# Ask the Wizard #200

"Anonymous" .

Welcome to my 200th “Ask the Wizard” column. I had no idea when I started this column on February 5, 2000, that I would get this far. It took four years and two months to get to the 100th column, and another three years and nine months to this one.

These columns may not seem like much, but a great deal of my time goes into them. It isn’t easy opening the door to questions for the column, without the whole world flooding me with requests for math tutoring, deconstruction of past gambling trips, feedback on worthless betting systems, general gambling questions for those too lazy to search the site, advice on alleged cheating boyfriends, and entire life stories.

Unfortunately, I recently decided to close the doors to new questions. When I clear out the existing backlog I will temporarily open them again. In the future, I plan to have open seasons for questions from time to time. I hope those with good questions will hang onto them until the next open season, and those with bad questions will not be so patient.

Some of the questions that do make the column take a lot of time to answer. For example, my last comment in this column about the effect of separate decks for the player and dealer took hours of time to reprogram my blackjack simulator to handle. Other numbers have had much more time and effort go into them. On average, it takes about two days to write each column.

In April 2005 I was so fatigued from gambling questions I stopped the column for an indefinite period of time, perhaps forever. However there was a loud outcry to bring it back, so in August I did. As I write this I am going through a bad case of “Ask the Wizard” fatigue again. Rather than stop cold turkey again, I plan to cut down on the frequency of the column, to just one or two a month.

I would like to take a moment to thank my proofreader, Don Schlesinger for his invaluable help correcting my bad English, over the last several months. Not only has it made the column much nicer to read, but he has helped my own grammar in everything I write.

I would also like to thank my readers for your support over the years. Ultimately, it is the readers that put rice on my table and send my kids to private school. I hope that in return I have helped you win, or at least lose less, in the casinos, by trusting in math and not myths. Until next time, set your expectations high!

Eliot from Santa Barbara

First, for the sake of simplicity, I’m going to assume that for any given question, all players have the same probability of producing the correct answer. Also, questions range in difficulty, and this probability itself is random. Thus, the results among players will be correlated.

I know that the equal probability of answering correctly among players is an unrealistic assumption. For one thing the leader probably has a higher probability of answering correctly. For another, each player knows the category before wagering. Give me a category of gambling and I’m close to a lock, but if is poetry, I’m doomed. So don’t write to me about how unrealistic my assumptions are, because I acknowledge that already. This is meant to be more of an exercise in game theory than practical advice for future contestants.

Next, we will need some data to look at. From j-archive.com(no longer available) we find that in season 22, in 2005, Final Jeopardy was answered correctly 43.80% of players. We also learn that all three contestants got it right 14.92% of the time, and all three got it wrong 24.86% of the time. Let p_{n} be the probability n players get the question right. The first equation to work with is:

p_{0} + p_{1} + p_{2} + p_{3} = 1

Substituting the known values for p_{0} and p_{3}:

0.2486 + p_{1} + p_{2} + 0.1492 = 1.

p_{1} + p_{2} = 0.6022.

(1) p_{1} = 0.6022 - p_{2}.

We can also construct the probability of any given player getting the question right as follows.

p_{0}×0 + p_{1}×(1/3) + p_{2}×(2/3) + p_{3}×1 = 0.4380.

p_{1}×(1/3) + p_{2}×(2/3) + 0.1492 = 0.4380.

p_{1}×(1/3) + p_{2}×(2/3) = 0.2888.

Multiplying both sides by 3:

(2) p_{1} + p_{2}×2 = 0.8664.

Substituting the value for p_{1} in equation (1) we get

0.6022 - p_{2} + p_{2}×2 = 0.8664.

p_{2} = 0.2642

Solving for p_{1}

p_{1} = 0.6022 - p_{2}.

p_{1} = 0.6022 − 0.2642 = 0.3380.

Based on the three-player probabilities above, in a two-player game the probability both get the question right is:

p_{3} + (1/3)×0.2642 = 0.1492 + 0.2642×(1/3) = 0.2373.

The probability one gets it right is:

(2/3)×p_{2} + (1/3)×p_{1} = (2/3)×0.2642× + (1/3)×0.3380 = 0.4015.

The probability neither gets it right is:

(1/3)×p_{1} + p_{0} = 0.3380×(1/3) + 0.2486 = 0.3613.

