This week the #puzzle is: Can You Throw the Hammer? #probabilities #game

You and your opponent are competing in a golf match. On any given hole you play, each of you has a 50 percent chance of winning the hole (and a zero percent chance of tying). That said, scorekeeping in this match is a little different from the norm.

Each hole is worth 1 point. Before starting each hole, either you or your opponent can “throw the hammer.” When the hammer has been thrown, whoever did not throw the hammer must either accept the hammer or reject it. If they accept the hammer, the hole is worth 2 points. If they reject the hammer, they concede the hole and its 1 point to their opponent. Both players can throw the hammer on as many holes as they wish. (Should both players decide to throw the hammer at the exact same time—something that can’t be planned in advance—the hole is worth 2 points.)

The first player to reach 3 points wins the match. Suppose all players always make rational decisions to maximize their own chances of winning.

Good news! You have won the first hole, and now lead 1-0. What is your probability of winning the match?

And for extra credit:

Instead of playing to 3 points, now the first player to 5 points wins the match.

Good news (again)! You have won the first hole, and now lead 1-0. What is your probability of winning the match?

Highlight to reveal (possibly incorrect) solution:

Program.

My own way of saying the above:

• The players are Alice (playing for me) and Bob.
• For any given hole, for any player X, p(X wins) = 50%.
• Each win is worth 1 point, unless a hammer is thrown and accepted, in which case it’s worth 2 points.
• A player can decide to throw a hammer.
• A player can decide to accept or reject a thrown hammer. Rejecting a hammer is the same as losing, worth 1 point.
• If both players throw hammers, it counts as an accepted hammer.
• One player throwing a hammer and the other accepting is equivalent to both players throwing hammers at the same time.
• The winner of the whole match is the first to reach 3 points.
• Players choose the optimal strategy.
• Alice won the first hole, and is in front 1-0.
• What is probability that Alice will win?
• The notation P_x,y is adopted to mean: the probability Alice will win when the match is at x-y. So we’re looking for P_1,0.
• A strategy consists of these choices: Given x-y, should Alice throw the hammer, or, if the hammer is thrown by Bob, accept or reject?
• In a table for P_x,y, choosing the optimal strategy for Alice, I can also see P_y,x and the optimal strategy for Bob.

Trivially, P_3+,y = 1 and P_x,3+ = 0.

With this in mind, let’s look at a table for P_2,2. Alice has the choices in the left column, Bob has the choices in the top row.

(My tables don’t quite reflect, that accept/reject occurs before the hole is played. I try to capture this nuance in the text instead. It’s important, because a match may be decided right there.)

Bob
P2,2
Alice	accept	reject	hammer
accept	50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%	50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%	50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%
reject	50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%	50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%	P2,3 = 0
hammer	50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%	P3,2 = 1	50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%

First, let’s notice, that if player X throws a hammer, player Y should never reject, as that leads to instant defeat. Therefore the middle column and middle row drop out.

Alice has to choose the best row out of (50, 50) and (50, 50). As long as she accepts a thrown hammer, it doesn’t matter what she chooses otherwise.

P_2,2 = 50%.

Bob
P2,1
Alice	accept	reject	hammer
accept	50% * P3,1 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%	50% * P3,1 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%	50% * P4,1 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%
reject	50% * P3,1 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%	50% * P3,1 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%	P2,2 = 50%
hammer	50% * P4,1 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%	P3,1 = 1	50% * P4,1 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%

Bob is trying to choose the optimal (i.e. lowest) column from (75, 75, 50), (75, 75, 1) and (50, 50, 50). He will choose the third one and therefore throw a hammer.

P_1,2 = 50%.

Knowing Bob will throw a hammer, it doesn’t really matter what Alice chooses. All her choices give 50%.

P_2,1 = 50%.

Bob
P2,0
Alice	accept	reject	hammer
accept	50% * P3,0 + 50% * P2,1 = 50% * 1 + 50% * 50 % = 75%	50% * P3,0 + 50% * P2,1 = 50% * 1 + 50% * 50 % = 75%	50% * P4,0 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%
reject	50% * P3,0 + 50% * P2,1 = 50% * 1 + 50% * 50 % = 75%	50% * P3,0 + 50% * P2,1 = 50% * 1 + 50% * 50 % = 75%	P2,1 = 50%
hammer	50% * P4,0 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%	P3,0 = 1	50% * P4,0 + 50% * P2,2 = 50% * 1 + 50% * 50% = 75%

Alice wants the best row of (75, 75, 75), (75, 75, 50) and (75, 1, 75). This is the third row, so she throws a hammer.

Bob wants the best column of (75, 75, 75), (75, 75, 1) and (75, 50, 75). This is the third column, so he throws a hammer.

We end up in the situation where both throw a hammer.

P_2,0 = 75%.

P_0,2 = 25%.

