Entering the Same Team Multiple Times: Is it a Good Play?

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Recently there have been a lot of forum posts and discussion about whether or not a player should be allowed to enter the same line up multiple times into a contest. I am going to use some basic math to show the effects of entering the same team more than once in DFS contests. One term you are going to need to know is Expected Value. The expected value of an event is the value of all the possible outcomes, weighted by each probability. For example, let’s say you bet your friend on a coin toss, and you bet heads for $10. You both put your $10 in the middle, making the pot $20. With a fair coin you win the $20 50% of the time, and win nothing 50% of the time.

To find the expected value of this we multiply $20 by .5(the probability of the event) and $0 by .5. This comes out to $10. This makes sense, as betting even odds on a coin flip shouldn’t make you any money or lose you any money in the long run.
Alright now let’s get into DFS. Say we enter a $10, 10-man, rakeless tournament The payouts will be as follows:

1st Place=$50
2nd Place=$30
3rd Place=$20
4th-10th = $0

Let’s assume everyone has equal skill. Without rake and with all of the contestants entering just one team, each player is expected to break even. Now let’s say one entrant decides to enter the same team twice. There are 9 different outcomes this player can expect, and, as he is equal in skill to the rest of the field, each outcome is equally likely. This means each outcome happens 1/9 times, or 11.11% of the time. To figure out his expected return on his two teams we will have to do an expected value calculation and multiply the money he makes by finishing in each position by the probability of finish (.1111) then add up the totals.

This is the expected value run out:

Place(s) Expected Value
1st+2nd-Win $80(.1111)= $8.88
2nd+3rd- Win $50(.1111)=$5.55
3rd+4th-Win $20(.1111)=$2.22
4th+5th-Win $0(.1111)=$0
5th+6th-Win $0(.1111)=$0
6th+7th-Win $0(.1111)=$0
7th+8th-Win $0(.1111)=$0
8th+9th-Win $0(.1111)=$0
9th+10th-Win $0

Buy-ins=$20; Expected Value: $8.88+ $5.55+ $2.22=$16.65; Net Win/Loss: $20-$16.65=-$3.35

So what we see here is, in a neutral setting, the player who plays the same lineup twice is losing -$3.35 per game. Why? Well the most important factor is, with multiple entries, when your team has a big day, only one of your entries has the possibility of getting first. This hurts you because, in these top heavy formats, first place is where most of the money is (50% in our case).
In my forum post, I then did the math for what would happen if a player with a 20% ROI put 2 entries into this same tournament This player ends up losing $.03 which, when you factor in the standard 10% rake, comes to -$1.03 a game. Since I want to elaborate on my original forum post, I won’t write out the math, but you can look at it here

Recently, there has been a lot of debate over using the same team twice in double ups, so let’s look at what happens when someone submits the same roster twice into a $10, 10-man, double up.

Place(s) Expected Value
1st+2nd = $40(.1111)=$4.44
2nd+3rd = $40(.1111)=$4.44
3rd+4th = $40(.1111)=$4.44
4th+5th = $40(.1111)=$4.44
5th+6th = $20(.1111)=$2.22
6th+7th = $0(.1111)=$0
7th+8th = $0(.1111)=$0
8th+9th = $0(.1111)=$0
9th+10th = $0(.1111)= $0

Buy-ins: $20; EV: $4.44(4)+$2.22=$20; Net Win/Loss: $20-$20= $0

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Awesome! So if we are winning players we should pick a team and enter it as much as possible in double ups right? Actually it turns out, with rake included, this isn’t the case. The easiest way to demonstrate this is with a $10+$1, 10 man double up where we put in 9 teams and our opponent puts in 1. Let’s say we were way better than him, beating him a seemingly impossible 70% of the time.
So we would win the entire $100 prize pool 70% of the time, $100(.7) and our opponent would get 10th. But our opponent pulls off the upset 30% of the time, taking 1st and $20 and leaving us with $80.

This is the EV calculation of the situation:

$100(.7)+$80(.3)=$94.

This means we are making $4 a game without rake, as we pay $90 in buy-ins, but with rake we pay $99 in buy-ins and lose $5 per game.

Why? I think some players out there are looking at how they make money the wrong way. You only win money because you are better than those who you play against. Our personal ROI isn’t a pure reflection of our skill, but rather a reflection of the difference in skill between us and those who we play against. We need to have bad players in the game in order to win. With a capped field, every entry we put in takes an entry away from a less skilled player, in turn lowering our ROI.

So before you hit that “enter again button” first think, “Is this a top heavy tournament?” If the answer is yes, DON’T CLICK THE BUTTON!! Next ask yourself, “is the field big enough where I can afford to take a spot away from a recreational player?” Answering that question is more of an art, but with a good team, and especially with overlay, you can sometimes answer yes to this and profitably put another team in.

Thanks for reading,
Jshilling09

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