### 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!

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.

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.

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

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.

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!

2. Awesome. Thanks

4. What a delightful puzzle!

Spoilers: first shot, neatened up.

1. It is a delightful puzzle indeed. Thanks

2. Thanks for the images, btw :)

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

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 ?

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.

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

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

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

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