A Q onProbabilty

Pankaj infinity · Jul 4, 2009, 12:35 AM

3 Persons A,B & C throw a dice alternatively . The Person Getting 2 or 4 first wins. What is the probabilty of their winning

Perito · Jul 4, 2009, 04:52 AM

Stated more succinctly: What is the probability of throwing a single die and getting a 2 or a 4? The 3 in "3 persons" is irrelevant.

If you rephrased the question, "What is the probability that someone will win in each round of three throws", the problem becomes more complicated.

There are six possibilities when you throw a die. Therefore, there is 1/6 of getting each number or 2/6 (1/3) of getting any selected two numbers. Each throw is independent of the other throws. Therefore, each time someone throws the die (dice is plural), he has 1/3 chance (probability = 1/3) of getting a 2 or a 4.

Pankaj infinity · Jul 8, 2009, 02:46 AM

For your kind information , the probability of winning of a, b & c is different

Unknown008 · Jul 8, 2009, 05:35 AM

I think I know what you mean. The probability for A to win is as Perito suggested, that is 1/3.

The probability that the person B wins depends on what the person A has got. If A won, then, he won't win. Therefore, A must have lost, so that the person B has a chance to win. So, his probability is $\frac23 \times \frac13 = \frac29$

For C, the same thing happens;

$\frac23 \times \frac23 \times \frac13 = \frac {4}{27}$

ebaines · Jul 8, 2009, 05:58 AM

Unknown008 - your solution shows the probability of each player winning in the first round. But there is a chance that no one wins in the first round, in which case the die goes back to A to start a second round. Notivce that the sum of the probabilities as you calculated them do not add up to 1.

As I understand this problem, person A throws the die and if he gets a 2 or 4 he wins, game over. If he doesn't, then B gets to roll the die, and if gets a 2 or 4 he wins, and the game is over. If neither A nor B has won then C gets a chance. And if he doesn't roll a 2 or for the die goes back to A for his next chance, and so on.

Think of this in terms of each round of A, B, and C rolling the die. In the first round the probability of A winning is 1/3 (since any 2 out of 6 possible rolls are winners), the probability of B winning is 2/3 * 1/3 (since A has to lose and B has to roll 2 or 4 to win), and the probability of C winning is 4/9 * 1/3.

The probability of no one wining in the first round is (2/3)^3. Hence the probability that A will win in the second round is (2/3)^3 * 1/3; the probability that B will win in the second round is (2/3)^4*1/3, and the probability that C will win in the second round is (2/3)^5*1/3. If you keep going and add the probbailities for each player you will see that the probabilities add up to this:

$ P(A) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i} \\ P(B) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i+1} \ = \frac 2 3 \times P(A) \\ P(C) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i+2} = \frac 4 9 \times P(A) $

The sum of all these probabilities must equal 1. So you have:

P(A) + 2/3 P(A) + 4/9 P(A) = 1

Solve for P(A), and you find:
$ P(A) = \frac 9 {19} \\ P(B) = \frac 6 {19} \\ P(C) = \frac 4 {19} $

Unknown008 · Jul 8, 2009, 08:33 AM

I know ebaines. I was in a rush and couldn't continue. I would have suggested that the probability of one winning was

$( \normal{\frac23} \large)^{\normal{n-1}} \times \frac13$

where n was the nth throw.

However, the extension you provided, that is of the sum to infinity, I was not aware of that.. and don't quite understand it (haven't done that at school yet... :()

Pankaj infinity · Jul 10, 2009, 12:22 AM

Originally Posted by ebaines

unknown008 - your solution shows the probability of each player winning in the first round. But there is a chance that no one wins in the first round, in which case the die goes back to a to start a second round. Notivce that the sum of the probabilities as you calculated them do not add up to 1.

As i understand this problem, person a throws the die and if he gets a 2 or 4 he wins, game over. If he doesn't, then b gets to roll the die, and if gets a 2 or 4 he wins, and the game is over. If neither a nor b has won then c gets a chance. And if he doesn't roll a 2 or 4 the die goes back to a for his next chance, and so on.

Think of this in terms of each round of a, b, and c rolling the die. In the first round the probability of a winning is 1/3 (since any 2 out of 6 possible rolls are winners), the probability of b winning is 2/3 * 1/3 (since a has to lose and b has to roll 2 or 4 to win), and the probability of c winning is 4/9 * 1/3.

The probability of no one wining in the first round is (2/3)^3. Hence the probability that a will win in the second round is (2/3)^3 * 1/3; the probability that b will win in the second round is (2/3)^4*1/3, and the probability that c will win in the second round is (2/3)^5*1/3. If you keep going and add the probbailities for each player you will see that the probabilities add up to this:

$ p(a) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i} \\ p(b) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i+1} \ = \frac 2 3 \times p(a) \\ p(c) = \frac 1 3 \sum _{i=0} ^\infty (\frac 2 3 )^{3i+2} = \frac 4 9 \times p(a) $

the sum of all these probabilities must equal 1. So you have:

P(a) + 2/3 p(a) + 4/9 p(a) = 1

solve for p(a), and you find:
$ p(a) = \frac 9 {19} \\ p(b) = \frac 6 {19} \\ p(c) = \frac 4 {19} $

thank u very much

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