# Baseball betting and probablity

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.

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Moving backwards is a good idea. Applying probability is probably not a good idea, since the problem doesn't make any assumptions about probabilities, so the solution should work regardless of whether the outcomes are probabilistic or one team is certatin to beat the other one. – joriki Oct 1 '11 at 20:33
Ok, So, something like enumerating all the cases would be better? – user16983 Oct 1 '11 at 20:38

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.

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You beat me to it by $9$ seconds :-) – joriki Oct 1 '11 at 20:50
@joriki, this means you are punished for the first sentence of your post... – Did Oct 1 '11 at 20:55
Why would I be punished for that? I think it means I got punished for putting all those dollar signs around all those numbers and then escaping the actual dollar signs inside them :-) – joriki Oct 1 '11 at 21:06
Thanks makes sense now, but how do I mark two answers as right? – user16983 Oct 3 '11 at 2:21

This is more interesting than I thought at first.

As you said, it makes sense to move backward. If you draw the possible non-final scores, with a win for your team moving down and right and a win for your uncles' team moving down and left, you get a nice appropriately diamond-shaped graph. Along the lower edges are the non-final scores containing a $3$, for which the next game could decide the series. There's only one score where the next game is certain to decide the series, $3:3$. That's where you can get the deduction going. The result has to be $-100$ or $+100$ depending on the outcome of this game, so the balance before the game must be $0$ and the bet on the game must be $100$. That allows you to go back to $3:2$ (and $2:3$, which must have the same bet and the opposite balance by symmetry). Now you know that if your team wins the game, the balance should be $+100$, and if it loses, changing the score to $3:3$, the balance should be $0$. it follows that the balance before the game must be $50$ and the bet on the game must be $50$. You can fill in the whole diamond like this until you get to the top and determine the bet required for the initial score $0:0$. I get $\$31.25$(that's$5/16$of$\$100$).

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