C r
= P(r,s)

wow!

Answer 3

P (r,s) = 1/2P(r-1,s) + 1/2P(r,s-1)

= P(head) * P(r - 1 heads, s tails) + P(tail) * P(r heads, s -1 tails)

this is the law of total probability. the 1/2 comes from the fact the coin is a fair coin.