Bob
P1,1
Alice	accept	reject	hammer
accept	50% * P2,1 + 50% * P1,2 = 50% * 50% + 50% * 50% = 50%	50% * P2,1 + 50% * P1,2 = 50% * 50% + 50% * 50% = 50%	50% * P3,1 + 50% * P1,3 = 50% * 1 + 50% * 0 = 50%
reject	50% * P2,1 + 50% * P1,2 = 50% * 50% + 50% * 50% = 50%	50% * P2,1 + 50% * P1,2 = 50% * 50% + 50% * 50% = 50%	P1,2 = 50%
hammer	50% * P3,1 + 50% * P1,3 = 50% * 1 + 50% * 0 = 50%	P2,1 = 50%	50% * P3,1 + 50% * P1,3 = 50% * 1 + 50% * 0 = 50%

Everything is 50%. Either player can do anything.

P_1,1 = 50%.

Bob
P1,0
Alice	accept	reject	hammer
accept	50% * P2,0 + 50% * P1,1 = 50% * 75% + 50% * 50% = 62.5%	50% * P2,0 + 50% * P1,1 = 50% * 75% + 50% * 50% = 62.5%	50% * P3,0 + 50% * P1,2 = 50% * 1 + 50% * 50% = 75%
reject	50% * P2,0 + 50% * P1,1 = 50% * 75% + 50% * 50% = 62.5%	50% * P2,0 + 50% * P1,1 = 50% * 75% + 50% * 50% = 62.5%	P1,1 = 50%
hammer	50% * P3,0 + 50% * P1,2 = 50% * 1 + 50% * 50% = 75%	P2,0 = 75%	50% * P3,0 + 50% * P1,2 = 50% * 1 + 50% * 50% = 75%

The best row for Alice is the third, and she throws a hammer.

P_1,0 = 75%.

The best columns for Bob are those where he doesn’t throw a hammer. Given that Alice does throw one, he ends up in the bottom row.

P_0,1 = 25%.

Oh! We’ve actually found the value, we were looking for. P_1,0 = 75%.

Just to make sure, I also wrote a program to confirm this (and the next) result.

And for extra credit:

Everything is the same, we’re just going for a win of 5 instead of 3.

Trivially, P_5+,y = 1 and P_x,5+ = 0.

If we read P_x,y in the fiddler as “Alice is 3-x away from having 3 points and Bob is 3-y away”, we can translate all of this into P_x+2,y+2 for this extra credit puzzle.

P4,4 = 50%	P2,4 = 25%
P3,4 = 50%	P3,3 = 50%
P4,3 = 50%	P3,2 = 75%
P4,2 = 75%	P2,3 = 25%

Having established that, let’s move forward.

Bob
P4,1
Alice	accept	reject	hammer
accept	50% * P5,1 + 50% * P4,2 = 50% * 1 + 50% * 75% = 62.5%	50% * P5,1 + 50% * P4,2 = 50% * 1 + 50% * 75% = 62.5%	50% * P6,1 + 50% * P4,3 = 50% * 1 + 50% * 50% = 75%
reject	50% * P5,1 + 50% * P4,2 = 50% * 1 + 50% * 75% = 62.5%	50% * P5,1 + 50% * P4,2 = 50% * 1 + 50% * 75% = 62.5%	P4,2 = 75%
hammer	50% * P6,1 + 50% * P4,3 = 50% * 1 + 50% * 50% = 75%	P5,1 = 1	50% * P6,1 + 50% * P4,3 = 50% * 1 + 50% * 50% = 75%

The best row for Alice is the 3rd, throwing a hammer.

Given this, and rejecting a hammer would be a losing strategy, the best column for Bob is the 1st, accepting a hammer.

P_4,1 = 75%. P_1,4 = 25%.

Bob
P4,0
Alice	accept	reject	hammer
accept	50% * P5,0 + 50% * P4,1 = 50% * 1 + 50% * 75% = 87.5%	50% * P5,0 + 50% * P4,1 = 50% * 1 + 50% * 75% = 87.5%	50% * P6,0 + 50% * P4,2 = 50% * 1 + 50% * 75% = 87.5%
reject	50% * P5,0 + 50% * P4,1 = 50% * 1 + 50% * 75% = 87.5%	50% * P5,0 + 50% * P4,1 = 50% * 1 + 50% * 75% = 87.5%	P4,1 = 75%
hammer	50% * P6,0 + 50% * P4,2 = 50% * 1 + 50% * 75% = 87.5%	P5,0 = 1	50% * P6,0 + 50% * P4,2 = 50% * 1 + 50% * 75% = 87.5%

The best row for Alice is 3rd, she throws a hammer.

The best column for Bob is the 3rd, he throws a hammer.

P_4,0 = 87.5%. P_0,4 = 12.5%.

Bob
P3,1
Alice	accept	reject	hammer
accept	50% * P4,1 + 50% * P3,2 = 50% * 75% + 50% * 75% = 75%	50% * P4,1 + 50% * P3,2 = 50% * 75% + 50% * 75% = 75%	50% * P5,1 + 50% * P3,3 = 50% * 1 + 50% * 50% = 75%
reject	50% * P4,1 + 50% * P3,2 = 50% * 75% + 50% * 75% = 75%	50% * P4,1 + 50% * P3,2 = 50% * 75% + 50% * 75% = 75%	P3,2 = 75%
hammer	50% * P5,1 + 50% * P3,3 = 50% * 1 + 50% * 50% = 75%	P4,1 = 75%	50% * P5,1 + 50% * P3,3 = 50% * 1 + 50% * 50% = 75%

Everything is 75%. Either player can do anything.

