[Math] Baseball betting and probablity

financeprobability

Here is a question that came up during class discussions on Friday:

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).

I am thinking to apply probability. For example, knowing 100 is the result, moving backward but not exactly sure how to proceed.

Best Answer

31.25.

As you said, draw the graph of the possible scores, linking a given score to the two possible scores after one more game is played. Place a +100 tag on every vertex where you won (these are 4-0, 4-1, 4-2 and 4-3) and a -100 tag on every vertex where you lost (these are 0-4, 1-4, 2-4 and 3-4) and compute the tags of the other vertices backwards.

For example, 3-3 gives 4-3 and 3-4 hence the tag of the vertex 3-3 should be half the sum of the tags of the vertices 4-3 (+100) and 3-4 (-100), that is, 0. Likewise, the tag of the vertex 3-2 should be half the sum of the tags of the vertices 4-2 (which you knew from the beginning to be +100) and 3-3 (which you just computed to be 0), that is, +50. And so on, crawling back until you reach the vertex 0-0 at the beginning of the graph.

Your bet on the opening game should be the tag of the vertex 1-0 and the opposite of the tag of the vertex 0-1. This is 31.25.

Related Solutions

Why will repeated betting on a fair coin toss lose you money in the most cases? And how can you calculate it

In fact your starting capital is multiplied by $(1/2)^i(3/2)^{n-i}$ in $n$ rounds by probability of $\binom{n}{i}/2^n$,
disregarding the order of loses and wins.

Since it can never become $1$, you can never return to any of your previous positions, including
"quit with no loss or profit".

Suppose $j = \underset{(1/2)^i(3/2)^{n-i}>1}{\max}(i)$, so you profit by chance of $p_j = \sum_{i=0}^j \binom{n}{i}/2^n$, and lose by chance of $1-p_j = \sum_{i=j+1}^n \binom{n}{i}/2^n$.

Also you can use $\min$ function similarly, if you like:

Suppose $k = \underset{(1/2)^i(3/2)^{n-i}<1}{\min}(i)$, so you lose by chance of $p_k = \sum_{i=k}^n \binom{n}{i}/2^n$, and profit by chance of $1-p_k = \sum_{i=0}^{k-1} \binom{n}{i}/2^n$.

Betting on game of the World Series

Step 0: Work backwards from the end

Step 1: If the score is $3-3$ you want the net position so far to be $0$, and you bet $1000$

Step 2: If the score is $3-2$ you want the net position so far to be $+500$, and you bet $500$; if the score is $2-3$ you want the net position so far to be $-500$, and you bet $500$

Step 3: If the score is $3-1$ you want the net position so far to be $+750$, and you bet $250$; if the score is $1-3$ you want the net position so far to be $-750$, and you bet $250$; if the score is $2-2$ you want the net position so far to be $0$, and you bet $500$ etc.

You want the position at a particular score to be the average of the two possible positions after the next game. You want to bet half the difference of the two possible positions after the next game. You should end up with this:

Bets:
=====
Score     0       1       2      3  
                        
0       312.5   312.5   250    125  
1       312.5   375     375    250  
2       250     375     500    500  
3       125     250     500   1000  

Positions:
==========                  
Score      0       1       2       3       4
                        
0          0     312.5   625     875    1000
1       -312.5     0     375     750    1000
2       -625    -375       0     500    1000
3       -875    -750    -500       0    1000
4      -1000   -1000   -1000   -1000

Best Answer

Related Solutions

Why will repeated betting on a fair coin toss lose you money in the most cases? And how can you calculate it

Betting on game of the World Series

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