Ask The Wizard #291
What is your advice for suicide pools?
For those who don't know, a suicide pool is a last-man-standing contest, generally based on the NFL, although any sports league will do. Here are the fundamental rules.
- Every player pays the same amount to play.
- Every week every player must predict the winner of one game that week. All games are scored straight up, not against the spread.
- If a player is ever wrong he is immediately eliminated from the contest.
- No player may choose the same team more than once.
- The last man standing wins the pot.
There may be additional rules, like the person running the pool getting a cut, and splitting the pot once it gets down to a few people, if and only if by mutual agreement.
Before going further, here are the results from last season from officefootballpool.com.
Office Football Pool — 2014 Season
Week | Team | Spread | Picks |
---|---|---|---|
1 | Phil | -10.5 | 54 |
1 | Chic | -7 | 29 |
1 | Pit | -6.5 | 14 |
1 | Det | -6 | 6 |
1 | KC | -3 | 5 |
1 | NYJ | -5.5 | 5 |
1 | NE | -4 | 1 |
1 | SF | -4.5 | 1 |
1 | StL | -3.5 | 1 |
2 | Den | -12.5 | 38 |
2 | GB | -8 | 23 |
2 | NO | -6.5 | 7 |
2 | SF | -7 | 5 |
2 | TB | -5.5 | 3 |
2 | Wash | -6 | 2 |
2 | Cin | -5.5 | 1 |
2 | NE | -5.5 | 1 |
3 | NE | -14 | 44 |
3 | NO | -10 | 16 |
3 | Atl | -6.5 | 3 |
3 | Cin | -6.5 | 1 |
3 | Ind | -6 | 1 |
4 | SD | -13 | 34 |
4 | Ind | -7.5 | 20 |
4 | Pit | -7.5 | 10 |
4 | Bal | -3 | 1 |
5 | NO | -10 | 21 |
5 | GB | -8 | 13 |
5 | Den | -7.5 | 6 |
5 | Det | -6.5 | 5 |
5 | Phil | -6.5 | 3 |
5 | Dal | -6.5 | 2 |
5 | Pit | -6 | 2 |
5 | SD | -6.5 | 1 |
5 | Sea | -7 | 1 |
5 | SF | -5 | 1 |
6 | Sea | -8.5 | 21 |
6 | Den | -9.5 | 13 |
6 | SD | -7 | 7 |
6 | Cin | -6.5 | 5 |
6 | Ten | -4 | 2 |
6 | Atl | -3 | 1 |
6 | SF | -3 | 1 |
7 | NE | -9.5 | 9 |
7 | Sea | -6.5 | 9 |
7 | GB | -6.5 | 3 |
7 | Bal | -6.5 | 2 |
7 | Buf | -5.5 | 2 |
7 | Wash | -5 | 2 |
7 | Dal | -6.5 | 1 |
8 | Dal | -9.5 | 12 |
8 | KC | -7 | 5 |
8 | Clev | -7 | 1 |
8 | Mia | -6 | 1 |
9 | SF | -10 | 3 |
9 | Cin | -10.5 | 2 |
9 | Sea | -14.5 | 2 |
10 | Bal | -9.5 | 2 |
10 | Cin | -6 | 1 |
10 | Sea | -9 | 1 |
11 | GB | -5.5 | 1 |
11 | Mia | -6 | 1 |
11 | NO | -7 | 1 |
12 | Ind | -14 | 2 |
13 | Bal | -6.5 | 1 |
13 | Det | -7 | 1 |
14 | NO | -10 | 1 |
The "picks" column refers to the number of players who picked that team.
Note that not once did anybody ever pick an underdog. In the 497 picks made, the average point spread was 9.3.
I do not know why they recorded a pick in week 14, because there was only one player left at that point.
Based on this study of other player behavior, here is my basic strategy for suicide pools. These are all just general guidelines and not hard and fast rules that you must follow every week.
- Remember that your enemy is the other players. Don't focus too much on handicapping the NFL but on beating your competition.
- Don't pick the biggest favorite. That is what most other people will do. Hopefully that big favorite will lose and the field will be significantly thinned down. You want to be a survivor rather than a victim when a huge favorite loses.
- Don't forget the rule that you can't pick the same team twice. That said, you can't always pick great teams. At times you should pick average teams playing awful teams, to preserve the easy wins on the best teams.
- Don't bother handicapping. Just use the point spread as an indicator of your chances to win any given pick.
- If a week has exactly two huge favorites, I would avoid both of them. Remember that you're trying to balance picking games likely to win while not following the masses. As a very general rule, I think picking the third highest spread is a good idea, especially if you're not wasting one of the best teams.
This question is raised and discussed in my forum at Wizard of Vegas.
You have two fair six-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a two."
What is the probability that both dice are showing a two?
This problem is easily solved with the Bayesian formula of conditional probability, which goes:
Pr (A given B) = Pr (A and B)/Pr(B), where "Pr(x)" means the probability of x."
To apply it to this case:
There is a very similar problem, known as the "boy-girl problem," that goes, "You're told a mother of two has at least one boy. What is the probability she has two boys?" The answer to that one is 1/3.
This question is raised and discussed in my forum at Wizard of Vegas.
According to your review of Gamesys N.V. software, the Player bet in baccarat pays 1.0282 to 1. You note the player advantage is 0.02%. If we ignore the 24-hour time limit rule, is there a way to have an advantage on this bet, even after the 10% commission on net gambling session wins?
Yes! Keep playing, betting the same amount every time, until you are up any amount of money. Then quit, wait 24 hours, and repeat.
To be specific, the advantage per bet is 0.0233341%. The overall advantage following this strategy is 90% of that, or 0.0210007%.
There may be other equally good strategies but if anyone has a superior strategy, I'm all ears.
Suppose you choose two ranks, for example kings and queens. What is the probability that at least once in a randomly-shuffled 52-card deck a king and queen will be next to each other?
I get 48.6279%. If you're looking to make a bet on it, fair odds on the "yes" would be 1.0564 to 1.