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Sports Gambling Puzzle

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Sports Gambling Puzzle

#1

Sports Gambling Puzzle

Note for non-Americans and non baseball fans: The American "World Series" is a best of seven playoff. Substitute your own sport, or even coin tossing, as long as it is best of seven, with no ties.

Your team makes it to the world series, playing against the team of one of your friends. You agree to bet $1,000 on the series, winner-take-all. However, you ask a third party to hold the bet money. He agrees, but only one game at a time. So you and your friend need to come up with a strategy of betting (even money) on each game, so that over all games combined, the winner of the series will win $1,000. Does the bet on game one matter, and how much is bet if it does?

Part 2 for probability geeks: After game one, it appears there are 15 possible combinations of future games in which the loser of game one wins the series (i.e., win games 2,3,4,5; 2,3,4,6; 2,3,4,7; 2,3,5,6; etc), while there are 20 possible ways for the winner of game one to win the series. Consider your answer to part 1, and it would appear that the winner of the first game has a negative expectation on subsequent games, even though all bets are even money. What is wrong with that analysis?

Re: Sports Gambling Puzzle

#2

Hint

Work backwards.

Re: Sports Gambling Puzzle

#3

Hint 2 and restatement

Was I unclear in the statement of the problem? It is to design a scheme for betting on individual games that will assure that the first one to 4 victories will have net winnings of exactly $1,000. I asked about the first game, only because you have to develop the whole scheme to determine that amount, and giving the answer to that shows you have gotten the whole betting scheme without giving way the scheme to those still working on it.

Now for hint#2: You may feel instinctively that you and your friend should be at net zero winnings whenever the series is tied. Trust your gut on that one.

Re: Sports Gambling Puzzle

#4

still noodling ;-)

Re: Sports Gambling Puzzle

#5

I was beginning to feel ignored! :-)

Re: Sports Gambling Puzzle

#6

Hint 3

Consider the net cumulative winnings at each possible series score in games.

Re: Sports Gambling Puzzle

#7

Re: Hint 3

I am still working on it, but haven't had much time. What you are suggesting is what I was doing. Will work on it some more tonight. It really should not be that hard, just getting my brain to function is a problem. ;-)

Re: Sports Gambling Puzzle

#8

Last Bet ...

would have to be $500.

Re: Sports Gambling Puzzle

#9

Could be

specifically, if someone wins the series 4-2. But otherwise, no. If the series goes to a 7th game, how can a $500 bet assure both of them that they will win $1,000 if their team wins?

Re: Sports Gambling Puzzle

#10

Answer and solution

The bets on the first two games have to be $312.50 each game.

How in the world did I come up with a weird number like that?!?

In explaining this, I will use a picture, since I'm a visual thinker. But hopefully I'll use enough words that you verbal thinkers will follow as well.

The attached picture is a graphical representation of the World Series, with the scores in games of each city along the axes.

Each dot in this graph represents a series standing, e.g., the circled dot represents the situation where city X holds a 2-1 lead. Each of these dots also has associated with it a dollar amount of where the bets so far stand. It is not apparent at this point that that dollar amount is unique (the bets at 0-0, 1-0, and 2-0 might result in a different standing at 2-1 than those at 0-0, 0-1, and 1-1 create) but we will see later that they are unique.

Each line in this graph is really an arrow going toward the right or up, representing the results of a game. Each arrow is associated with a bet, and the dollar amount of the arrow going up or right from a given point is the same (even-money bets).

At this point, we know the dollar standing associated with 9 of these points, $0 at 0-0, $1,000 at the four points in the right most column (x winning the series 4-3, 4-2, 4-1, or 4-0, and ($1,000) at the four points on the top row (We'll look at all dollar amounts from X's perspective.)

Now consider when the series is tied 3-3. If x loses, we go up to ($1,000), while if x wins, we go to the right, to $1,000. The only way this can happen is if the bettors' positions were even at 3-3, and the bet was $1,000. This is the principle used to fill in the whole grid. For any unknown point, if we know the standing that needs to exist at the points above and to the right of that point, then the point in question is midway between those two points, and the bet is half the difference between those two points.

Let's apply that to the point where the score is 3-2. We know that if X wins, he needs to be up by 1,000, and from the previous step, we know that if he loses, he and y need to be at even money. Halfway between $0 and $1,000 is $500,so x must be ahead by $500 at 3-2. Also, ($1,000-$0)/2 = $500 so the wager at 3-2 has to be $500.

One more time. If the score in games is 3-1, the two possible outcomes are X up by $1,000 at 4-1, or X up by $500 at 3-2. So at 3-1, x must be up by $750, and they bet $250 on the subsequent game.

Filling in the rest of the grid is just following this example, which I just did on the lower right, since there is obvious symmetry with Y ahead.

The odd-looking results are that you bet $312.50 on the first two games (0-0 or 1-0), $375 if the series is at 1-1 or 2-1, $500 if the series is at 2-2 or 3-2, $250 if the series stands at 2-0 or 3-1, $125 if the series stands at 3-0, and $1,000 on a game seven.

Or find someone else to escrow your bet, who will do the whole series at once!!


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Re: Sports Gambling Puzzle

#11

And part two restated

After game one, if x won that game, he has $312.50 in the bank. So from that point forward, the bet is that if his team wins the series, he wins another $687.50, while if Y wins, he will lose $1,312.50. But the odds are greater for him to win, so you would expect some difference such as this.

In fact, there are 15 ways for Y to win the series after losing game one, by winning games 2,3,4,5; 2,,3,4,6; 2,3,4,7; 2,3,5,6; 2,3,5,7; etc. Similarly, there are 20 ways for X to win. But even though all subsequent bets are at even money, 20 chances to win $687.50 versus 15 ways to lose $1,312.50 is a pretty poor bet.

What's wrong with this analysis?

P.S., I'm going to let this one stand, unless anyone want to pursue it, maybe by eamil. I'm an actuary, and I have to recognize that we are an odd lot finding questions such as the above of interest!

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