Challenge Archive Problem (Funball) 06/16/03

[exl2]

In the game of Funball, each batter can either hit a home run, hit a single, or strikeout, and the likelihood of each outcome is completely determined by the opposing pitcher. A Funball batter scores a point for their team by advancing sequentially through each of four "bases", according to the following rules:

Home run: The batter and any players already on a base advance through all four bases.
Single: The batter advances to first base, and any players already on a base advance one base each.
Strikeout: No one advances any bases, and the batter loses his/her turn.
If the batting team has a runner on first base, which pitcher (Roger or Greg) is more likely to allow a point before recording a strikeout?
(1) Greg is twice as likely as Roger to allow a single, and four times as likely as Roger to record a strikeout.
(2) Greg is twice as likely as Roger to allow a single, and one fourth as likely as Roger to allow a home run.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

The answer explanation says that the probability of a strikeout occurring before a point is recorded is
S= single
K= strikeout
H= Home run

S*S*K + S*K + K

That makes sense to me. So wouldn't the probability of a point recorded before a strikeout be:

S*S*S+ S*S*H + S*H + H

And if that was the case, and say hypothetically both pitchers had equal probability of allowing a home run and allowing a strikeout (H and K are equal for both pitchers), but Greg was twice as likely to allow a single, wouldn't that make both equations above greater (greg is more likely to allow a strikeout before a point and greg is more likely to record a point before a strikeout)? Isn't that impossible?

I must be looking at something wrong here right? To quote the Answer explanation

" This means that Greg is more likely than Roger to record a strikeout before allowing a point (which, in turn, means that Rorger is more likely than Greg to allow a point before recording a strikeout.) "

But if I was to think about it, we have three mutually exclusive probabilities S,K,H, and 3 levels in a decision tree being considered. So technically, Greg should be able to have a higher probability of striking out before a point, and recording a point before a strike than Roger because the total probability should be:

point before a strikeout + Strikeout before a point + X
SSS +SSH +SH +H +SSK +SK +K + (X)

with X being all the other possibilities like KSK. Thus shouldn't the answer be E?

As an addendum, I understand that my question is not the question asked, but I would like to know why I am wrong here or if the answer explanation is wrong.

[mschwrtz]

Let rH=the probability of a home run when Roger pitches, etc., and let gH=the probability of a home run when Greg pitches, etc. Since H, S, and K are the only possible outcomes, rH+rS+rK=1=gH+gS+gK.

So your imagined scenario, both pitchers had equal probability of allowing a home run and allowing a strikeout ... but Greg was twice as likely to allow a single, is impossible. If rH=gH, rK=gK, 2rS=gS, and rH+rS+rK=1 then gH+gS+gK=1+rS. But that can't be.