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Beat The Book (Posted on 2021-03-10) Difficulty: 3 of 5
An online sportsbook is advertising a “Risk Free” promotion for new users: If your first wager loses, you will be awarded a “free bet” in the amount of your wager (up to $1,000).

In reading the terms and conditions of the promotion, you note that unlike a normal wager, a “free bet” does not return the stake along with any winnings. e.g. if you wager $1,000 cash on a selection with odds of 1.95 and win, you’d get back $1,950 (your $1,000 stake is returned, plus $950 in winnings); if you’d placed a $1,000 “free bet” on that same selection, you’d only be paid the $950 in winnings.

For this puzzle make the following simplifying assumptions:
- The sportsbook reduces the fair odds of all of their selections by 2.5%. For example, the true odds on selection with a 50% probability of occurring should be 2.00. However the sportsbook pays only 2 * (1 - 0.025) = 1.95.
- The sportsbook knows the true probabilities of all events occurring, and uniformly prices all selections with the same 2.5% margin.
- The sportsbook offers a large enough variety of markets that any odds you seek are available for you to bet on.

a) Determine a strategy to maximize the expected value of this promotion.

b) Assuming you wanted to make this a truly risk-free proposition, maximize the amount of guaranteed profit you can get out of this promotion.

See The Solution Submitted by tomarken    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts computer calculations | Comment 1 of 10
clc; clearvars
max=0;
for p=.01:.14:.99
  fprintf('%5.2f  ',p)  
  for p2=.01:.14:.99  
    q=1-p; q2=1-p2;
    bet=1000;
    nomPayoff=bet/p-bet;
    nomPayoff2=bet/p2-bet;
    payFrac=.975;
    expVal=-bet+p*(nomPayoff*payFrac+bet)+q*p2*nomPayoff2*payFrac;
    fprintf('    %6.2f', expVal)
    if expVal>max
        max=expVal;
        mp=p; mp2=p2;
    end
  end
  disp(" ")
end
 disp([mp mp2 max])   
 
finds the highest expected value when the probability of winning either bet is the lowest, that is has the highest payoff. That's at p=.01 of winning for each game and an expected value overall of 930.8475. The program produces a brief table of expected value:

          p2
  p1        .01       .15       .29       .43       .57       .71       .85       .99
 0.01      930.85    795.71    660.58    525.44    390.31    255.17    120.04    -15.10 
 0.15      799.21    683.19    567.16    451.14    335.11    219.09    103.06    -12.96 
 0.29      667.58    570.66    473.75    376.83    279.92    183.00     86.09    -10.83 
 0.43      535.94    458.14    380.33    302.53    224.72    146.92     69.11     -8.69 
 0.57      404.31    345.61    286.92    228.22    169.53    110.83     52.14     -6.56 
 0.71      272.67    233.09    193.50    153.92    114.33     74.75     35.16     -4.42 
 0.85      141.04    120.56    100.09     79.61     59.14     38.66     18.19     -2.29 
 0.99        9.40      8.04      6.67      5.31      3.94      2.58      1.21     -0.15 

So play the higest odds both times and have the best chance of winning almost a hundred thousand dollars, although it's still more likely you'll lose the $1000. 

b) I can't see any way of getting a guaranteed profit. Both bets can lose.
 

Edited on March 10, 2021, 12:41 pm
  Posted by Charlie on 2021-03-10 12:39:51

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