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Pascal and Fermat bet on the World Series

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NOTE: This post uses advanced math formatting, which may not be rendered correctly in syndicated versions of the article. For correct formatting, see the version on my web site.

A fun problem of applied probability says:

Your team, the Phillies, is playing the Qankees in the World Series. You want to bet $100 on the Phillies to win the Series, but your bookie will only take bets on individual games. (Because he has a heart of gold, he pays even money on all bets.) Describe a betting strategy for each game whose cumulative effect is the same as a single bet of $100 on the Phillies to win the entire series.

(For non-USA readers, I should explain that the World Series lasts until one of the two teams has won four of the seven games.)

Suppose the Phillies and the Qankees had each already won three games. Then, assuming that you still have your $100, it would be easy to bet on the Phillies to win the series: just bet $100 on the outcome of the final game. If you have any amount other than $100, it is impossible. For suppose you have !!X!! dollars and bet !!B!! dollars. Then if the Phillies win you end with !!X+B!! dollars, and if the Qankees win you end with !!X-B!! dollars. The problem requires that !!X+B=200!! and !!X-B=0!! and the only solution is to start with !!X=100!! and to bet !!B=100!!.

Now suppose the Phillies have won three games and the Qankees only two, which I will write as !!\def\pq#1#2{\langle P=#1,Q=#2\rangle}\pq32!!. Suppose you currently have !!X!! dollars. You want a series of bets that will leave you with $200 if the Phillies win either of the next two games, and with $0 if the Qankees win both. Suppose you bet !!B!! dollars on the next game. If the Phillies win, you will then have !!X+B!! dollars, and you want !!X+B=200!!. If the Qankees win, you will have !!X-B!! dollars, and you want !!X-B=100!! so that you can make a $100 bet on the final game, as in the previous paragraph. The solution here is that you must have an initial bankroll of !!X=150!! and you must bet !!B=50!!. If the Phillies win, you end with $200, as required, and if the Philies lose you end with !!X=100!!, which sets you up properly for the $100 bet on the final game.

What if the Qankees have won three games and the Phillies only two? (This is !!\pq23!!.) Then you want !!X+B = 100!! and !!X-B = 0!!, so you must start with !!X=50!! and bet the entire fifty dollars. If the Qankees win you are left with nothing (because the Qankees have won the series four games to two) and if the Phillies win you are left with $100, which sets you up for your $100 bet on the final game, as required.

We can tabulate the required bankrolls as follows:

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & & & & & \$200 \\ 1 & & & & & \$200 \\ 2 & & & & & \$200 \\ 3 & & & & & \$200 \\ \hline 4 & \$0 & \$0 & \$0 & \$0 & \\ \end{array} $$

The cell in row !!Q!! and column !!P!! will give your required bankroll when the Phillies have won !!P!! games and the Qankees have won !!Q!!. When !!P=4!! the Phillies have won the Series and you want to have accumulated a total of $200; when !!Q=4!! the Qankees have won and you should have lost your original stake, leaving you with $0. We can fill in the three values from the three previous paragraphs:

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & & & & & \$200 \\ 1 & & & & & \$200 \\ 2 & & & & 150 & \$200 \\ 3 & & & 50 & 100 & \$200 \\ \hline 4 & \$0 & \$0 & \$0 & \$0 & \\ \end{array} $$

The cell in row 2, column 3 corresponds to the !!\pq32!! case; it says that you must make a bet that could result in your final bankroll being $200 (if the Phillies win; this is the cell to the right) or $100 (if the Qankees win; this is the cell below.) The only way this is possible is if your starting amount is exactly halfway between $100 and $200, namely $150; then you evidently must bet $50 of your $150 to reach the desired conclusions.

