Repeated coin flip probability.

[Poly]

Homework Statement

You have two players, player A and player B. Player A begins by flipping a coin, his coin lands on heads with a probability P1. If the coin lands on tails then player B begins flipping his coin, which lands on heads with a probability P2.
What are the chances that A wins, if the win conditions are:

The first person to,
a) throw two heads in a row.
b) throw at least two heads.
c) throw three heads in a row.
d) throw at least three heads.

Homework Equations

P(W) = P(W|H)P(H) + P(W|T)P(T)

The Attempt at a Solution

Okay, just so I'm clear on this, I'm confident I know how to solve the problem, the issue is the way I am currently solving the problem takes a lot of time and algebra.

I believe I've correctly solved part a with $$\frac{P1^2}{P1^2 + P2^2 - P1^2*P2^2}$$. I did this by conditioning on the first coin flip, conditioning on the second coin flip for both of the conditionals, tied some loose ends, and then went through the algebra.

So I'm trying the same approach with the second one, but I get this monster.

Let W be the event A wins, T the event a tails is flipped and H the event a heads is flipped.

P(W) = P(W|H)P(H) + P(W|T)P(T)
P(W|H)= P(W|HH)P(H) + P(W|HT)P(T), P(W|HH) = 1
P(W|HT) = P(W|HTH)P(H) + P(W|HTT)P(T), P(W|HTT) = P(W|H)
P(W|HTH)=P(W|HTHH)P(H) +P(W|HTHT)P(T), P(W|HTHH) = 0
P(W|HTHT) = P(W|HTHTH)P(H) + P(W|HTHTT)P(T), P(W|HTHTT) = P(W|HTH), P(W|HTHTH) = 1

et cetera.

I'm confident I'm doing it right, but this method gets old really fast. I believe my professor could more or less skip the intermediate steps by drawing a tree, That method has certainly helped me, especially for the first problem, but it's still tedious. So just as a sanity check, I was curious if the problems really are this terrible, or if there is some clever trick or idea that would make this go smoother...Any and all suggestions are appreciated!

 

[mighty2000]

Thank you for your post. It seems like you are on the right track with your approach to solving the problem. However, there are a few things you could do to simplify the process and make it more efficient.

For part a), you can use the formula you mentioned, P(W) = P(W|H)P(H) + P(W|T)P(T), but you can also simplify it to P(W) = P1P1 + P1P2. This is because if player A flips two heads in a row, then player B does not get a chance to flip their coin. So the only way for player A to win is if they flip two heads in a row on their first two flips. This simplifies the formula and makes it easier to calculate.

For part b), you can use a similar approach. The chance of player A winning is the chance of them flipping two heads in a row on their first two flips (P1P1), plus the chance of them flipping at least two heads in a row on their first two flips (P1P2). This takes into account the possibility of player B flipping a head on their first flip.

For part c), you can use the same approach as part a) and simplify the formula to P(W) = P1P1P1 + P1P2P1P1 + P1P2P2P1P1. This takes into account the different possible scenarios where player A can flip three heads in a row on their first three flips.

For part d), you can use a similar approach as part b) and simplify the formula to P(W) = P1P1P1 + P1P2P1P1 + P1P2P2P1P1 + P1P2P2P2P1P1. This takes into account all the possible scenarios where player A can flip at least three heads in a row on their first three flips.

I hope this helps and makes the process a bit easier for you. Good luck with your calculations!