Revisiting a prior puzzle -- betting on best of 7

Re: Revisiting a prior puzzle -- betting on best o

david weaver

Thinking out loud - the guy who is going to win the series never wants to have 1000 or more in a series where he's already got three wins.

When a series is 3-3, you want the pot for both guys to be the same, otherwise one team wins, and it's OK, but if the other team does, then it won't hit $1000.

Will think some more.

On the right track here

Alex Y

You say

When a series is 3-3, you want the pot for both guys to be the same, otherwise one team wins, and it's OK, but if the other team does, then it won't hit $1000.

Remember they are betting against each other, so I'm not sure exactly what a "pot" would refer to here. This is definitely the beginning of the correct solution, though.

Re: On the right track here

Gary Smyth

Am I doing this right? Each game is an unknown dollar value bet. The dollar amount makes no difference as long as whoever selects and bets first, the other accepts the exact opposite bet. Since we know that the series is 3-3 the net result up until game seven is zero for both betters. The seventh game is for $1000. One team wins and the victor takes home $1000.

Re: On the right track here

David Weaver

The trouble in not betting any money prior is that what happens in an 0-3 series when the team with the 0 wins their first game. Then they're up $1000, and the team with 3 is at minus $1000.

Then if the winning team is to be at 1000 after that, they need to bet 2000, and if the team who has only won one game is at 3000, they're in a pickle because if they don't lose again, they have more than $1000.

I'm leaning toward a strategy that puts teams at even stakes any time the series draws even (1-1, 2-2 or 3-3). I haven't thought a lot about this one today, but I don't want to forget it, nor do I want to google anything.

Re: On the right track here

Alex Y

Am I doing this right?

So far, so good.

Each game is an unknown dollar value bet. The dollar amount makes no difference as long as whoever selects and bets first, the other accepts the exact opposite bet.

Yes, they have a common objective of having the winner win (from the loser) $1,000, and will cooperate in the strateg that will bring that result. I don't agree with the "dollar amount makes no difference" part of your statement, though--the key to solving this is to get the right dollar amounts bet on the individual games.

Since we know that the series is 3-3 the net result up until game seven is zero for both betters. The seventh game is for $1000. One team wins and the victor takes home $1000.

Yes, that is the bet on game seven, for a series that goes seven games

A little more insight ...

Larry Barrett

For a one game series the solution is simple: both players agree to bet $1000 on the outcome of the one game, one wins exactly $1000.

For a three game series (best 2 of 3) the solution is also straightforward (I think): both players agree to bet $500 on the first game. The winner is up $500, loser down $500. They agree to bet $500 on second game. If same team wins, then the series is over and that player wins exactly $1000 ($500 + $500). If the winner of first game loses the second, the series stands at 1-1 and the players are at net $0. They agree to bet $1000 on the third game, and the winner will be up exactly $1000.

One observation from this is that when one team is one game from winning the series or the series being all even (ie 1-0 in a 3 game series, 2-1 in a 5 game series, 3-2 in a 7 game series), one player should be up $500, the other down $500, and the bet should be $500 on next game. The series will either end with one player at +$1000 or the series will be all even (3-3 in best of 7), the players will be at net $0, and they agree to bet $1000 on final game.

In a best of 5 series, if the series stands at 2-0, I think we want one player to be at +$750 and the bet to be $250. In the series goes to 3-0 then one player is up +$1000 and the series is over; but if the series goes to 2-1 then the players are at +$500, -$500 and they agree to bet $500 on the next game. If the series stands at 1-1, I think we want the players to be at net $0 and the bet to be $500. Continuing backward, if series is at 1-0, I think we want one player to be at +$750/2 and the bet to be $750/2. So on the first game the bet should be +$750/2.

This seems a little complicated, so need to think it through; if correct, it can be extended to best 4 of 7, I think.

Re: A little more insight ...

David Weaver

The complication on the 3-0 is that one player is up 1000, but there is no guarantee that player will win. If they make zero bets and the opponent wins four in a row, they are at 1000, but they lost.

if they bet to win and do win, then they win something other than 1000.

That's it, Larry

Alex Y

Your reasoning is exactly on target, and if you carry it just one more step, you will be there!

Betting strategy

Larry Barrett

Here is my betting strategy for the best of 7 series. I constructed a chart of all possible paths to win for player A; I will just list the series states that can lead to a win for player A, the amount player A has won (or lost) at that state and the amount of the bet for the next game. Still complicated so not sure about this.

Series...Player A $...Next Bet $

0-0........0..............312.5

1-0........312.5.........312.5

0-1.......-312.5.........312.5

2-0........625...........250

1-1........0...............375

0-2.......-625...........250

3-0........875...........125

2-1........375...........375

1-2.......-375...........375

0-3.......-875...........125

4-0........1000..........game over

3-1........750...........250

2-2........0..............500

1-3.......-750...........250

4-1........1000..........game over

3-2........500...........500

2-3.......-500...........500

4-2........1000..........game over

3-3........0...............1000

4-3........1000..........game over

 

Alex Y

That's it!!!

I think it is easier to see in a semi-graphical form, with the standing (amount won) at each possible game status. Here it is in spreadsheet form, treated as a World Series, with AL on the y axis and NL on the x axis, and the dollar amounts being the standing of the player who bets on the AL. Filling out this chart, we realize that every position with y=4 has to be $1,000, every position with x=4 has to be -1,000. Then realizing that each game results in moving one place to the right or one place up in this chart, you can see that each position has to be the average of the positions above it and to its right (and the corresponding bet is 1/2 of the difference between the up one and over one positions).

Whoops! Just noticed that I rounded on the $312.50 positions, but not worth redoing.

Re: Betting strategy

David Weaver

Nice!

Re: 

Larry Barrett

Your chart does make it easier to understand. It is an interesting puzzle. Thanks.