800guy wrote:A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

1- Find the probability that the person who tosses first will win the game?

2- What are the odds against A's losing if she goes first?


oa coming when people respond with explanations. from diff math doc

ok this is a tuff q....

im not sure abt the solution... but here it is.... there will be more than one case in which the one who throws first will win

1st case... he wins in the first try itself
2nd case.... he throws a tail, then the next guy throws a tail and then the first guy throws a head ..... so it wuld be something like TTH

3rd case... TTTTH
4th case.... TTTTTTH
5th case....TTTTTTTTH

so the probability shuld be 1/2+(1/2)^3+(1/2)^5+(1/2)^7 +(1/2)^9..... again i am really not sure abt the solution...


now i did not exactly understand what the 2nd case means... i am assuming it wants us to find the probability that the guy who tosses second ( in this case B ) wins...

if that is what it means then...

1st case ... TH
2nd case... TTTH
3rd case...TTTTTH
4th case .... TTTTTTTH
5th case.....TTTTTTTTTH

therefore the probability shuld be ...(1/2)^2+(1/2)^4+(1/2)^6+(1/2)^8+(1/2)^10.... this was indeed a very good q....

here's the oa from the doc:

look at the conditions; it says that the first person who tosses a head wins.

Let’s say A tosses first.

what is the probability that he wins

H + TTH + TTTTH + TTTTTTH + TTTTTTTTH

i.e. either the first toss is head,
or the first time A tosses the coin he gets a tail and B also gets a tail , n in the second throw A gets a head.....

This continues for a max till 5 throws, because the game is for 5 throws only.
So, 1. 1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9

2. (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10

agreed, this is really tough