Ask The Wizard #316
At the beginning of a Jeopardy round, why do some players, like James Holtzhauer, start picking from the bottom? Wouldn't it make more sense to warm up with the easier questions at the top, in part to ensure proper understanding of the category, which are sometimes tricky?
The reason is the Daily Doubles are placed in the bottom three rows 91.5% of the time. The following table shows their locations on the board over 13,660 Daily Doubles found.
Daily Double Location
| Row | Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | Column 6 | |
|---|---|---|---|---|---|---|---|
| 1 | 5 | - | 3 | 3 | 2 | 3 | 16 |
| 2 | 280 | 137 | 216 | 167 | 207 | 140 | 1,147 |
| 3 | 820 | 442 | 677 | 658 | 643 | 472 | 3,712 |
| 4 | 1,095 | 659 | 982 | 907 | 895 | 627 | 5,165 |
| 5 | 787 | 403 | 670 | 671 | 613 | 476 | 3,620 |
| Total | 2,987 | 1,641 | 2,548 | 2,406 | 2,360 | 1,718 | 13,660 |
Source: J! Archive.
Here is the same data in the form of how often a Daily Double is found in each cell of the board.
Daily Double Probability
| Row | Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | Column 6 | |
|---|---|---|---|---|---|---|---|
| 1 | 0.0% | 0.0% | 0.0% | 0.0% | 0.0% | 0.0% | 0.1% |
| 2 | 2.0% | 1.0% | 1.6% | 1.2% | 1.5% | 1.0% | 8.4% |
| 3 | 6.0% | 3.2% | 5.0% | 4.8% | 4.7% | 3.5% | 27.2% |
| 4 | 8.0% | 4.8% | 7.2% | 6.6% | 6.6% | 4.6% | 37.8% |
| 5 | 5.8% | 3.0% | 4.9% | 4.9% | 4.5% | 3.5% | 26.5% |
| Total | 21.9% | 12.0% | 18.7% | 17.6% | 17.3% | 12.6% | 100.0% |
The reason for searching for Daily Doubles is they are a good way to double your score. Most contestants will have a probability of about 80% to 90% of getting any given clue right. It is a great value to get even money on a wager you have an 80% to 90% chance of winning. A major reason James Holtzhauer won as much as he did was aggressive Daily Double searching and then going "all in" most of the time when he found them. It is also how he lost against Emma, when she employed the same strategy against him.
With this subject, I would like to try and quantify skill in backgammon. For consideration, take two good players but one of them is just 1% better (consider this as fact and exact figure) than the other. So, statistically, out of 1000 match games player A should win 505 and player B should win 495.
I have a double question:
- What is the minimum amount of match games that player A should play against player B to be 90% sure to come out as a winner on overall ?
- What is the minimum amount of match games, where a match is the first player to win five games, that player A should play against player B to be 99% sure to come out as a winner on overall ?
Story behind this is that many backgammon players (including me) seem to have no idea what "long run" really means. It is just generally agreed that the better player will overcome the luck factor and will win in the long run. OK, but when the level is that close ?
I would see this 1% like a biased coin flip but really don't know the answers.
I'm going to ignore the doubling cube and assume every game results in a simple win or loss.
That said, if every game counted as one point, it would take 16,221 games to ensure you had a 90% chance of winning more than half of them, assuming a 50.5% chance of winning each game.
At a 50.5% probability of winning each game, I show a 51.23% chance of winning a match. You would need to play 8,853 matches to have a 90% probability of winning more than half of them.
These answers can be found using either the binomial distribution or the Gaussian curve approximation.
This question is raised and discussed in my forum at Wizard of Vegas.
Assume I'm playing craps at a table with 100x odds. I'm debating whether to make a place bet on the 6 or 8 or a put bet. How much odds would I have to put on a put bet to have a better value than the place bet.
Good question. The house edge on a place bet on the 6 or 8 is 1.52%. At 5x odds, the overall house edge is exactly the same on a put bet on the 6 or 8 at 1.52%. At 6x odds, it drops to 1.30%. So, it takes 6x odds to become a better value.
In video poker, how often will a player have 0 up to 5 cards to a royal after the deal?
The answer is rather involved, because there are a number of ways the player can have a royal possibility, after the deal, in multiple suits. I assume the player always keeps the cards in the suit with the greatest chance at a royal and picks arbitrarily in the event two or more suits are tied for the most cards to a royal. That said, let me define some abbreviations:
- Royal cards = cards of rank 10 to ace.
- H = Royal cards in hearts.
- S = Royal cards in hearts.
- C = Royal cards in hearts.
- D = Royal cards in hearts.
- x = Non-royal card
The following table shows the number of combinations of every possible situation. A row will include all mathematically equivalent cases. For example, Hxxxx will include having one card to a royal only in any suit (not just hearts).
Combinations to Royal after Deal
| Hand | Cards to Royal | Combinations |
|---|---|---|
| HHHHH | 5 | 4 |
| HHHHS | 4 | 300 |
| HHHHx | 4 | 640 |
| HHHSS | 3 | 1,200 |
| HHHSC | 3 | 3,000 |
| HHHSx | 3 | 19,200 |
| HHHxx | 3 | 19,840 |
| HHSSC | 2 | 6,000 |
| HHSSx | 2 | 19,200 |
| HHSCD | 2 | 5,000 |
| HHSCx | 2 | 96,000 |
| HHSxx | 2 | 297,600 |
| HHxxx | 2 | 198,400 |
| HSCDx | 1 | 20,000 |
| HSCxx | 1 | 248,000 |
| HSxxx | 1 | 744,000 |
| Hxxxx | 1 | 719,200 |
| xxxxx | 0 | 201,376 |
| Total | 2,598,960 |
The next table shows the overall probability of having 0 to 5 cards to a royal after the deal.
Cards to Royal Probabilities
| Cards to Royal | Probability |
|---|---|
| 5 | 0.0002% |
| 4 | 0.0362% |
| 3 | 1.6637% |
| 2 | 23.9403% |
| 1 | 66.6113% |
| 0 | 7.7483% |
| Total | 100.0000% |
Not that you asked, but if a player followed a "royal or nothing" strategy, his probability of a royal per hand would be 1 in 23,162.