The same rule applies to every other cell in the table: your starting amount in each cell !!\pq pq!! must be halfway between the amount in the cell below (!!\pq p{q+1}!!) and the cell to the right (!!\pq{p+1}q!!). It's easy to fill in the rest of the cells according to that rule:

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 100\hphantom{.00} & 131.25 & 162.50 & 187.50 & \$200 \\ 1 & 68.75 & 100\hphantom{.00} & 137.50 & 175\hphantom{.00} & \$200 \\ 2 & 37.50 & 62.50 & 100\hphantom{.00} & 150\hphantom{.00} & \$200 \\ 3 & 12.50 & 25\hphantom{.00} & 50\hphantom{.00} & 100\hphantom{.00} & \$200 \\ \hline 4 & \$0\hphantom{.00} & \$0\hphantom{.00} & \$0\hphantom{.00} & \$0\hphantom{.00} & \\ \end{array} $$

From this table you can read off the bets. Consider the upper-left position !!\pq00!!. You start with $100, as required. If the Phillies win the first game, you should end with $131.25. Therefore you should bet $31.25 on the outcome of the first game of the series. The next square to the right says $162.50$, so you should be $31.25 on the next game as well.

Suppose you would like to end the series with $200 if the Phillies win, and $100 if the Qankees win. Using the same method, we can generate a table that says how much money you should have at each stage of the World Series in order to guarantee that:

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 150\hphantom{.00} & 165.62 & 181.25 & 193.75 & 200\hphantom{.00} \\ 1 & 134.38 & 150\hphantom{.00} & 168.75 & 187.50 & 200\hphantom{.00} \\ 2 & 118.75 & 131.25 & 150\hphantom{.00} & 175\hphantom{.00} & 200\hphantom{.00} \\ 3 & 106.25 & 112.50 & 125\hphantom{.00} & 150\hphantom{.00} & 200\hphantom{.00} \\ \hline 4 & 100\hphantom{.00} & 100\hphantom{.00} & 100\hphantom{.00} & 100\hphantom{.00} & 0\hphantom{.00} \\ \end{array} $$

You can have that, but you must start with $150; your first bet should be $15.62½.

What if you want to place a bet that pays off only if the Phillies sweep the series four to nothing?

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 6.25 & 12.50 & 25\hphantom{.00} & 50\hphantom{.00} & 100\hphantom{.00} \\ 1 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ 2 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ 3 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \hline 4 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \end{array} $$

You need to start with $6.25; the bet effectively pays off at fifteen to one odds. If the bookie is paying even money on all bets, then he is paying fifteen to one on the Phillies to sweep the series.

What about stranger bets? Suppose you want to make a bet that the Phillies will win the series four games to two?

$$ \begin{array}{l|rrrr|c} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 15.62 & 18.75 & 18.75 & 12.50 & 0\hphantom{.00} \\ 1 & 12.50 & 18.75 & 25\hphantom{.00} & 25\hphantom{.00} & 0\hphantom{.00} \\ 2 & 6.25 & 12.50 & 25\hphantom{.00} & 50\hphantom{.00} & 100\hphantom{.00} \\ 3 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \hline 4 & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \end{array} $$

Now you need to start with $15.62½ and your first bet is $3.12½. If the Phillies win the first game, you must not bet on the outcome of the second game!

Let's say that a 5×5 matrix is a bankroll matrix if it has this property that !!a_{pq} = \frac12(a_{p+1,q} + a_{p,q+1})!! for each !!0\le p,q \lt 4!!. Notice that the set of bankroll matrices forms a vector space under matrix addition and scalar multiplication. In fact it is an 8-dimensional space, whose basis elements are matrices of the type:

$$ \left[\begin{array}{rrrr|c} \frac5{32} & \frac5{32} & \frac18 & \frac1{16} & 0 \\ \frac5{32} & \frac3{16} & \frac3{16} & \frac18 & 0 \\ \frac18 & \frac3{16} & \frac14 & \frac14 & 0 \\ \frac1{16} & \frac18 & \frac14 & \frac12 & 1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \end{array}\right] $$

where exactly one element in the last row or the last column is nonzero.

