Question

Adam, Brian, Chris and Donald are taking goal shots in soccer. They take shots in the order, Adam, Brian, Chris, Donald, then repeat. By going in this order they have an equal chance of making a goal first. Donald makes a goal 50% of the time. What would be the probability of Adam winning if he went last instead of first?

Answer 1

here first we have to find the probability of each one goal making percentage

Here

Pr(donald) = 0.50

so here

if Pr(Donald winning the game) = 0.25

so here if Adam winning probability is A, Brian's B, Chris' C

so

Pr(A winning the game) = 0.25

A + (1 - A) (1-B)(1-C) (1-D) A + ...... = 0.25

A + (1 - A) (1 - B) (1-C) * 0.5 * A + ... = 0.25

A [1 /1 - {(1-A)(1-B)(1-C)*0.5}] = 0.25

A = 0.25 - 0.125 (1-A)(1-B)(1-C) ....(i)

similarly for Brian

(1-A)B + (1-A) (1-B)(1-C)(1-D)(1-A)B + .... = 0.25

(1 -A)B = 0.25 - 0.125 (1-A)(1-B)(1-C) ..............(ii)

for Chris

(1 - A) (1-B) C = 0.25 - 0.125 (1-A)(1-B)(1-C) .....(iii)

and for Donald

(1-A) (1-B) (1-C) D = 0.25 - 0.125 (1 -A) (1-B) (1-C) ..........(iv)

dividing here (iv)/(iii)

(1-C)D/C = 1

C = 0.5 * (1 - C)

C = 0.5 - 0.5 C

1.5 C = 0.5

C = 1/3

now (iii/(ii)

(1-B)C/B = 1

(1-B)C = B

(1-B)* 1/3 = B

1-B = 3B

B = 1/4 = 0.25

similary

A = 1/5 = 0.20

so here now it is asking that if Adam goes last instead of first than

Pr(Adam winning if he go last) = (1 - 1/2) * (1 - 1/3) * (1-1/4) * 1/5 + [(1 - 1/2)² * (1 - 1/3)² * (1-1/4)²(1-1/5)] * 1/5 + ..........

= (1 - 1/2) * (1 - 1/3) * (1-1/4) * (1/5)/ [1 - (1 - 1/2) * (1 - 1/3) * (1-1/4) * (1-1/5)]

= 0.05/ 0.8 = 0.0625