Your favorite baseball team is playing against your uncle's favorite team in the World Series. At the beginning of each game, you and your uncle bet on the game's outcome. Your uncle, being wealthy and carefree, always lets you choose the amount of the bet. If you bet b dollars and your team wins the game, your uncle gives you an IOU for b dollars. But if they lose the game, you give him an IOU for b dollars. When the series is over, all outstanding IOUs are settled in cash. You would like to walk away with $100 in cash if your team wins the series, and lose $100 if your team loses the series. How much should you bet on the opening game? (For non baseball fans, the first team to win a total of four games wins the series).

This is the queston from general forum, Athletico said he have seen this problem in Brainteasers some where, but I cann't find it, please help.

Solve it as a binomial tree. Answer is 31.25.

QuoteOriginally posted by: liuliudadaThis is the queston from general forum, Athletico said he have seen this problem in Brainteasers some where, but I cann't find it, please help.The thread might be gone, and even if it's not I doubt a full solution was posted.As I said in the other thread, a truncated seven-step binomial tree gets you to the result - see attached (but you should try to solve it yourself before looking). It's a best-of-seven so the process is stopped when one of the teams wins its 4th game -- this gives the tree a diamond shape.In the tree you start at the left node, value 0. If your team wins you take a step up and your position gains value. Obviously the tree is symmetric (assume p = 1/2). You get the values at the intermediate nodes by taking expectations of the two subsequent nodes -- start with the (3,3) state, which must be a $100 bet because it decides the series.

QuoteOriginally posted by: AthleticoQuoteOriginally posted by: liuliudadaThis is the queston from general forum, Athletico said he have seen this problem in Brainteasers some where, but I cann't find it, please help.The thread might be gone, and even if it's not I doubt a full solution was posted.As I said in the other thread, a truncated seven-step binomial tree gets you to the result - see attached (but you should try to solve it yourself before looking). It's a best-of-seven so the process is stopped when one of the teams wins its 4th game -- this gives the tree a diamond shape.In the tree you start at the left node, value 0. If your team wins you take a step up and your position gains value. Obviously the tree is symmetric (assume p = 1/2). You get the values at the intermediate nodes by taking expectations of the two subsequent nodes -- start with the (3,3) state, which must be a $100 bet because it decides the series.I got the tree and the right answer 31.25 yesterday. Thank you very much

Fine - Your first bet should be $31.25 based on a "truncated" binomial tree, but what about the second bet, third bet, etc.? Obviously, you can't bet $31.25 on each of the first four games because if the series went only four games, you would not have a +100/-100 as a final payoff. Yet, you must make at least four bets.The probabilty that the series lasts exactly four games = 0.125, i.e. of 16 possibilities, WWWW or LLLL ends the series, so P(more than 4 games) = 0.875. What is P(exactly 5 games)? Six? Seven?I contend that at t=0 there is no single bet amount that solves the problem. Now, if I can change the bet amount based on the results of the prior game(s), there may be a correct amount to bet at t = 1,2,3,.. but no one in this thread has provided a solution to this.Mike

QuoteOriginally posted by: mensa0 I contend that at t=0 there is no single bet amount that solves the problem. Now, if I can change the bet amount based on the results of the prior game(s), there may be a correct amount to bet at t = 1,2,3,.. but no one in this thread has provided a solution to this.MikeThis is simply not true, and there is no need to change bet amounts of past games.The amount to bet at each stage is pre-defined in the tree, given by the difference of the two subsequent nodes. We know we'll be betting $31.25 on Game 2 regardless of the result of Game 1. But we'll have a $37.50 bet on for Game 3 if the series is tied 1-1, otherwise a $25 bet. The idea behind using the tree is to replicate a desired final payoff (+/- 100) with a prescription of what to bet in every possible state of the series.

