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Forced Win, Forced Loss, or Neutral? (Posted on 2008-02-08) Difficulty: 4 of 5
Professor Paradoxicus has invented a new game. A single card is dealt to each of two players. The cards are consecutive and come from a 9 card deck, each card bearing a distinct integer between 1 and 9. Each player then holds his card (number facing outward) against his forehead; ie, each player can see his opponentís card, but not his own. The players are then asked whether they want to bet, and if both agree, they examine their own card, with the player who drew the lower card paying his opponent the amount on the higher card.

Professor Paradoxicus has invited three students to analyse the game.

Simplicimus notes that the game is symmetric and zero sum. He asserts that players will be indifferent as to whether or not to bet.

Optimisticus points out that whenever a player sees his opponent holding a 9 he can count on losing. On the other hand, Optimisticus reasons, if the opponent is holding any card N<9, then that player has a 50% chance of winning N+1 and a 50% chance of losing N. Optimisticus asserts that a player will quit when he sees his opponent holding a 9, and choose to play otherwise.

Finally, Sceptisimus expresses the view that a player will refuse to bet unless he sees his opponent holding a 1. He justifies his peculiar opinion as follows: Suppose a player sees his opponent holding a 9. He will certainly refuse to bet. Now suppose a player sees his opponent holding an 8. Clearly, he himself will be holding either a 9 or a 7. In the former case, he can be sure that his opponent will refuse to bet. He concludes that he has nothing to gain by agreeing to bet, but could have something to lose. He therefore should refuse to bet.

Sceptisimus next reapplies this argument repeatedly, with 7 (,6,5..) taking the role of 8 (,7,6..).

Whatís your view of these three arguments and how do things change if the players, while still aware that the deck is finite, donít know what are the lowest and highest numbers?

See The Solution Submitted by FrankM    
Rating: 1.5000 (2 votes)

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Solution But game theory says (part II) ... | Comment 6 of 16 |
Sceptimus is right!

I am guilty of misunderstanding that the cards must be consecutive, but my analysis (slightly recast) still stands.

Taking a game theory approach, each player has 10 basic strategies.  I assert (without proof) that any other strategy is dominated by one of these:

The basic 10 strategies are:

A -- Always bet
8 -- Bet only if the opponent has 8 or less
7 -- Bet only if the opponent has 7 or less
6 -- Bet only if the opponent has 6 or less
5 -- Bet only if the opponent has 5 or less
4 -- Bet only if the opponent has 4 or less
3 -- Bet only if the opponent has 3 or less
2 -- Bet only if the opponent has 2 or less
1 -- Bet only if the opponent has 1
N -- Never bet

The expected results matrix is calculated as follows:

a) If either player adopts strategy N, then no betting occurs, and the game is Neutral
b) If both players adopts strategy 1, then no betting occurs, and the game is Neutral
c) If both players adopt the same strategy that is higher then 1, then betting occurs occasionally, and the game is neutral (things average out)
d) If players adopt different strategies that are not strategy N, then betting occurs occasionally, and the player who is less willing to bet has the advantage.  For instance, I adopt strategy 4, betting only if you have 4 or less.  You adopt strategy 5, betting if I have 5 or less.  1 time out of 18, I have a 5 and you have a 4 and I win 5.  6 times out of 18 we both have 4 or 3 or 2 or 1 and we bet and things average out.  My expected profit is +5/18 per game, and your expected win is - 5/18 per game.

So, quantities aside, the results matrix for different strategy combinations is as follows:

   A  8 7  6 5 4  3  2 1 N
   ------------------------
A|0  -  -  -  -  -  -  - -  0
8|+ 0  -  -  -  -  -  - -  0
7|+ + 0  -  -  -  -  - -  0
6|+ + + 0  -  -  -  - -  0
5|+ + + + 0  -  -  - -  0
4|+ + + + + 0  -  - -  0
3|+ + + + + + 0  - -  0
2|+ + + + + + + 0 -  0
1|+ + + + + + + + 0 0
N|0 0  0  0 0  0 0 0  0 0

For each player, strategy A is inferior to (dominated by) Strategy 1, so neither player should Always Bet, and this strategy can be removed from the results matrix.  The reduced matrix looks like:

    8 7  6 5 4  3  2 1 N
   ----------------------
8|0  -  -  -  -  -  - -  0
7|+ 0  -  -  -  -  - -  0
6|+ + 0  -  -  -  - -  0
5|+ + + 0  -  -  - -  0
4|+ + + + 0  -  - -  0
3|+ + + + + 0  - -  0
2|+ + + + + + 0 -  0
1|+ + + + + + + 0 0
N|0  0  0 0  0 0 0 0 0

Now strategy 8 (bet unless the opponent has a 9) is inferior to (dominated by) Strategy 1, so neither player should use strategy 8, and this strategy can be removed from the results matrix.  Repeating this process results in the following matrix:

   1 N
   ----
1|0 0
N|0 0

The only strategies that remain are to Never Bet, or to bet only if the opponent shows a 1.  Since both players cannot have a 1, betting never takes place. 
  Posted by Steve Herman on 2008-02-10 23:14:01
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