View Full Version : A Darts Problem

SPritchard

04-25-2010, 03:47 AM

In a game of darts the person that throws first in a leg has an advantage. The players take it in turns to go first. So if we assume Player A has a probability of 0.6 of winning a leg when he throws first, and 0.4 when he throws second, how do you calculate the probability of Player A winning by a certain score (eg 3-2 in a best of 5 leg match)?

If someone win by x : y, (i.e. x > y)

then he have to win (x - 1) legs out of the first (x - 1 + y) legs

and win the (x + y)th leg

If each leg have the same probability of winning and they are independent

then we can use the negative binomial distribution to calculate the probability

http://en.wikipedia.org/wiki/Negative_binomial_distribution

However, now in each leg the probability of winning are different.

Maybe I just calculate the example for you.

For the scores 3 : 2, player A win 2 legs and lose 2 legs in the first 4 legs

and he win the 5th leg.

The probability of winning are 0.6, 0.4, 0.6, 0.4, and 0.6

in the 1st, 2nd, 3rd, 4th, and 5th leg respectively.

In the first 4 legs, you have 4C2 = 6 possible cases, i.e.

{WWLL, WLWL, WLLW, LWWL, LWLW, LLWW}

So the probability of having 2 wins and 2 loses in the first 4 legs

is just the sum of the probability of these 6 possible cases.

But you can partition the 4 legs into 2 part: the odd legs (1st and 3rd)

with 0.6 probability of winning and the even legs (2nd and 4th)

with 0.4 probability of winning

And you can count the cases like

winning (2, 0), (1, 1) and (0, 2) in the odd legs and the even legs

Then the probability

= (2C2)(0.6)^2(0.4)^0(2C0)(0.4)^0(0.6)^2

+ (2C1)(0.6)^1(0.4)^1(2C1)(0.4)^1(0.6)^1

+ (2C0)(0.6)^0(0.4)^2(2C1)(0.4)^2(0.6)^0

= 0.6^4 + 6(0.6)^2(0.4)^2 + (0.4)^4

= 0.1296 + 0.3456 + 0.0256 = 0.5008

The since he has to win the 5th leg

So the probability = 0.5008*0.6 = 0.30048

SPritchard

04-25-2010, 12:26 PM

Thanks for your answer. I don't quite follow the bottom part though. If I work it out the long way I get

WWLL = 0.6*0.4*0.4*0.6 = 0.0576

WLWL = 0.6*0.6*0.6*0.6 = 0.1296

WLLW = 0.6*0.6*0.4*0.4 = 0.0576

LWWL = 0.4*0.4*0.6*0.6 = 0.0576

LWLW = 0.4*0.4*0.4*0.4 = 0.0256

LLWW = 0.4*0.6*0.6*0.4 = 0.0576

Adding all these up gives me 0.3856.

Multiply that by the probability of winning the 5th leg 0.3856*0.6 = 0.23136.

Where am I going wrong?

You are correct.

Sorry I have made a mistake that (2C1)(2C1) = 4, but not 6

You can count the cases easier if you count all the possible cases

that player A win in the odd legs and the even legs respectively

as the probability of winning in the odd/even legs are the same.

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