World Series Puzzle


Source: Discussion with Ashwin Rao (http://zlemma.com, Ex-MD Morgan Stanley) and Gaurav Kumar (Credit Suisse IB, Ex-MS, GS, IITB CSE 2006, Booth 2012)



Problem:
Suppose teams A and B play in the world series of up to 7 games in which the first team to win 4 games wins the series and then no more games are played. Suppose that you want to bet on each individual game in such a way that when the series ends you will be ahead by exactly $100 if your team wins the series, or behind by exactly $100 if your team loses the series, no matter how many games it takes. How much would you bet on the first game?

Update (22 June 2014):
Solution posted by Alex, JDGM, Arnab in comments!


Comments

  1. Can you bet $0 ? I think not, but if so, then the answer is to bet $100 on the first game and $0 in the remaining.

    ReplyDelete
    Replies
    1. There is a problem with this strategy. Suppose you support team A and team A wins the first game. Then you are ahead by $100. Now suppose that B wins the next 4 rounds. So A loses but you have made $100.

      Delete
  2. Are the bets decided before the contest starts or they can be modifies as it goes on ?

    ReplyDelete
  3. I get $31.25. You should bet that amount on the second game as well. (Then it gets more complicated.)

    I'm interested in hearing other people's methods, but I just turned it into a matrix equation and solved.

    ReplyDelete
    Replies
    1. I went straight to the n=1 case looking for clues. It was trivial and gave me a good feeling about this problem because I am an optimist (!!). I moved onto n=2 and AHA! realised that whatever the bets are here, they have to reduce to the n=1 case if the first two games go one each way. That insight expands and generalises to build the tree backwards from known bets, with the values easy to calculate as I drew it. While I was doing that, I thought "I couldn't work these out in my head so easily if there weren't the right number of equations and unknowns - I bet it can be done with a matrix" - hahaha!

      Delete
  4. The answer is $31.25.

    What a delightful puzzle!

    Spoilers: first shot, neatened up.

    ReplyDelete
    Replies
    1. It is a delightful puzzle indeed. Thanks

      Delete
  5. Suppose the person bets for Team A.
    Let the states be defined as (# wins for A,# wins for B,Money at hand)
    We need to work backwards from Match 7 considering only the valid states.
    For Match 7, valid states are (4,3,+100) and (3,4,-100)
    For Match 6, the only valid state which can lead to Match 7 is (3,3,_). It is simple to see that this state comes to (3,3,0) and the amount to bet is 100.

    For Match 6, valid states are (4,2,+100), (3,3,0) and (2,4,-100). From Match 5, the valid states which can lead to Match 6 are (3,2,_) and (2,3,_). They work out to be (3,2,+50) and (2,3,-50) and amount bet is 25 for each state.

    Carrying this on, we get the betting amount for Match 1 to be 31.25

    ReplyDelete
    Replies
    1. How did you come up with numbers +50/-50 for (3,2)/(2,3) state? Is it some random numbers which eventually gets us to the correct answer ?

      Delete
    2. Kiran, Let the bet be y after (3,2,x) . So, win leads to (4,2,x+y) and loss leads to (3,3,x-y)
      But the valid states are (4,2,100) and (3,3,0) , x is 50 and y is 50.

      Delete
  6. It should be $37.5, since one should have $100 at the end if Team A wins in 4, 5, 6 or 7 matches. Similarly, one should have -$100 if Team B wins. Going backward through the square, we get the betting amount to be $37.5

    ReplyDelete
    Replies
    1. Wrong solution. What square are you referring to?

      Delete
  7. The point of this puzzle is to understand the concept of backward induction in simple, intuitive terms.

    ReplyDelete
  8. @ Pratik : Does this solution assumes equal probability of winning for both teams or is the answer same with any prior probs ??

    ReplyDelete

Post a Comment