# Daily Fantasy 50/50 Variance

This article is part 2 of my series on variance. Last time we took a look at Daily Fantasy Tournament Variance, and this time we will explore 50/50s. The question once again is what does variance look like and many games do we have to play until we have a good idea of our real ability?

### Daily Fantasy 50/50 Variance: Example 1

I will use a \$22 50/50 game with a \$40 prize as an example. I created a hypothetical player to play in this game. This person cashes 60% of the time for a long-term ROI of 9.1%. Again whether or not you agree with this number is not that important – what is important is the variation around the number, which I will demonstrate.

Methodology:

I simulate my hypothetical player in the 50/50 as follows:

• Generate a random number from 0 to 1
• If this number is below .60, consider it a “cash,” if above .60, a loss
• Repeat this process one million times

I did this and got the following graph. The blue line is cumulative ROI across all trials up to that point, and the red line is the 9.1% long-term ROI that I referenced earlier.

ROI stabilizes at around 500 trials. If you recall when I did this for tournaments things didn’t stabilize until around 1,300 to 6,000 trials depending on what “stable” meant to you.

[It is worth noting that this assumes all trials are completely uncorrelated with each other. If you are entering multiple lineups with similar core players across many lineups, than these volatility figures would be higher for you.]

### Daily Fantasy 50/50 Variance: Risk of No Profits

Although in statistics it is not really defined this way, when people talk about variance in DFS, they typically are referring to a risk of losing money. Therefore, I approached the analysis from another angle. I broke my sample into intervals and analyzed profitability (or lack thereof) over these smaller samples. You can then ask yourself the following question: given a certain number of 50/50s, what is the percent chance of losing money? For example, for a range of 100 consecutive games, the picture looks like this (ROI on the x-axis):

In this case there is a 23.4% chance of losing money over any 100 game period. If we extend the sample period out to 1,000 trials the picture looks like this:

Here there is only a 1.0% chance of losing money. In fact at 550 trials is when I first had less than 5% chance of losing money. Recall from the tournament example that the 5% threshold occurred at 2,600 trials.

Thanks for reading and let me know in the comments if there is anything you would like to see from me in the future.

• ### bripc23 (bripc23)

• BriPC23 is a GrindersU instructor and was the 2013 StarStreet PFBC Runner-Up.

• jshilling09

2012 DSBC Finalist

Good article, I think it should be noted that if you are putting one team a day in 50/50s for baseball you are getting around 200 trials. So you still have a decent chance of losing money. If you play a bunch of different sites it should even out more but still this article shows that variance is indeed our bitch lover.

• Galante118

• ### Blogger of the Month

any articles involving the math behind the game really intrigues me. Very similar to poker in that regard and don’t think its been explored enough, or at the very least discussed enough.

• phrygia

Applying the normal distribution approximation, a player’s standard deviation of W% is (p*q/n)^0.5. If this player wants to reduce his losing chance to less than 5%, his true-talent W% must 1.645SD higher than the break even percentage(in this case, it’s 55%). Given the true-talent W% is 60%, we know that 260 games are enough, much fewer than your simulation.

• bripc23

2013 PFBC Finalist

Sorry you are going to have to break that down for me. You are using the p-value to find number of entries? Also how do we know standard deviations of winning percentages?

• phrygia

If a player with p winning percentage won x games in n double-up contests, then x will follow the binomial distribution with expectation E(x)=n*p and variance V(x)=n*p*(1-p). By the Central Limit Theorem(CLT), when n goes larger and larger, binomial distribution will converge to the normal distribution with the same expectation and variance. Thus, we can use the normal distribution to approximate the binomial distribution. In practice, n equals to 30 or more usually large enough to provide a good approximation.

In our case, p=0.6, E(x)=0.6*n and V(x)=0.24*n. Since we set the break even W%=0.55, the player losing money means he won less than 0.55*n games. Let’s using P(xï¼œ0.55*n) to denote the probability of x smaller than 0.55*n. To applying the normal approximation, we should normalize the both sides of xï¼œ0.55n first.

xï¼œ0.55n
(x-E(x))/SDï¼œ(0.55n-E(x))/SD
(x-0.6n)/square-root(0.24n)ï¼œ(0.55n-0.6n)/sqrt(0.24n)
(x-0.6n)/sqrt(0.24n)ï¼œ(-0.05n)/sqrt(0.24n)

The left side of the inequality(name it y) is a binomial variable with E(y)=0, V(y)=1. By CLT, we could use standard normal variable Z to approximate it.

Now we want to know how large n needed such that the losing chance will less than 5%. That is,

P(Zï¼œ(-0.05n)/sqrt(0.24n))ï¼œ0.05

Check the normal distribution table, Z must smaller than -1.645 to let the probability less than 0.05. So we have

(-0.05n)/sqrt(0.24n)ï¼œ-1.645

Solving the inequality, we find nï¼ž259.78.

• phrygia

I’m still trying to figure out why your simulations are so different to the math. Even in the 100 games period, the binomial probability table gives only 0.131 probability for the player in question won 54 or fewer games, much less than 23.4% the simulation suggested. (There is only 1/1000 chance a simulation would go north to 23.4% or more, when given an a priori probability 0.131.)

I suspect that your simulations may set the break even percentage as 56%. If this was true, the probability of losing money in a 100/500/1000 games period would be 0.237, 0.05, 0.005, respectively. Much close to the simulations suggested.