# Ask The Wizard #346

Yes. It refers to a change in strategy for the second-player player if he has more than 2/3 the score of the first-place player.

Let's simplify the situation to a two-player game, as follows:

- Situation A: Second place has less than half of first place.
- Situation B: Second player has between 1/2 and 2/3 of first place.
- Situation C: Second place has more than 2/3 of first place.

Before going further, let me reminid the reader of a Jeopardy rule change pertaining to ties after Final Jeopardy. No longer do both players advance, but there is now a sudden death tiebreaker question. Here is such a situation.

Situation A

Let A=$10,000 and B=$4,000

Player A should not risk losing by betting no more than A-2B-1. If he doesn't feel confident in the category, he can bet $0. Either way, he ensures winning. In this case, A should bet between $0 and $1,999.

Player B has no hope, unless A bets too much and misses. Here, B should consider the score of third place and try to stay above him, if he can, winning $2,000 for second place, as opposed to $1,000 for third place.

Situation B

Let A=$10,000 and B=$6,000

The strategy for A is to expect B to bet everything and to bet enough to cover 2B if correct. However, to be safe, he shouldn't bet too much as to fall below B if incorrect. In this case, he should bet at least 2B-A+1 and A-B-1. In this case, the range is $2,001 and $3,999.

The strategy for B is to get at least enough to pass A if correct and up his entire score. In this case, $4,001 and $6,000.

If both players do as expected and follow this strategy, the only way player B can win is if A is wrong and B is right. The probability of this is about 19%.

Situation C

Here things get more complicated and involve more game theory and randomization.

Let A=$10,000 and B=$7,000.

Before going further, it is important to estimate the probability of the Final Jeopardy clue being answered correctly. Based on seasons 30 to 34, the first place player was correct 52% of the time and second place 46%. However, these probabilities are positively correlated. Here is a breakdown of all four possibilities:

- Both correct: 27%
- First place correct, second place incorrect: 25%
- First place incorrect, second place correct: 19%
- Both incorrect 29%.

Despite a Jeopardy average of 49% for the first two players, the probability of both being right or both being wrong is 56%.

Of course, these can change based on the category, but let's keep things simple and use the probabilities above.

In this situation, player B does not have to depend on A being wrong and B being correct. He can bet low, say $0, ensuring a win if A is wrong. In other words, if A bet enough to cover B, if correct, then he would risk falling below B if wrong and B bet $0.

However, if A predicted B would bet low, say $0, then A could lock in a win by also betting $0. Both player basically have a choice to make, to go low or high. A should wish to wager the same way as B and B should wish to wager the opposite way of B. If both players were perfect logicians, they would randomize their decisions.

In this case, a high bet by A should be 2B-A+1 to A-B-1, the same as for situation B. In this case $2,999 and $4,001. A low bet by A should be $0.

A high bet by B should be the same as for situation B, bet enough to pass A if correct. In this case, $3,001 and $7,000. A low bet by B should be $0.

Forgive me if I skip the math and get right to the randomization strategies for both players.

Player A should go high with probability 62.3% and low with probability 37.7%.

Player B should be high with probablity 61.2% and low with probability 38.8%.

Assuming both players following this randomization strategy and the probability pairings of being correct stated above, the probability of player A winning is 65.2%.

If player A had more than 2/3 the score of player B, his probability of winning would go up to 81.0%.

Both players should keep in mind the significance of the 2/3 rule when wagering on Double Jeopardy.

In your video poker programming tips, you explain how that although there are 2,598,960 possible starting hands in video poker, with a 52-card deck, there are only 134,459 classes of hands necessary to analyze. My question is if someone were playing a game where the order of the cards matter, like Ace$ Bonus Poker or one with a jackpot for a sequential royal, how many different classes of hands would be needed to analyze?

For this one, I turned to my esteemed colleague, Gary Koehler, who is an expert at video poker math. His answer is 15,019,680.

A six-sided die is rolled until either of the following events happen:

A) Any side has appeared six times.

B) Every side has appeared at least once.

What is the probability event A occurs first?

To answer this one as I did, using calculus, I recommend an integral calculator like the one at integral-calculator.com/.

Here is my solution (PDF).

This problem is asked (in slightly different words) and discussed in my forum at Wizard of Vegas.