In suicide pools, each entrant picks one team to win each week, with no point spreads. If the team wins, the entrant continues to the next week. Once a team has been picked, that team cannot be picked again during the season. If the team loses, the entrant is eliminated from the pool. These pools are generally winner-take-all but, typically, when there are a few entrants remaining late in the season, they will agree to split the pool money.

What is the statistical benefit of having multiple entries in suicide pools?

For purposes of this question, assume that the odds of one (1) entry winning* is .5%.

* "winning", means getting some money either by winning the pool outright or dividing the pool money with other entrants.

If the odds of one entry winning are .5%, what would be the odds of winning if a player had three (3) entries? Would the odds of three be 1.5% (3 x .5 = 1.5)?

In other words, are the odds of winning with multiple entries greater/or the same as the sum of the odds for individual entries?

My sense, which is intuitive and without any statistical basis, is that multiple entries have a greater advantage that the sum of the individual entries. Thus, in my example, the odds of winning for the multiple entries would be greater than 1.5%.

Hoping there's a mathematician or statistician out there who has an answer.

What is the statistical benefit of having multiple entries in suicide pools?

For purposes of this question, assume that the odds of one (1) entry winning* is .5%.

* "winning", means getting some money either by winning the pool outright or dividing the pool money with other entrants.

If the odds of one entry winning are .5%, what would be the odds of winning if a player had three (3) entries? Would the odds of three be 1.5% (3 x .5 = 1.5)?

In other words, are the odds of winning with multiple entries greater/or the same as the sum of the odds for individual entries?

My sense, which is intuitive and without any statistical basis, is that multiple entries have a greater advantage that the sum of the individual entries. Thus, in my example, the odds of winning for the multiple entries would be greater than 1.5%.

Hoping there's a mathematician or statistician out there who has an answer.

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