[Math] Math riddle – Forehead numbers

puzzle

Three players, Annie, Billy and Katie, each have a natural number written on their foreheads. Each player can only see the foreheads of the other two. The following two things are common knowledge among the players:

One of the numbers is the sum of the other two.

One of the numbers is equal to exactly two thirds of one of the other numbers.

Annie, Billy and Katie tell the truth the whole time without exception. Each player is also a perfect logician: anything that they could infer from the information and observations available to them, they instantly will.

The following truthful conversation may or may not take place:

Annie: I don’t know my number.
Billy: I don’t know my number.
Katie: I don’t know my number.
Annie: I now know my number, and it is 150.

Is the above conversation plausible or not? Why?

My answer is NO. There are $2$ cases where one term of addition is $2/3$ from the other term or is $2/3$ from sum. The first player (Annie) doesn't know her number only if she sees two numbers which are in ratio $2:3$ and doesn't know in what case she is. For example Annie sees $300$ and $450$ and can't tell if her number is $150$ or $750$ or she sees $60$ and $9$0 and her number can be $150$ or $30$. Because of this, the second player (Billy) can know his number by computing sum and difference and see which one is in ratio $2:3$ with one of the numbers he sees.

I want to know if what I said is true and if not why. But please I don't want the right answer of the riddle. Thanks

Best Answer

Your logic is fine. The common knowledge says that the numbers are in either the ratio $1:2:3 \text{ or } 2:3:5$. As you say, Annie must see two numbers in the ratio $2:3$ to make her statement. Then Billy will not see numbers in the ratio $2:3$ and (by the same argument we made about Annie) knows his number without referring to Annie's statement at all.

Related Solutions

[Math] Dot on forehead riddle

If the villain isn't sufficiently demented to tell you things you already knew, you could reason like this: Since the villain had to tell you that some of you may have an ink dot on your forehead, you don't both have one; thus if you both guess that you don't have one, one of you must be right.

[Math] Bidding Tic Tac Toe

The paper by Develin and Payne referenced in the comments gives a very thorough treatment of this type of game, and in particular addresses the complicated nuances of breaking ties that arise during the bidding. However, in the Richman games they treat, the winning bid is given to the other player... this problem is slightly different. Glossing over the nuances of tied bids, we can see the following. (Assume that $B$ has one unit to bid with, and that both players will bid and play optimally.)

Every game state $G$ has a critical ratio $x(G)$ such that player $A$ can win if he has more than $x(G)$ to bid, and cannot win if he has less than $x(G)$ to bid.
A game state that is already won by player $A$ has $x(G)=0$; a game state already drawn or lost by player $A$ has $x(G)=\infty$.
Player $A$ can win with $x$ to bid when he has a winning bid, i.e., an amount $0 \le y \le x$ such that player $B$ loses if he bids more than $y$ (because $x(G_B) \le x/(1-y)$ for any state $G_B$ that player $B$ can move to; this applies only for $y<1$), and also loses if he bids less than $y$ (because $x(G_A) \le x - y$ for some state $G_A$ that player $A$ can move to).

We see two constraints on a winning bid for player $A$. First, $y \le x - x(G_A)$ for some state $G_A$ that player $A$ can move to; i.e., $y \le x - \min_{G_A} x(G_A)$. Second, either $y \ge 1$ (so player $B$ can't outbid it), or else $y\ge 1 - x / x(G_B)$ for all states $G_B$ that player $B$ can move to; i.e., $y \ge 1 - x / \max_{G_B} x(G_B)$. The latter constraint is never stronger than the former, so we have: $$ 1-x/\max_{G_B}x(G_B) \le y \le x-\min_{G_A}x(G_A). $$ This interval of winning bids shrinks as $x$ decreases, and vanishes (by definition) when $x=x(G)$. So setting the two endpoints equal to one another lets us solve for $x(G)$: $$ 1-x(G)/\max_{G_B}x(G_B) = x(G)-\min_{G_A}x(G_A) \implies x(G)=\frac{1+\min_{G_A}x(G_A)}{1+1/\max_{G_B}x(G_B)}. $$

At this point we can calculate. I'm pasting Python code to evaluate this recursion below. The result that I get for Tic-Tac-Toe is $x(G_0)\approx 1.018375$. I have modified the code to give an exact (rational) result (code not shown here, as it's not worth the extra space): it is $$x(G_0)=\frac{396925}{389763}=1+\frac{7162}{389763}.$$


g0 = '---|---|---'
NO = '-'
DRAW = 'x'

def legalMoves(g, player):
  ret = []
  for i in xrange(0, len(g)):
    if g[i]==NO: ret.append(g[:i] + player + g[(i+1):])
  return ret

def winner(g):
  for i in xrange(3):
    if g[i]==g[i+4] and g[i+4]==g[i+8] and g[i]!=NO: return g[i]
    if g[4*i]==g[4*i+1] and g[4*i+1]==g[4*i+2] and g[4*i]!=NO: return g[4*i]
  if g[0]==g[5] and g[5]==g[10] and g[0]!=NO: return g[0]
  if g[2]==g[5] and g[5]==g[8] and g[2]!=NO: return g[2]
  if NO in g: return NO
  return DRAW

def value(g, infty=10**12, cache={}):
  if cache.has_key(g): return cache[g]
  w = winner(g)
  if w == NO:
    amin = min(map(lambda(gg):value(gg, infty, cache), legalMoves(g, 'A')))
    bmax = max(map(lambda(gg):value(gg, infty, cache), legalMoves(g, 'B')))
    val = (1.0 + amin) / (1.0 + 1.0 / bmax)
  elif w=='A': val = 0
  else: val = infty
  cache[g] = val
  return val

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