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sonpham
 
Reply Wed 21 Sep, 2011 02:30 pm
players A and B play a match consisting of a series of points. each point is independently won by A with probability p and by B with probability 1-p. they stop when the number of points by one player is two more than that of the other player; the player with the most points is the match winner.
a/ find the probability (in term of p) that a total of 4 points are played.
b/ find the probability (in term of p) that A is the match winner.
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Type: Question • Score: 0 • Views: 1,163 • Replies: 6
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engineer
 
  1  
Reply Wed 21 Sep, 2011 04:06 pm
@sonpham,
The first two points must be split. The probability of that happening is

1 - p^2 - (1-p)^2

which is just 100% minus the probability that A wins twice minus the probability that B wins twice.

The next two points must go to the same person. The probability of that is

p^2 + (1-p)^2

The overall chance of a four point match win is the product of those two probabilities.

The probability of A winning in four is the first part times the probability of A taking the last two points

(1 - p^2 - (1-p)^2 ) p^2
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markr
 
  1  
Reply Wed 21 Sep, 2011 07:58 pm
@sonpham,
If in part (b), you want the probability of A winning a match of any length (call it P), then it can be solved recursively with:

P = p^2 + 2p(1-p)P

which works out to:

P = p^2 / (2p^2 - 2p + 1)

p^2 is the probability that A wins the first two games and ends the match.
2p(1-p) is the probability that they split the first two and, essentially, start over (which is why it is multiplied by P).
sonpham
 
  1  
Reply Thu 22 Sep, 2011 08:00 am
@markr,
where is the P that you multiplied from. is it the probability that A win two point when the game start over. i don't get it why you go from P=p^2 +2p(1-p)P which is multiplication and addition to P=p^2/(2p^2 -2p+1) which is division
engineer
 
  1  
Reply Thu 22 Sep, 2011 09:55 am
@sonpham,
If you want the chance that A wins regardless of the number of matches, you get a geometric series. The change that they win in two is p^2. The chance that they win in four is (p^2)(2p(p-1)). The chance they win in six is (p^2)(2p(p-1))^2 and so on. Use the formula for an infinite series to get the total probability as

P = p^2 / (1- 2p(p-1) )
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markr
 
  1  
Reply Thu 22 Sep, 2011 01:31 pm
@sonpham,
engineer explained the geometric series associated with the problem, although his (p-1) should be (1-p).

p is the probability that A wins any point.
P is the probability that A wins any match (unknown length).

The first equation is a recursive implementation of what engineer explained. The second equation is derived by solving the first equation for P.
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sonpham
 
  1  
Reply Fri 23 Sep, 2011 09:20 pm
get it, thanks a lot
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