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Grand Anticipation (Posted on 2008-12-24) Difficulty: 3 of 5
Two teams play against each other in a tournament. The rules of the tournament do not allow any drawn games and the first team to win four games wins the tournament. Therefore, a minimum of four, and a maximum of seven games must be played in the tournament to determine the winner.

Riley supports one of the teams, and wants to end up having lost $1000 if his team loses the tournament, and having won $1000 if his team wins the tournament. Riley wants to wager an amount on each game, and in terms of betting rules he will win that amount of money if the team he bets on wins, and lose it if the team loses. The amount wagered by Riley may not necessarily be the same for any two games.

How much money should Riley bet on the first game?

Note: For the purposes of the problem, assume that Riley will bet on every game until the tournament ends, so that he cannot bet $0 on any game.

No Solution Yet Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

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Solution Anticipating the end by working back. | Comment 3 of 4 |

Let's call this tournament "the series".

We can work from the end back.

If the series gets to 3-3 (3 wins and 3 losses), at the end of the next game, the series will have been won or lost, meaning that this punter wants to have winnings of +/-1000. That further means that coming into that game, with the series tied at 3-3, he wants to have zero winnings (the average between +1000 and -1000), and needs to bet 1000 (half the difference between these values).

So whatever situations could lead to 3-3 must have a betting scheme that will result in zero net winnings if it goes to 3-3. Of course those situations are 3-2 and 2-3.  In these instances, the alternative to it going to 3-3 is a win or a loss, respectively, of the series as a whole. So the bettor wants the net result after such to be either zero winnings, in the case of going to 3-3, or +/-1000 in the case of a win/loss of the series. Thus in these cases, he needs to have +/-500 net winnings coming into the 3-2 or 2-3 situations (from 2-2 or 3-1/1-3) and bet 500. That's because 500 is the average of the two outcomes he needs and because 500 is half the difference between the two needed outcomes.

Originally I worked this out backward in this manner, ignoring the signs, assuming the symmetry makes it work the same for negative outcomes as for positive, eventually arriving at a bet of 312.50 for the first game. This indeed is correct as verified (as also the numbers) by the computer program below.

The table of bets for the various possible states of the series is given below (read from bottom up to understand the logic):

wins    losses          bet        needed prior
                                   net winnings
 0         0          312.50            0.00
 0         1          312.50         -312.50
 0         2          250.00         -625.00
 0         3          125.00         -875.00
 1         0          312.50          312.50
 1         1          375.00            0.00
 1         2          375.00         -375.00
 1         3          250.00         -750.00
 2         0          250.00          625.00
 2         1          375.00          375.00
 2         2          500.00            0.00
 2         3          500.00         -500.00
 3         0          125.00          875.00
 3         1          250.00          750.00
 3         2          500.00          500.00
 3         3         1000.00            0.00

So, to repeat, the bet on the first game should be 312.50.

The program, being recursive, goes through all the possibilities when figuring out what 0-0 should be, and indeed all the cases below any given line on the chart, and so does a given line many times, but it is the simplist way of programming it. It averages the needs of the two possible outcomes of the next games to figure the line's own need, and takes half the difference as its own bet value.

DECLARE SUB figure (w!, l!, bet!, need!)
FOR w = 0 TO 3
FOR l = 0 TO 3

  figure w, l, bet, need
  PRINT USING "##        ##         ####.##         ####.##"; w; l; bet; need


SUB figure (w, l, bet, need)
  IF w = 4 THEN need = 1000: bet = 0: EXIT SUB
  IF l = 4 THEN need = -1000: bet = 0: EXIT SUB
  figure w, l + 1, b1, n1
  figure w + 1, l, b2, n2
  need = (n1 + n2) / 2
  bet = ABS(n1 - n2) / 2



Edited on December 24, 2008, 5:43 pm
  Posted by Charlie on 2008-12-24 17:41:10

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