Probability of winning four holes in golf

[akhanijow]

--------------------------------------------------------------------------------
Hi All,
I just wanted to be sure I am calculating this correctly:

In a golf game, Player A and Player B are competing in 4 holes of golf. The probabilites for 4 holes are below:

Player A Prob. Draw Player B Prob.
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 5 0.3453 0.4102 0.2444

I am trying to find the probability that Player A wins, the probability of a tie, and the probability that player B wins. would this calculation be correct?

Probability of A wins 0.2375*0.5117+0.2375*0.5117+0.2375*0.5117+0.3753*0 .4102 =0.5062

Probability of B wins 0.2509 *0.5117+0.2509 *0.5117+0.2509 *0.5117+0.2444 *0.4102 =0.4854

Probability of tie will be 0.0208 (difference of A abd B wins)

 

[sjb-2812]

--------------------------------------------------------------------------------
Hi All,
I just wanted to be sure I am calculating this correctly:

In a golf game, Player A and Player B are competing in 4 holes of golf. The probabilites for 4 holes are below:

Player A Prob. Draw Player B Prob.
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 5 0.3453 0.4102 0.2444

I am trying to find the probability that Player A wins, the probability of a tie, and the probability that player B wins. would this calculation be correct?

Probability of A wins 0.2375*0.5117+0.2375*0.5117+0.2375*0.5117+0.3753*0 .4102 =0.5062

Probability of B wins 0.2509 *0.5117+0.2509 *0.5117+0.2509 *0.5117+0.2444 *0.4102 =0.4854

Probability of tie will be 0.0208 (difference of A abd B wins)

No, this seems wrong, as the probability of a tie is at least 0.5117³ * 0.4102 (they tie all holes) = 5.5%.

What is the probability of A winning 4 holes? of winning 3 holes and a tie?...

 

[akhanijow]

Player A Prob. Draw Player B Prob.
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 3 0.2375 0.5117 0.2509
Par 5 0.3453 0.4102 0.2444


So the probability of A winning 4 holes would be:

.2375^3 * .3453

?

A winning 3 holes:
.2375^3 * (1-.3453)
.2375^2 * (1-.2375) * .3453
...and so on?

 

[willem2]

Yes, and so on. It's not much fun to do it by hand.

The easiest way to let a computer do it, is to use a program that can multiply polynomials. (mathematica or gp/pari (free) )

If you compute (0.2372 x^2 + 0.5117 x + 0.2509)^3 * (0.3453 x^2 + 0.4102 x + 0.2444)
then the coefficient of x^n will be the probability that player A scores n points (2 for a win, 1 for a draw). You'll need to add all the coefficients where a has more than 4 points for the final answer.

This works because all the 81 ways that the games can go, correspond to the 81 terms you get when multiplying those four polynomials.
The probability that a wins 4 matches is the coefficient of x^8, and that's equal to the product of all the terms with x^2. The probability that a gets 7 points. You get the sum of four products, where you multiply 3 terms with x^2 and one with x.
etc.

You'll end up with 0.003860*x^8 + 0.03010*x^7 + 0.1042*x^6 + 0.2101*x^5 + 0.2695*x^4 + 0.2245*x^3 + 0.1179*x^2 + 0.03539*x + 0.004626

and need to add the coefficients of x^5 up to x^8 to get 0.003860+ 0.03010+ 0.1042 + 0.2101 = 0.3843