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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 opponent’s 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..).

What’s your view of these three arguments and how do things change if the players, while still aware that the deck is finite, don’t know what are the lowest and highest numbers?

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

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Ingeniousus Remarks.. | Comment 7 of 16 |
(In reply to But game theory says (part II) ... by Steve Herman)

Professor Emeritus Ingeniosus' happened to be passing through the university on day and was delighted to read the student analyses.

"Steve Herman was really the only one on the right track. Steve was right on the mark when he said that no betting ever takes place. Let's see if I can help him along in identifying some Nash equilibria.."

Ingeniosus' solution:

Let Wk = probability player W agrees to bet whenever he sees card k.

Then player A’s expected payoff after participating in N rounds is

RA = (N/16) S Ak QBk where:

QB1 = 2 B2

QB9 = -9 B8 and

QBk = [k+1] Bk+1 – k Bk-1 for 1 < k < 9

It follows that player A should choose

Ak = 1 if QBk > 0

Ak = 0 if QBk > 0

Ak = any if QBk = 0

Considering the sign of QB9 we get B8 A9 = 0 ( = A8 B9, by symmetry). This leads to a cascade of similar relations thru the QBk and QAk equations, so that

A2k+1 B2k = 0 and A2k B2k+1 = 0.

In other words, betting never takes place.

Some of Nash equilibria are:

Both players offer to bet only when they see an even card

Both players offer to bet only when they see (1,3,5,8)

Edited on February 13, 2008, 8:19 pm
  Posted by FrankM on 2008-02-13 20:16:45

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