Question

Jack and Jill are playing Bean Bag Toss. Jack throws into the hole 50% of times and Jill throws into the hole 75% of the times. If they start a turn-basedmatch, what is the probability that Jill gets the first hole shot if:

a) Jack starts the game ?

b) Jill starts the game?

Answer 1

Probability that Jack throws into the hole =1/2

Probability that Jill throws into the hole =3/4

Probability that Jack fails to throw into the hole =1/2

Probability that Jill fails to throw into the hole =1/4

a)

So, if Jack starts, to make Jill gets the first hole shot, Jack should fail in first shot, then Jill can succeed in 2nd throw,

or Jack fails, then Jill fails,Jack fails then again Jill succeeds,

or Jack fails, Jill fails, Jack fails, Jill fails, Jack fails, Jill succeeds and so on.

So, required probability

1/2*3/4+1/2*1/4*1/2*3/4+1/2*1/4*1/2*1/4*1/2*3/4+...

=3/8 *(1+1/8+1/64+...)

=3/8 *(1/(1-1/8)) [infinite gp]

=3/7

b)

So, if Jill starts, he can succeed in first throw,

or he fails, then Jack fails, then again he succeeds,

or he fails, Jack fails, he fails, Jack fails, he succeeds ans so on.

So, required probability

=3/4+1/4*1/2*3/4+1/4*1/2*1/4*1/2*3/4+...

=3/4 *(1+1/8+1/64+...)

=3/4 *(1/(1-1/8)) [infinite gp]

=6/7

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