A and B alternately toss a coin. The first one to turn up a

Here is what I think how this problem should be worked upon.

The first question asks to find the probability that the person who tosses first will win the game. Any one of A or B can tosse first. Let us say that A tosses first. So basically we need to find the probability that A will win.

The game can follow in the following possible ways:

A B A B A B A B A B

H....end of game A wins
T H ......end of game B wins
T T H ..... end of game A wins
T T T H.....end of game B wins
T T T T H .....end of game A wins
T T T T T H .....end of game B wins
T T T T T T H ....end of game A wins
T T T T T T T H ....end of game B wins
T T T T T T T T H ....end of game A wins
T T T T T T T T T H ....end of game B wins
T T T T T T T T T T ....NOBODY wins ...tie

First question:

ONLY 1, 3, 5, and 7th options above are applicable for the first question. Because these are the only options where A wins.

Probability for the firts option = 1/2
Probability for the 3rd option = 1/8 and so on

The probability that A wins (the one who tossed first) = 1/2 + 1/8 + 1/32 + 1/128 + 1/512 = 341/512.

I know this answer is different from what araspai has mentioned. But it is approximately ~ 21.3/32. May be it is correct answer.

SECOND QUESTION

Here the condition is given that A tosses first. The above configuration will work for this condition. So we do not need to create a new configiration.

Here I really have concern about the answer araspai has given (probability > 1?)

IMHO, the "odds against A's loosing" = Probability of A wining OR probability of occuring a tie

we already found the probability of A wining = 341/512

based on the above configuration of all the possible game options,

the probability of occuring a TIE = 1/1024 (The last option)

Hence, the "odds against A's loosing" = 341/512 + 1/1024 = 643/1024.

araspai,

can you tell us if these answers seem reasonable?