It is obvious that if the leader has more than double the amount of the follower, he should bet less than the difference, to ensure victory. However, what should he do if his balance is less than double that of the follower? Let’s draw up a specific example where player A has $10,000 and player B has $8,000. To make matters easy, let’s restrict A’s options to making a big bet of $6001 (enough to guarantee victory with a correct answer) or a small bet of $1999 (leaving enough to guarantte victory if B answers incorrectly). Let’s restrict B’s options to a betting all or nothing.

At first glance, it would seem the right thing to do for player A to bet small, forcing player B to get the question right to win. That would give player A a 1 − 43.8% = 56.2% chance of winning. Assuming that strategy, player B would have to bet big to win. However, because of the correlation between players, if player B gets the question right, player A probably would too. The exact probability that player A gets the question right, given that player B got it right, by Bayes’ Theorem, is:

Probability(A and B correct)/Probability(B correct) = 0.2373/0.438 = 0.5417.

So, if player A knew player B would bet big, he should too. However, if player B knew player A would bet big, then he should bet small, and hope player A gets it wrong, ensuring a 56.2% chance of winning. Going further, if player A knew player B would be small, he would too, and have a 100% chance of winning. Assuming player A bet small, player B would of course bet big. And so we go around and around.

The optimal strategy for both players is to randomize their bet.

The following table shows the probability player A will win according to all four combinations of bets.

### Probability Player A Wins

Player A | Player B | |

High | Low | |

High | 0.7993 | 0.438 |

Low | 0.562 | 1 |

The next stop is hard to explain why, but for either player the optimal probability of either option is proportional to the absolute difference in values of the other option. Player A should bet high with probability proportional to abs(0.5620 − 1) = 0.4380. He should bet low with probability proportional to abs(0.799267- 0.4380) = 0.3613. So, the actual probability of A betting high should be 0.4380/(0.4380 + 0.3613) = 0.548002. The probability of going low is obviously 1 − 0.548002 = 0.451998.

By the same logic, player B should go high with probability proportional to abs(0.438000 - 1) = 0.562000, and low with probability proportional to abs(0.799267 − 0.562000) = 0.237267. The actual probability of going high should be 0.562000/(0.562000 + 0.237267) = 0.703145. Thus, the probability B should go low is 1 - 0.703145 = 0.296855.

The next table shows the probability of all four outcomes in strategy.

### Strategy Probabilities

Player A | Player B | ||

High | Low | Total | |

High | 0.385325 | 0.162677 | 0.548002 |

Low | 0.31782 | 0.134178 | 0.451998 |

Total | 0.703145 | 0.296855 | 1 |

David from New York City

I think you’re over-tipping. I think a good range is 2% to 5%, the greater the win, the lower the percentage.

Geng from Palm Bay

I feel your pain. 9/6 Jacks or Better is getting harder to find, even at the locals casinos, but they still definitely exist. Some of the MGM/Mirage properties have 9/6 Jacks in their high limit rooms. The Wynn is king of 9/6 Jacks; they have them all over the place. For information about current video poker offerings in Las Vegas, I highly recommend VP Free 2.

**Update**: Since this answer was published, the Wynn removed all their 9-6 Jacks machines except at the $5 denomination and higher.

Matthew from Norman, OK

I’m assuming the player must put up the 50 cents. My blackjack house edge calculator says the house edge under those rules, without the tax, is 0.36%. Add to that 0.5/10 = 5%. So the total house edge is 5.36%. In my opinion, players should refuse to play this game, on principle alone.

Diana from Albuquerque

It is probably not a real video poker machine, but a "pull tab." In jurisdictions with heavy regulation, like New Mexico, players should be careful that they understand what they are playing. With a pull tab the win is predestined. The cards are just for show.

Chuck from Mountain Top, PA

Yes, you do. I’m told by Shufflemaster that to meet the definition of a slot machine, one player’s actions can not affect the other players, as is the case in live blackjack. To get around this law, each player and the dealer are dealt cards from a unique six-deck shoe. So, you are in control of your own fate, but not that of the other players or the dealer. I understand that the game is programmed with six-deck shoes. According to my simulations, using separate shoes for the player and dealer adds 0.06% to the house edge.