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For the PerplexusBowl match between the Pascal Probabilities and the Random Results, a bookie was offering the following payoffs:

• PP to win in normal time, 3 to 2
• RR to win in normal time, 2 to 1
• PP to win in overtime, 7 to 1
• RR to win in overtime, 9 to 1

• (The first line means that if you bet \$2 on PP to win in normal time, and it does, you get your money back plus \$3.)

Without knowing anything about football or the involved teams or the actual probabilities, can you show why these payoffs are illogical?

 See The Solution Submitted by Federico Kereki Rating: 4.0000 (5 votes)

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 No Subject | Comment 2 of 8 |

Beting "a" in PP, youŽll gain 5a/2.
Beting "b" in RR, youŽll gain 3b/1.
Beting "c" in PPO, youŽll gain 8c/1.
Beting "d" in RRO, youŽll gain 10d/1.

Total bet: TB = a + b + c + d.
Total gain in normal time: TGNT = 5a/2 or 3b
Total gain in overtime: TGOT = 8c or 10d.

Hyp 1) Normal time:
Netgain = TGNT - TB = 5a/2 - (a + b + c + d) = 3a/2 - (b + c + d) or 3b - (a + b + c + d) = 2b - (a + c + d).

Hyp 2) Overtime:
Netgain = TGOT - TB = 8c - (a + b + c + d) = 7c - (a + b + d) or 10d - (a + b + c + d) = 9d - (a + b + c).

3a/2 - (b + c + d) > 0
2b - (a + c + d) > 0
7c - (a + b + d) > 0
9d - (a + b + c) > 0

3a > 2(b+c+d)
2b > (a+c+d)
7c > (a+b+d)
9d > (a+b+c)

3a > 2b + 2(c+d) > (a+c+d) + 2(c+d) ===> a > 3(c+d)/2

3a + 2b + 7c + 9d > 3a + 4b + 4c + 4d
3c + 5d > 2b ====> b < (3c+5d)/2.

7c > a+b+d ===> d < 7c - a - b

d < 7c - 3(c+d)/2 - b = 11c/2 - 3d/2 - b
5d/2 < 11c/2 - b
5d < 11c - 2b

The worst case is b = (3c+5d)/2, then 5d < 11c - 3c - 5d ===> d = 4c/5.

To be continued.....

Edited on November 26, 2005, 5:58 pm
 Posted by pcbouhid on 2005-11-26 12:14:25

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