P_3,1 = 75%. P_1,3 = 25%.

Bob
P2,2
Alice	accept	reject	hammer
accept	50% * P3,2 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%	50% * P3,2 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%	50% * P4,2 + 50% * P2,4 = 50% * 75% + 50% * 25% = 50%
reject	50% * P3,2 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%	50% * P3,2 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%	P2,3 = 25%
hammer	50% * P4,2 + 50% * P2,4 = 50% * 75% + 50% * 25% = 50%	P3,2 = 75%	50% * P4,2 + 50% * P2,4 = 50% * 75% + 50% * 25% = 50%

The best row for Alice is the 3rd. She throws a hammer.

The best column for Bob is the 3rd. He throws a hammer.

P_2,2 = 50%.

Bob
P3,0
Alice	accept	reject	hammer
accept	50% * P4,0 + 50% * P3,1 = 50% * 87.5% + 50% * 75% = 81.25%	50% * P4,0 + 50% * P3,1 = 50% * 87.5% + 50% * 75% = 81.25%	50% * P5,0 + 50% * P3,2 = 50% * 1 + 50% * 75% = 87.5%
reject	50% * P4,0 + 50% * P3,1 = 50% * 87.5% + 50% * 75% = 81.25%	50% * P4,0 + 50% * P3,1 = 50% * 87.5% + 50% * 75% = 81.25%	P3,1 = 75%
hammer	50% * P5,0 + 50% * P3,2 = 50% * 1 + 50% * 75% = 87.5%	P4,0 = 87.5%	50% * P5,0 + 50% * P3,2 = 50% * 1 + 50% * 75% = 87.5%

For Alice the 3rd row is best. She throws a hammer.

Knowing this, it doesn’t matter what Bob does.

P_3,0 = 87.5%. P_0,3 = 12.5%.

Bob
P2,1
Alice	accept	reject	hammer
accept	50% * P3,1 + 50% * P2,2 = 50% * 75% + 50% * 50% = 62.5%	50% * P3,1 + 50% * P2,2 = 50% * 75% + 50% * 50% = 62.5%	50% * P4,1 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%
reject	50% * P3,1 + 50% * P2,2 = 50% * 75% + 50% * 50% = 62.5%	50% * P3,1 + 50% * P2,2 = 50% * 75% + 50% * 50% = 62.5%	P2,2 = 50%
hammer	50% * P4,1 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%	P3,1 = 75%	50% * P4,1 + 50% * P2,3 = 50% * 75% + 50% * 25% = 50%

For Bob the best column is the 3rd. He throws a hammer.

Given this, it doesn’t matter what Alice chooses.

P_2,1 = 50%. P_1,2 = 50%.

Bob
P2,0
Alice	accept	reject	hammer
accept	50% * P3,0 + 50% * P2,1 = 50% * 87.5% + 50% * 50% = 68.75%	50% * P3,0 + 50% * P2,1 = 50% * 87.5% + 50% * 50% = 68.75%	50% * P4,0 + 50% * P2,2 = 50% * 87.5 + 50% * 50% = 68.75%
reject	50% * P3,0 + 50% * P2,1 = 50% * 87.5% + 50% * 50% = 68.75%	50% * P3,0 + 50% * P2,1 = 50% * 87.5% + 50% * 50% = 68.75%	P2,1 = 50%
hammer	50% * P4,0 + 50% * P2,2 = 50% * 87.5 + 50% * 50% = 68.75%	P3,0 = 87.5%	50% * P4,0 + 50% * P2,2 = 50% * 87.5 + 50% * 50% = 68.75%

For Alice the best row is the 3rd. She throws a hammer.

Knowing this, Bob just avoids rejecting it.

P_2,0 = 68.75%. P_0,2 = 31.25%.

Almost there!

For symmetry reasons, P_1,1 = 50%.

Bob
P1,0
Alice	accept	reject	hammer
accept	50% * P2,0 + 50% * P1,1 = 50% * 68.75% + 50% * 50% = 59.375%	50% * P2,0 + 50% * P1,1 = 50% * 68.75% + 50% * 50% = 59.375%	50% * P3,0 + 50% * P1,2 = 50% * 87.5 + 50% * 50% = 68.75%
reject	50% * P2,0 + 50% * P1,1 = 50% * 68.75% + 50% * 50% = 59.375%	50% * P2,0 + 50% * P1,1 = 50% * 68.75% + 50% * 50% = 59.375%	P1,1 = 50%
hammer	50% * P3,0 + 50% * P1,2 = 50% * 87.5 + 50% * 50% = 68.75%	P2,0 = 68.75%	50% * P3,0 + 50% * P1,2 = 50% * 87.5 + 50% * 50% = 68.75%

Alice throws a hammer.

Bob doesn’t reject.

P_1,0 = 68.75%.