By Josh Feldman

While I don’t play poker nearly as much as I did 15 years ago, I still enjoy playing and watching the greatest card game ever created. The pandemic took that away from me, but now poker is back, nearly as popular as ever. Heck, there is even a 24-hour poker streaming service that I subscribe to. The little joys in life!

Many have named the summer of 2022 as the summer of cheating, with big controversies hitting the world of chess and professional fishing over the past few weeks. Poker and cheating are unfortunately synonymous, yet somehow poker nearly escaped the cheating headlines this past year … until a few days ago. That’s when all heck broke loose in the poker world when a novice player won a $270,000 pot by calling an all-in turn bet with only jack-high.

Everyone has an opinion of whether something nefarious took place with that hand at the Hustler Casino, and at the time of publication the casino currently has undergone an investigation to determine whether cheating took place. My brother and I discussed the famous hand yesterday, and we still don’t know what to think. But because poker is back in the spotlight, my brother asked me a simple question: How come I don’t see any good poker puzzles anywhere? Alas, I didn’t have a good answer for him, but figured I would do my small part to help fix that.

For the following questions, assume that the game Texas hold’em is played strictly between exactly two players. And please don’t cheat when submitting your solution; we have had enough of that already this year!

One can define a “player’s equity” as simply the size of the pot, multiplied by the share of the pot the player would win, times the probability (s)he wins the pot. So if there is $100 in the pot and I win the entire pot 5% of the time, my equity would be $5. If there is $100 in the pot, what is the smallest non-zero amount of equity can one have after the flop? What is one example of seven cards (two cards in your hand, two cards in your opponent’s hand, and three cards shared by both) that would create such a situation?

In poker, “the nuts” is a slang term meaning that one holds the best possible hand that cannot be beat. For example, on a run-out of {2, 3, 3, 9, Jack}, the best hand possible is to have four-of-a-kind, which occurs if you got dealt pocket 3’s. Can having three-of-a-kind 10’s ever be considered “the nuts”? Why or why not? What is the worst possible hand one can have that can be considered “the nuts”?

In this year’s most-talked-about poker hand, an amateur player called an all-in turn bet with just jack-high. This created one of those rare situations where although the novice player had the lead after the turn, in reality she was an underdog to win the hand after the river. Given that you currently have the best hand after the turn, what is the smallest possible probability you end up winning the hand after the river card gets dealt? Give an example of how this could occur.

Solutions may be emailed to **cont.puzzles@gmail.com**.

In order to make the solver list, your solutions must be received

by December 15, 2022.

**Solutions to Previous Puzzles—Playing With Dice**

Problem 1. Can probabilities be assigned to a pair of two-sided dice so that the three possible face sums will have equal probabilities?

**Answer: **No.

**Solution: **Assume 1/3 = probability(face sum is 2) = Pr(2) = p(1)q(1) → q(1)=1/3p(1). Similarly, assume 1/3 = Pr(4) = p(2)q(2) → q(2)=1/3p(2). Then Pr(3) = p(1)/3p(2)+p(2)/3p(1) = f(p(1)/p(2))/3 where f(r)=r+1/r which has a local minimum at r=1, f(1)=2 → Pr(3) ≥ 2/3.

**Problem 2. **Can probabilities be assigned to a pair of six-sided dice so that the 11 possible face sums will have equal probabilities?

**Answer: **No.

**Solution:** As in Problem 1, q(1)=1/11p(1), q(6)=1/11p(6), and Pr(7) = p(1)/11p(6)+p(6)/11p(1)+… ≥ f(p(1)/p(6))/11 ≥ 2/11. (Note: If instead of reassigning probabilities, we revise the labels on one die and allow blanks, we can get face sums 1 to 12 with equal probability.)

**Problem 3.** Can probabilities be assigned to a pair of two-sided dice so that the sums 2, 3, and 4 will have probabilities 1/7, 4/7, and 2/7, respectively?

**Answer: **Yes.

**Solution:** See Problem 4.

**Problem 4. **Find a set S of conditions on the probabilities A, B, and C, that are satisfied if, and only if, probabilities can be assigned to a pair of two-sided dice so that the three face sums 2, 3, and 4 will have probabilities A, B, and C, respectively. For example, the set of conditions, S, should contain A + B + C = 1.

**Answer: **S should also contain that A, B, and C need to be non-negative. One additional condition is needed. Variations on each of the following were submitted: B ≥ 2(A – A) and A + C ≤ 1 and (B/A)2 ≥ 4(C/A). You might enjoy showing that they are equivalent.

**Solution:** p(1)q(1)=A and p(1)(1–q(1)) + (1–p(1))q(1) = B **↔** p(1)+q(1) = B+2A. Solving for q(1), inserting it into the first equation, and rearranging gives the quadratic p(1)2– p(1)(B+2A)+A = 0, which has a real solution if the discriminant is non-negative and that is the first of the three conditions given in the answer. The quadratic is concave up with value A at p(1)=0 and C at p(1)=1 so its roots p(1) and q(1) lie in the interval [0,1].

**Solvers:** Bob Byrne; Bob Conger; Rui Guo; Clive Keatinge; Alex Kozmin; Don Onnen; David Promislow; Noam Segal; John Snyder; Al Spooner; and Ron Stokes