A and B alternately toss a coin

Question

A and B alternately toss a coin. the first one to turn up a heads wins. if no more than five tosses each are allowed for a single game, find the probability that the person who tosses first will win the game. what are the odds against A's losing if she goes first

Answer 1

Assuming all coins used in this game are fair (i.e. there is a 50% chance that "Heads" will occur), then the probability that the person who goes first will win is given by the following:

 

1/2 + 1/8 + 1/32 + 1/128 + 1/512 = 341/512 or approximately 0.666

 

The person who flips 2nd wins  (exactly half as often) 

 

1/4 + 1/16 + 1/64 + 1/256 + 1/1024 = 341/1024 or approximately 0.333

 

What is the probability of a "tie" ? 

 

Exactly 1 - 341/512 - 341/1024 = 1/1024 or approximately 0.001

 

For the "tie" to occur, that means each player flipped 5 consecutive "heads". There is a probability of 1/32 for each of them to do that; (1/32) (1/32) = 1/1024.

 

The player going first has a 2 to 1 advantage. However, if the players alternate who goes first for the start of each game, then it becomes no advantage over the long run provided they agree to play some even number of games.

 

 

Answer 2

Mcwiills:

 

You say that the coins are "two-headed" (both sides "Heads").  In that case the answers are:

 

/ Find the probability that the person who tosses first will win the game: Answer 100 per cent.  No need to play after the first toss.

 

/ What are the odds against A's losing if she goes first  ZERO.  A always wins, because her first toss is a Head with probability 1    !!!!!!

 

What do you really really really mean by "two-headed"?  Explain.

 

Dr. G