QuoteOriginally posted by: AthleticoQuoteOriginally posted by: mensa0 I contend that at t=0 there is no single bet amount that solves the problem. Now, if I can change the bet amount based on the results of the prior game(s), there may be a correct amount to bet at t = 1,2,3,.. but no one in this thread has provided a solution to this.MikeThis is simply not true, and there is no need to change bet amounts of past games.The amount to bet at each stage is pre-defined in the tree, given by the difference of the two subsequent nodes. We know we'll be betting $31.25 on Game 2 regardless of the result of Game 1. But we'll have a $37.50 bet on for Game 3 if the series is tied 1-1, otherwise a $25 bet. The idea behind using the tree is to replicate a desired final payoff (+/- 100) with a prescription of what to bet in every possible state of the series.I think you misunderstood me. I'm not saying that we go back and change the amount of a prior bet. I'm just saying that we have to place a different bet after each game depending upon the outcome of the most recent (prior) game. So the question remains, what are the bet amounts at each time "t"? If we bet $31.25 on the first game, then after the game we have either +31.25 or -$31.25. What is our bet on game #2? And, after game 2, depending upon whether our record is 2-0, 1-1, or 0-2 what is our next bet? My point is that there is no single (equal) bet amount at each time period, i.e. the bet we place at time "t" depends upon the prior outcomes (our won-lost record.)Mike

>> My point is that there is no single (equal) bet amount at each time period, i.e. the bet we place at time "t" depends upon the prior >> outcomes (our won-lost record.)Agreed but was that (equal bet amounts at each stage) a requirement of the original problem? The solution I posted does solve the problem and I claim it's unique, the only possible solution.>> What is our bet on game #2? And, after game 2, depending upon whether our record is 2-0, 1-1, or 0-2 what is our next bet?Reread earlier posts; this has already been explained.

QuoteOriginally posted by: Athletico>> My point is that there is no single (equal) bet amount at each time period, i.e. the bet we place at time "t" depends upon the prior >> outcomes (our won-lost record.)Agreed but was that (equal bet amounts at each stage) a requirement of the original problem? The solution I posted does solve the problem and I claim it's unique, the only possible solution.>> What is our bet on game #2? And, after game 2, depending upon whether our record is 2-0, 1-1, or 0-2 what is our next bet?Reread earlier posts; this has already been explained.Athletico = Agreed! Your tree is the same as mine. Bets are: 31.25, 31.25, 25, 12.5, 25, 50, 100 for t=0 (before the first game) to t=6 (before the 7th game, if the series goes that far.)I didn't see that bet sequence enumerated in this post, hence my earlier questions.Mike (mensa0)

Let's generalize this. Say that you are playing a "best of 2n+1" series, in which the first team to win n+1 games wins the series. Now how much should you bet on the opening game, to guarantee that you win 100 if your team wins the series, and you lose 100 if your team loses the series? Also, how much should you bet on the next game if you're at the point where your team has won p games and the other team has won q games, with 0 <= p,q < n+1, to ensure the same aforementioned payoffs at the end?

QuoteOriginally posted by: gentinexLet's generalize this. Say that you are playing a "best of 2n+1" series, in which the first team to win n+1 games wins the series. Now how much should you bet on the opening game, to guarantee that you win 100 if your team wins the series, and you lose 100 if your team loses the series? Also, how much should you bet on the next game if you're at the point where your team has won p games and the other team has won q games, with 0 <= p,q < n+1, to ensure the same aforementioned payoffs at the end?Solution using binomial trees is the least elegant way to solve the problem, although it best exposes the underlying technique for solving it. A more elegant method is using state variables to capture nodes on the tree. Even more elegant is using basic probability. I'll show how probability lends itself to generalization in one case. For instance, consider a seven game series. The question really is how much money will you have after one game? Since risk neutral probabilities are 1/2, we can solve as such:Payoff(Win) = 100. How can you win? You have to win at least 3 of the remaining six (ie. 3,4,5,6 of six) So 3 choose six, 4 choose six, 5 choose six, 6 choose 6Payoff(Loss) = -100. You have to lose 4 of the remaining 6. So ways to loose 0 choose six = 6 choose six, 1 choose six= 5 choose six, 2 choose six = 5 choose sixHow many total nodes on the left side of tree? 2^7/2=124/2=64So E = 3 choose six/64 * 100=31.25It's pretty easy to abstract this:E = (n Choose 2n)/2^(2n) * 100I can't explain the theory behind this as well as the practicality of the solution, unfortunately. Essentially, if you're read Shreve on discrete finance, you're talking about the value of a derivative at time N =1, where the first toss has been favorable. You're using the fact that the risk neutral probabilities are 1/2 (that the value of the derivative is the average of the two values before it), since your wager is two-way.