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Ask The Wizard #291

What is your advice for suicide pools?

anonymous

For those unfamiliar, a suicide pool is a last-person-standing challenge typically associated with the NFL, although it can be based on any sports league. Here’s a breakdown of the key rules involved.

  1. All participants contribute an equal entry fee to join the game.
  2. Each week, players must select the winning team from one game that week. The scoring is done based on actual game outcomes, without considering point spreads.
  3. If a contestant makes an incorrect prediction, they are instantly removed from the competition.
  4. Players are prohibited from selecting the same team on multiple occasions.
  5. The last man standing wins the pot.


Additional guidelines may exist, such as the organizer taking a share and distributing the remaining pot once a few players are left, but this requires mutual consent.

Before we go deeper, let's refer back to the outcomes from the previous season, sourced 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 indicates the number of contestants that selected that particular team.

Interestingly, throughout the 497 choices made, there wasn’t a single instance where an underdog was picked. The average point spread calculated was 9.3.

It’s puzzling that there was a recorded pick in week 14, given that only one player remained at that stage.

From analyzing the choices of other players, here’s my suggested approach for navigating suicide pools. These are more of general principles rather than strict regulations to adhere to every week.

  1. Keep in mind that your main competition is the other participants. Instead of focusing solely on evaluating NFL matchups, concentrate on outsmarting your opponents.
  2. Avoid selecting the most dominant teams, as this is often the path taken by most players. The aim is to have that top favorite lose, which would lessen the competition significantly; you want to be the survivor when that happens.
  3. Bear in mind the rule against picking the same team more than once. With that said, you sometimes have to choose mid-level teams facing very weak opponents to protect your chances with the stronger teams for later.
  4. Disregard complex handicapping. Instead, let the point spread guide your likelihood of success for any pick.
  5. If you find two major favorites in a week, it would be wise to steer clear of both. Always aim to strike a balance between picking likely winners and setting yourself apart from the group. Generally, opting for the third highest spread can be advantageous without burning one of the elite teams.


This topic has been raised and is under discussion on my forum at Wizard of Vegas .

Imagine you have two standard six-sided dice in a cup. You shake the cup and slam it down on the table, concealing the outcome. Your partner peeks under, stating truthfully, 'At least one of the dice shows a two.'

What are the odds that both dice are displaying the number two?

Dween

This question can be resolved easily using Bayesian conditional probability, represented by the formula:

Pr (A given B) = Pr (A and B)/Pr(B), where 'Pr(x)' signifies the probability of occurrence x.

To apply it to this case:

Pr(Both dice showing a two if at least one is a two) = Pr(Both dice are a two)/Pr(At least one die is a two) = Pr(Both dice showing a two given that one is a two) = [(1/6) × (1/6)] / [1 - (5/6) × (5/6)] = (1/36) / (11/36) = 1/11.

A related scenario known as the 'boy-girl problem' states, 'A mother with two children has at least one son. What are the odds that both children are sons?' The solution to this particular question is 1/3.

This topic has been raised and is under discussion on my forum at Wizard of Vegas .

According to your review of Gamesys N.V. software In baccarat, a player bet pays 1.0282 to 1. You observe that the player’s edge is 0.02%. If we disregard the 24-hour limit, is there a strategy to gain an advantage on this wager after accounting for the 10% commission on net winnings?

anonymous

Absolutely! Continue placing bets of the same amount until you achieve any profit. Then stop, wait 24 hours, and repeat the process.

Specifically, the advantage per bet amounts to 0.0233341%. When employing this strategy, your overall edge is about 90% of that figure, rounding to 0.0210007%.

While there could be other effective strategies available, I am open to hearing about any that might be even better.

Let’s say you decide on two ranks, like kings and queens. What are the chances that at least once in a randomly shuffled 52-card deck a king will sit next to a queen?

anonymous

The probability comes out to roughly 48.6279%. If you're interested in placing a bet, fair odds for a 'yes' response would be approximately 1.0564 to 1.