By composing an appropriate combination of bankroll matrices, we can obtain the bankroll matrix that describes a betting strategy for any possible bet on any outcome of the series that you want, and the !!0,0!! entry of the resulting matrix will tell us how much money you need to start with to obtain the desired payoffs; or we can take this result and scale it so that the !!0,0!! entry matches your actual starting funds, and the result will tell us the payoffs you can expect to receive for your favored outcomes. For example, what if you start with $100 and you want a bet that pays off if the series ends in a four-to-two result for either team? We should add:

$$ \left[\begin{array}{cccc|c} \frac5{32} & \frac3{16} & \frac3{16} & \frac18 & 0 \\ \frac18 & \frac3{16} & \frac14 & \frac14 & 0 \\ \frac1{16} & \frac18 & \frac14 & \frac12 & \color{darkred}{1} \\ 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \end{array}\right] + \left[\begin{array}{cccc|c} \frac5{32} & \frac18 & \frac1{16} & 0 & 0 \\ \frac3{16} & \frac3{16} & \frac18 & 0 & 0 \\ \frac3{16} & \frac14 & \frac14 & 0 & 0 \\ \frac18 & \frac14 & \frac12 & 0 & 0 \\ \hline 0 & 0 & \color{darkblue}{1} & 0 & 0 \\ \end{array}\right] = \left[\begin{array}{cccc|c} \frac5{16} & \frac5{16} & \frac14 & \frac18 & 0 \\ \frac5{16} & \frac38 & \frac38 & \frac14 & 0 \\ \frac14 & \frac38 & \frac12 & \frac12 & \color{darkred}{1} \\ \frac18 & \frac14 & \frac12 & 0 & 0 \\ \hline 0 & 0 & \color{darkblue}{1} & 0 & 0 \\ \end{array}\right] $$

and then we multiply every element of the sum by !!\$100 \div \frac5{16} = \$320!! to scale the upper-left entry to be exactly $100:

$$ \left[ \begin{array}{rrrr|r} \$100\hphantom{.00} & 100\hphantom{.00} & 80\hphantom{.00} & 40\hphantom{.00} & 0\hphantom{.00} \\ 100\hphantom{.00} & 120\hphantom{.00} & 120\hphantom{.00} & 80\hphantom{.00} & 0\hphantom{.00} \\ 80\hphantom{.00} & 120\hphantom{.00} & 160\hphantom{.00} & 160\hphantom{.00} & \color{darkred}{\$320}\hphantom{.00} \\ 40\hphantom{.00} & 80\hphantom{.00} & 160\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \hline 0\hphantom{.00} & 0\hphantom{.00} & \color{darkblue}{\$320}\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \end{array}\right] $$

This bet pays off at odds of 11 to 5. Notice that in some cases the bankroll in a space is less than the bankroll in the space to the right; in these cases you have to bet against the Phillies. (Heresy!)

Bookies that pay off bets at even money are hard to find. If the payoff for a winning bet is only ninety cents on the dollar, we can compute a different set of eight basis matrices and add and scale them the same way. Now when we start with $100 and bet on the Phillies to win the series we get:

$$\left[ \begin{array}{rrrr|r} 100\hphantom{.00} & 126.34 & 151.36 & 170.37 & 179.40 \\ 70.74 & 98.54 & 130.23 & 160.34 & 179.40 \\ 39.85 & 63.33 & 96.78 & 139.15 & 179.40 \\ 13.77 & 26.16 & 49.70 & 94.42 & 179.40 \\ \hline 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} & 0\hphantom{.00} \\ \end{array} \right] $$

The bet on the first game is $29.26. If you lose, you are left with $70.74, and if you win, it pays off only $26.33, leaving you with $126.34. (There is roundoff error in the last place.) The entire bet on the series pays off a little under 4 to 5.

I had been meaning to write this up for a long time—since at least January 2006—but I never got around to it, and then Michael Lugo got to it last October, which was kind of a bummer. Michael's math blog is a lot of fun, but while the overlap between his subject matter isn't too great, this is squarely in the middle. So I thought I might give up on it, but then yesterday I found something new to say.

