»  Solutions to puzzles in my VDARE.com monthly Diary

  October 2018


My October diary included the following brainteaser.

My friend Joe Shipman has patented a new casino game, which he will present at a trade show in Las Vegas next month. He's given me permission to publicize it here.

The game is based on the following pure-math puzzle, which is remarkably hard to solve.

Players 1 and 2 ante one unit and are each "dealt" a random real number between 0 and 1. If there is a showdown (bets called, no one folds) the higher number will win.

Player 1 checks or raises one unit.
If Player 1 raised, Player 2 must call or fold.
If Player 1 checked, Player 2 checks or raises one unit.
If Player 2 raised, Player 1 must call or fold.

Who has the advantage, and by how much?


• Solution

As I said, the math here, although not advanced — no motivic cohomology, no contravariant functors — is nontrivial. We are in the world of Game Theory, a subtle and fascinating branch of math that calls for a certain orientation, an ability to think a certain way, that I confess I don't much have.

A key concept here is expected return, hereinafter "ER," best illustrated by an example.

You and I agree to roll two dice. If the total that comes up is seven or more, I'll pay you $36; if the total's less than seven, you pay me $48. Should I go ahead and play?

There are 36 equally-probable ways two dice can come up, from 1+1, 1+2, 1+3, … to 6+4, 6+5, 6+6. Fifteen of those results give a total less than seven; twenty-one give seven or more.

Put it another way: The probability I lose $36 on the game is 21/36; the probability I win $48 is 15/36.

So overall, if I play the game many-many times, my expected return is (−36) times 21/36 plus (+48) times 15/36. That works out to −21 plus 20, a net loss of one dollar to me. Don't play!

(Unless, of course, you enjoy the excitement of playing, of perhaps beating the slight negative odds, and don't mind the other party — whose name is usually "The House" — having a small overall edge.)


OK, to the problem as stated.

Joe actually offered it to me as the middle problem of three, which he characterizes mathematically as "easy," "hard," and "extra hard."

I'll change "Player 1," "Player 2" to "Player A," "Player B," if you don't mind. It's best to avoid numbers in this sort of thing unless they refer to actual quantities.

•  Easy:  Player A checks or bets (i.e. raises one unit).
Player B must call or fold.
•  Hard:  Player A checks or bets.
If Player A bet, Player B must call or fold.
If Player A checked, Player B checks or bets.
If Player B bet, Player A must call or fold.
•  Extra hard:  Player A checks or bets.
If Player A bet, Player B must call or fold.
If Player A checked, Player B checks or bets.
If Player B bet, Player A can call, fold, or re-bet.
If Player A re-bet, Player B must call or fold.

The easy case illustrates the principles well enough, so I'll work through the math. The hard and extra hard cases use the same logic but involve way more math.



•  Easy:  Player A checks or bets (i.e. raises one unit).
Player B must call or fold.

Let's suppose A draws the real number a, B draws b. Once again, these are random real numbers between 0 and 1.

Some assumptions:

(I'm using only < and > signs, no ≤ or ≥. The equality cases are sets of zero measure and don't affect the main logic.)

Even without having figured out the value of x, both players can now work up the following table.

A B   a > b?     a > x?     b > x?     A wins     B wins     Probability  
check fold (never happen)
check call Y Y +1 −1 a
check call N Y −1 +1 1 − a
bet fold Y N +1 −1 x
bet call Y Y Y +2 −2 a − x
bet call N Y Y −2 +2 1 − a
bet fold N N +1 −1 x
bet call N N Y −2 +2 1 − x

From there it's a breeze to work out the ER values for A.

Notice an interesting thing about the arithmetic there. If, when we finally figure out the value of x, it turns out to be more than 1/3, the ER for A betting will be bigger than −1. If a is small enough, that ER may be bigger than 2a − 1, the ER for checking. If, for example, x turns out to be 0.4 and A draws the number 0.05, then the ER on betting is −0.8 — better than the ER on checking, which is −0.9. As Joe says: "This is why bluffing is essential in poker!"

Arithmetically: When a < x and x > 1/3 and 2a − 1 < 3x − 2, A should bet (i.e. bluff).

That last expression tidies up to a < 1.5x − 0.5. So when x > 1/3 and a is in the range 0 < a < 1.5x − 0.5, A should bet (i.e. bluff).

When a increases past that boundary he should check, until … until when? Why, until the ER for checking drops below the ER for betting, i.e. 2a − 1 < 4a − x − 2. That simplifies to a > 0.5x + 0.5. From there until a = 1 A is betting again with ER 4a − x − 2, topping off at 2 − x when a = 1.

Now you have to imagine — no, I'm not going to draw it — a graph with a along the horizontal axis and ER along the vertical. The value of a goes from 0 to 1 of course; the value of ER goes from 3x − 2 (which is some number not less than −1; we're working here with x > 1/3, remember) to 2 − x (some number no bigger than 5/3).

The actual graph is continuous, made of three straight line segments: one horizontal at height 3x − 2 for 0 < a < 1.5x − 0.5 (A's bluff-betting), then a slanting one up to height x for 1.5x − 0.5 < a < 0.5x + 0.5 (A's checking), then another slanting one to take us from 0.5x + 0.5 < a < 1 (A's regular-betting).

The area under that graph is the total ER for A if x > 1/3. It comes out to 2.5x² − 2x + 0.5, which has a minimum at x = 0.4. At that minumum A's expected return is 0.1.

All that assumes x > 1/3. You can chew through corresponding algebra for x < 1/3. It's actually easier: there is no bluff zone, so the graph is just two line segments. The total ER you come up with (area under the graph) is 0.25(1 − x)². Remembering that x < 1/3, the minimum for that is when x actually is 1/3, for an ER of 1/9.

That's a tad more than the minimum ER for x > 1/3. From the point of view of B, who wants to minimize A's ER, x = 0.4 is the way to go.

A can do the same arithmetic as B, so he'll assume B's settling on x = 0.4. A will bet (i.e. bluff) when he draws less than 0.1, check from 0.1 to 0.7, bet from 0.7 to 1. He'll have a slight advantage over B, expected return 0.1.


If you followed all that you'll understand why I picked Joe's easy case. The denominators in the solution are a neat 10. In the hard case, Joe tells me, they are 18; and in the extra hard case they are 1377. I'm taking his word for it.

The solution for the hard case — the case I posed in my October Diary — is in fact that Player B has an advantage of 1/18 of the original stake.

The casino game based on this logic is being promoted with videos here and here.