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A and B play game in whic they toss coin 3 times.The one obtaining heads first wins the game.If A tosses coin first and if total value of stake is Rs 20.How much should be contributed by B in order that game is fair?

2 Answers

1 votes
1 votes

I think A flips 3 time if loose then B flips three time and it continues till anyone wins.

First, calculate the probability that A does not win on his first turn. That is simply the probability that a coin flipped three times turns up no 'heads'. 
That probability is (1/2)^3 = 1/8. So, that means A has a 7/8 chance of winning on his first turn. 

Assuming A doesn't win (which happens 1/8 of the time), B then flips three times. 
The probability that B doesn't win these three flips is again 1/8, so the probability of B winning on his first turn (if B gets a first turn) is 7/8, 
But B only gets a first turn 1/8(if A loose first turn) of the time. 
So the actual probability of B winning on his first turn is: 1/8 * 7/8 = 7/64. 

The A's second turn, with the probability of winning  7/8 again, if A gets a second turn. 
But the probability of getting to A's second turn is 1/8 * 1/8 = (1/8)^2. So the total probability of A winning on his second turn is (1/8)^2 * 7/8 = 7/512. 

The pattern repeats. The total probability of B winning on his second turn is (1/8)^3 * 7/8 = 7/4096. 

The total probability of A winning on his third turn is (1/8)^4 * 7/8 = 7/32768. 

final probability that A wins the game, we get an infinite series, which evaluates to: 
Probability that A wins = 7/8 + 7/8*(1/8)^2* + 7/8*(1/8)^4 + 7/8*(1/8)^6 + ... 
Probability that A wins = 7/8 * (1 + (1/8)^2* + (1/8)^4 + (1/8)^6 + ...) 
Probability that A wins = 7/8 * (64/63) 
Probability that A wins = 8/9 

So, A wins the game with a probability of 8/9. We can then deduce that B wins the game with a probability of 1/9. 
For a fair game, B must put of 1/9 of the stakes, or: 
B's stakes = 1/9 * 20 
B's stakes = 2.222

0 votes
0 votes
By 'fair game', they mean that if the game were played many, many times, you would expect each player to win just as much money as they lost. In other words, neither player is favored in the long run (though of course some player always wins a particular game).

I'll assume you mean player A flips a coin three times, then player B flips the coin three times, then player A flips three times, and this repeats until someone obtains 'heads'. If so, here's how to solve it.

First, calculate the probability that A does not win on his/her first turn. That is simply the probability that a coin flipped three times turns up no 'heads'. That probability is (1/2)^3 = 1/8. So, that means A has a 7/8 chance of winning on his/her first turn.

Assuming A doesn't win (which happens 1/8 of the time), B then flips three times. The probability that B doesn't win these three flips is again 1/8, so the probability of B winning on his/her first turn (if B gets a first turn) is 7/8, just as it was for A. But B only gets a first turn 1/8 of the time. So the actual probability of B winning on his/her first turn is the product of these two probabilities: 1/8 * 7/8 = 7/64.

The same argument then goes for A's second turn, with the probability of winning being 7/8 again, assuming A gets a second turn. But the probability of getting to A's second turn is 1/8 * 1/8 = (1/8)^2. So the total probability of A winning on his/her second turn is (1/8)^2 * 7/8 = 7/512.

The pattern repeats. The total probability of B winning on his/her second turn is (1/8)^3 * 7/8 = 7/4096.

The total probability of A winning on his/her third turn is (1/8)^4 * 7/8 = 7/32768.

Summing up all the terms for player A to find the final probability that A wins the game, we get an infinite series, which evaluates to:
Probability that A wins = 7/8 + 7/8*(1/8)^2* + 7/8*(1/8)^4 + 7/8*(1/8)^6 + ...
Probability that A wins = 7/8 * (1 + (1/8)^2* + (1/8)^4 + (1/8)^6 + ...)
Probability that A wins = 7/8 * (64/63)
Probability that A wins = 8/9

So, A wins the game with a probability of 8/9. We can then deduce that B wins the game with a probability of 1/9. For a fair game, A must put of 8/9 of the stakes, or:
A's stakes = 8/9 * $20
A's stakes = $17.77777777...

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