Probability theory is generally held to have begun in a series of letters between Pierre de Fermat (yes, that Fermat) and Blaise Pascal (yes, that Pascal) in 1654. The problem they considered, the so-called “problem of points”, is as follows: two gamblers, say the Chevalier des Philadelphiens and the Marquis des Querelleurs, will throw a die. If it comes up !!1,2,!! or !!3!!, then !!P!! wins and scores a point, and if it comes up !!4,5,!! or !!6!!, then !!Q!! wins and scores a point. The gamblers have each staked 32 pistoles on the outcome of the game, and whichever of them scores some pre-agreed number of points first is the winner, and takes the whole pot of 64 pistoles. But suppose the tavern is closing and the gamblers are evicted before the game is complete. How should they split the pot?

Pascal, discussing a game to three points, takes for granted that if if the game ends with the score tied, the gamblers should split the pot evenly. He then considers the case of a !!\pq 21!! score, and reasons as follows:

They now play one throw of which the chances are such that if the first wins, he will win the entire wager that is at stake, that is to say 64 pistoles. If the other wins, they will be two to two and in consequence, if they wish to separate, it follows that each will take back his wager that is to say 32 pistoles.

Consider then, Monsieur, that if the first wins, 64 will belong to him. If he loses, 32 will belong to him. Then if they do not wish to play this point, and separate without doing it, the first should say “I am sure of 32 pistoles, for even a loss gives them to me. As for the 32 others, perhaps I will have them and perhaps you will have them, the risk is equal. Therefore let us divide the 32 pistoles in half, and give me the 32 of which I am certain besides.” He will then have 48 pistoles and the other will have 16.

(Translation from David Eugene Smith A Source Book in Mathematics (McGraw Hill, 1929) p.546–552)

That is to say, the payoff to gambler !!P!! in the event of an early cessation is:

$$ \begin{array}{l|rrr|r} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 \\ \hline 0 & & & & 64 \\ 1 & & & \color{darkblue}{48} & 64 \\ 2 & & & 32 & 64 \\ \hline 3 & 0 & 0 & 0 & & \\ \end{array} $$

with !!Q!! getting the remainder, and the corresponding value, !!16!!, in the !!\pq 12!! case. From this conclusion, Pascal then reasons similarly that in the !!\pq20!! case, gambler !!P!! should get 56 of the 64 pistoles:

$$ \begin{array}{l|rrr|r} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 \\ \hline 0 & & & \color{darkblue}{56} & 64 \\ 1 & & & 48 & 64 \\ 2 & \color{darkblue}{8} & 16 & 32 & 64 \\ \hline 3 & 0 & 0 & 0 & & \\ \end{array} $$

And then proceeding in the same way he finds the one remaining nontrivial entry, that if the gamblers quit at the !!\pq 10!! stage, !!P!! should get 44 pistoles and !!Q!! should get 20, or vice versa if the scores are reversed:

$$ \begin{array}{l|rrr|r} {Q\atop \downarrow}P\rightarrow & 0 & 1 & 2 & 3 \\ \hline 0 & 32 & \color{darkblue}{44} & 56 & 64 \\ 1 & \color{darkblue}{20} & 32 & 48 & 64 \\ 2 & 8 & 16 & 32 & 64 \\ \hline 3 & 0 & 0 & 0 & & \\ \end{array} $$

Note that the problem Fermat and Pascal are solving doesn't immediately appear to be the same as the one we were solving at the beginning of this article. They want to calculate, for each possible game position, the probability that the Chevalier des Philadelphiens will win the entire game; say this is !!P!!. Then upon quitting, Chevalier P should take the fraction !!P!! of the entire stake and Marquis Q should take the other !!1-P!!.

But the actual values !!P!! are also equal to the total bankroll we must have at each stage of the game

It should be clear that the technique for constructing the table is exactly the same as the technique for contructing the bankroll matrix earlier: Each entry is the mean of the two entries immediately below and to the right.

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