chaitanyareddy 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?

Please explain the solution for this.

Probability = (# of desired outcomes)/(total # of possibilities)

Fairly simple game, so let's brute force it by writing out strings of possible results:

1) H

game over, person 1 wins

2) TH

game over, person 2 wins

3) TTH

game over, person 1 wins

4) TTTH

game over, person 2 wins

5) TTTTH

game over, person 1 wins

6) TTTTT

game over, no winner

Question 1: probability that 1st person wins

# of scenarios in which she wins: 3
Total # of scenarios: 6

Prob(person 1 wins) = 3/6 = 1/2

Question 2: probably that 1st person doesn't lose

# of scenarios in which she doesn't lose: 4 (tie isn't a loss)
Total # of scenarios: 6

Prob(person 1 doesn't lose) = 4/6 = 2/3

Another way to look at question 2:

"Odds against her losing" = 1 - (odds of her losing)

# of scenarios in which she loses: 2
Total # of scenarios: 6

Odds of her losing = 2/6 = 1/3

Odds against her losing = 1 - 1/3 = 2/3

Hello Stuart,
I want to ask you, I see that you assumed that the total number of tosses all together is 5 while it should be that "EACH" has a maximum of 5 tosses.
So if we say we have person A & B

each has either an H or T to get

Toss 1)
A:H B:T A wins
A:T B:H B wins
A:T B:T no one wins
Toss 2) (reached because no one won)
A:TH B:TT A wins
A:TT B:TH B wins
A:TT B:TT no one wins
Toss 3) (reached because no one won)
A:TTH B:TTT A wins
A:TTT B:TTH B wins
A:TTT B:TTT no one wins
Toss 4) (reached because no one won)
A:TTTH B:TTTT A wins
A:TTTT B:TTTH B wins
A:TTTT B:TTTT no one wins
Toss 5) (reached because no one won)
A:TTTTH B:TTTTT A wins
A:TTTTT B:TTTTH B wins
A:TTTTT B:TTTTT no one wins


So the total of A winning is 5, total B winning is 5, total no one winning is 5
Total scenarios are 15.
Probability of A winning is 5/15 = 1/3

Probability of A not losing is either winning or a tie = 10/15 = 2/3
Odds against her losing = 1- odds of her losing = 1- 5/15 = 1-1/3 = 2/3

Isnt this the right way of getting the answer?

For 1, why can't we consider it this way? why is this wrong:

- first tosser tosses a head on 1st toss and wins
- or, first tosser wins on his second toss
- or, first tosser wins on his third toss
- or, first tosser wins on his fourth toss
- or, first tosser wins on his fifth toss

each of the above requires the second tosser to toss a tail when he gets a chance to toss

prob = 341/512