For example: Xeno can have 2475 painted on his forehead, Yvonne 2476 on hers . Xeno sees the 2476 on Yvonne's forehead ; he therefore knows that his number is either 2475 or 2477, but not which one. Whoever is the first to figure out the number on his or her own forehead is declared the winner .
The game proceeds in rounds. You are the referee:
If no hand is raised proceed to round 2.
The players are allowed to take as much time as they wish. Once both concede that they don't know their own numbers, proceed to round 3:
I know! I know! My number is 2476!!
The crowd roars with excitement as you, in some bewilderment, hand her the coveted Consecutive Integer Game trophy
Next, suppose that Yvonne's forehead has a 2 and Xeno's has a 3. After round 1, Neither knows their own number. However, Xeno knows that he has either 1 or 3. He also knows that Yvonne didn't win in the first round. He therefore concludes in round 2 that his number must be a 3 !
Next: suppose that Xeno's forehead holds a 3 and Yvonne's holds a 4. Yvonne sees the 3 on Xeno's forehead and concludes that she knows that she has either a 2 or a 4 on hers . She's also studied the previous paragraph of this article, and knows that, if her number were a 2 , Xeno would have figured out his number by round 2 and raised his hand. won in round 2. Since Xeno failed to do this in round 2, she immediately concludes her number is 4 and wins in round 3.
...and so forth and so on....!
What we appear to have going here is a nice little exercise in mathematical induction. We can state this conclusion in the form of a theorem:
Theorem I: If two numbers are N and N+1 , then the player marked with the number N+1 will win in round N .
Proof:Since both integers must be positive, this assertion is clearly true for N = 1 .
Assume that it is true for N= k . Suppose that Xeno's forehead holds the integer k+1 , Yvonne's holds the integer k. Assume also that round k+1 arrives and that neither of them has raised their hand. Yvonne knows that her own number is either k or k+2 . If it were k , then since , by the induction assumption, Theorem 1 is true for k , Xeno would have seen the k on her forehead in round k, and won. Since he didn't, Yvonne knows in round k+1 that her number must be k+2 . She therefore raises her hand and wins.
QED
Second, how do we know that there can be no winner before round k+1 , or that there may be some way by which Xeno might figure out the value of his number by round k+1 ? In order to be certain that Yvonne will win, it is necessary to know that there is no better strategy that Xeno might use to figure out his own number sooner.
It will be shown that it is possible to give a proof of Theorem 1, provided that one adds specific assumptions about the knowledge and abilities of each of the players.
SN(A) : If one player's forehead holds an N , and there is no win before round N, then the other player's must hold an N+1
SN(B) : Neither player can determine his or her number before round N. Specifically, the player holding the number N cannot determine that it is N, in round N.
We will prove this by establishing Tk : for each positive integer k, If round k arrives without a previous win, then any player seeing a number larger than k on the forehead of the other player, will not know his number in that round.
Since the collections of statements {SN} , and {Tk} concern the behavior and abilities of the two players, we find ourselves obliged to clarify the qualifications needed for playing the game.
For example, if Yvonne sees a 3 onXeno's forehead, and he has not raised his hand in round 2, then Yvonne must be able to do the reasoning that leads her to conclude that her number is 4. She cannot do this without making the assumption that Xeno is able to reason in a similar fashion about the number on her forehead, and in fact has already done so! Otherwise he should not qualify as a player.
What this means is that there is a hidden assumption in the proof of Theorem 1 , namely that it is not enough to know that SN is true (for all N) , but one must also know that Yvonne knows that it is true!
Yet one must also know that, unless Xeno possesses superhuman abilities, or has bribed the person who put the number on his forehead, he will not be able, in round 1, to see a 4 on Yvonne's forehead and know that his number is a 3!
These conditions for a qualified player are summarized in three Axioms:
Axiom 1 eliminates my dog as a potential player.
Axiom 2, which is just T1 , eliminates supernatural agencies, God, telepaths and swindlers as potential players.
Axiom 3 guarantees that, for each pair N,k each player will know SN or Tk once we've proven them to be true. There is, however, a strange self-referential component to Axiom 3 : it implies that the definition of a player includes a requirement that he be able to prove certain statements about players . Furthermore, until we prove SN and Tk for each pair N, k , we don't know whether Axiom 3 might not turn out to be inconsistent, in which case there are no qualified players.
The situation is not hopeless, as we will show with:
Lemma 1: SN is provable, and therefore true, for each positive integer N .
Proof: Since all of the integers used in the game are positive, S1 is provable. Assume that
Lemma 2: Tk is provable, and therefore true, for each positive integer k
Proof: T1 is just Axiom2 . Assume
Since there has been no win, Y didn't know her number in round k . One must conclude that she did not acquire enough information from round k that would have allowed her to deduce her number in round k+1. The only information she could have acquired in round k, was that Xeno did not know that his number was q . One must conclude that Yvonne knew by the end of round k-1 that Xeno did not yet know his number.
Since Xeno's number is q , the smallest number Yvonne can have is q-1.
Since q >k+1 , q-1 >k .
Since Tk is provable, Yvonne knows Tk , which tells her that Xeno did not know his number in round k . QED
Theorem II: If
This clears up the situation with respect to Axiom 3 . Which of the statements {SN } and { Tk } must a player be able to prove? The answer is all of them , since they are all provable. Is the definition of a "qualified player" consistent? Yes, because there are at least two people who can prove all of these statements and therefore qualify as players.
I am a qualified player, since I just proved Lemmas 1 and 2 .
You are a qualified player since you've just read this essay!