When You Should Re-Enter A Poker Tournament

I was having a conversation with a friend about whether or not re-buying into a poker tournament improves ones odds of a higher payout and argued that the overall expected value (EV) should be negative the more times you buy in. But as I ran the numbers through my head, I realized that something funky was happening. So I brought up a spreadsheet and pretended that I was running a tournament with 3 players:

Assuming that the rake is 15%, then the EV for each player is -$15. But what if someone rebuys into the tournament? Then it should look something like this:

While the payout is certainly higher, when calculating the EV, it goes down to -$20 as expected. If each person busts out on their first bullet (buy-in) and re-enters/re-buys into the tournament, then the resulting EV is -$30, which is twice the EV of the first situation where everyone buys in just once. At first I felt this definitively answered my initial question, but something bothered me. While the overall EV is on a downward trend, what would be the marginal EV for someone who only bought in once? If everyone else except for one person fired 10 or 100 bullets into a tournament, the prize pool would be much higher and should drastically improve the odds of the one who bought in once.

Let's take the second scenario where only one person re-bought. For the marginal expected value of having 1 or 2 buy-ins, we have the following expressions:

\[E[b=1] = \frac{2}{3}(340*\frac{1}{3}-100) ~ +9\]

\[E[b=2] = \frac{1}{3}(340*\frac{1}{3}-200) ~ -29\]

Notice that the EV for the single-entry players is now positive! By staying in there with just one bullet to their name, they're the real winners while the one who had decided to rebuy is in a worse spot EV-wise. Now let's formalize this.

Say we know at any point during the tournament the following information:

*\(n_{e}\), total number of entries

*\(n_{p}\), total number of players remaining

*\(n_{x}\), number of players who risked \(r_{x}\)

*\(c_{i}\), payout for rank \(i\)

*\(r_{x}\), risk amount for different \(x\) number of bullets

*\(a\), rake

To compute the expected value of any number of bullets \(x\), when mirroring our earlier calculations, we obtain the following expression:

\[E[x]=\sum_{x=1}^{k}\frac{n_{j}}{n_{e}}(\sum_{i=1}^{n}\frac{c_{i}}{n_{p}}-r_{j}) (1)\]

At first glance it can be a little overwhelming but we can start by easily simplifying the inner summation since we know that if \(c\) is the total prize pool, then:

\[c=\sum_{i}^{n}c_{i}=n_{e}b(1-a)\]

where b is the buy-in amount. Therefore, (1) reduces down to:

\[E[x] = \sum_{x=1}^{k}\frac{n_{x}}{n_{e}n_{p}}*\left(\sum_{i=1}^{n}c_{i}-n_{p}r_{j}\right)\]

\[= \sum_{x=1}^{k}\frac{n_{x}}{n_{e}n_{p}}*(n_{e}b(1-a)-n_{p}r_{j})\]

Since \(r_{j}=jb\)

\[E[x] = \sum_{x=1}^{k}n_{x}\left(\frac{b(1-a)}{n_{p}}-\frac{xb}{n_{e}}\right)\]

\[=b\sum_{x=1}^{k}n_{x}\left(\frac{1-a}{n_{p}}-\frac{x}{n_{e}}\right)\]

Notice that if we impose that the marginal expected value, \(E[X=x]\), is positive, then we get a nice inequality:

\[\frac{1-a}{n_{p}}-\frac{x}{n_{e}}>0\]

If we solve for x first:

\[x<\frac{n_{e}(1-a)}{n_{p}}\]

So what's going on here? For any given \(x\) buy-ins, the player's EV will be positive only if the right-hand side of the inequality is greater than said number of buy-ins. Remember the scenario where only one player bought back in? In that case, there were 4 entries with 3 players remaining, which means that the right-hand side is 1.133, which is indeed greater than 1 but not greater than 2. This matches exactly what we saw earlier.

Now let's solve for \(n_{e}\):

\[n_{e}>\frac{xn_{p}}{1-a}\]

Why is this useful for any poker player? Because this tells you exactly when it makes sense to enter a rebuy tournament. When you enter on time, then \(n_{e}=n_{p}\), which violates the inequality hence a negative EV. However, if you wait to enter until the number of entries is greater than \(n_{p}/(1-a)\), then your EV becomes positive which makes more sense to enter. But what if you bust out? Now you'll want to wait until the number of entries exceed \(2*n_{p}/(1-a)\). And so on and so forth. This is why theoretically your EV is maximized when you max-late register for a tournament because that is the point in which the difference between the number of entries and players is at its peak. Furthermore, if you do need to rebuy, you will also want to wait as long as possible before you reenter.

A key assumption I am making here is that everyone has an equal chance at cashing, which realistically is not true. Nevertheless, this should be a useful way of gauging when to enter a tournament and to help justify why late-registration is highly recommended if you want to maximize your EV in the long-run.