Interesting Hand from the WSOP Main Event Final Table

The fact that the players who made the WSOP final table used to have four months to prepare created an interesting dynamic that diligent players could exploit. If you happened to make the final table with a short stack, you had four months to study exactly which hands you should be willing to go all-in with when the action folds to you. This high-risk all-in or fold situation just so happened to occur on the second hand of the 2015 WSOP final table.

Without going into too much detail, there were two short stacks at the table, Chan and Butteroni, both with 15 big blinds. The next shortest stack had 30 big blinds. This typically implies that Chan and Butteroni should try to outlast each other. However, the WSOP Main Event has a bizarre payout structure:

1^st $7,683,346

2^nd $4,470,896

3^rd $3,398,298

4^th $2,615,361

5^th $1,911,423

6^th $1,426,283

7^th $1,203,293

8^th $1,097,056

9^th $1,001,020

10^th $756,897

11^th $526,778

12^th $526,778

There is a normal 240k jump from 11^th and 10^th then a 245k jump from 10^th to 9^th. Things get wonky at the final table, with a 96k jump from 9^th to 8^th, followed by a 104k jump from 8^th to 7^th, then a 223k jump from 7^th to 6^th. For reasons I will never understand, besides “wanting everyone at the final table to get a million dollars”, going from 11^th place to 10^th place is worth significantly more than going from 7^th place to 6^th place. This means the short stacks should be gambling much harder than normal because of the incredibly small payout increases. While I understand a $96,000 jump is a lot of actual money, in terms of buy-ins and the jumps near the end of the final table, the payout increase from 9^th to 8^th place can effectively be ignored. This means, at least initially, moving up the payout ladder is less important than it typically is.

On the second hand of the final table, everyone folded around to McKeehen, the big stack, with 150 big blinds, who went all-in for both Chan’s and Butteroni’s 15 big blind stacks. Chan woke up with K-Q in the small blind and had what I thought was a difficult decision. I asked my followers on twitter (@JonathanLittle) what they thought about this situation and I got a wide range of answers. Some said Chan should fold because he had to wait four months for this final table, which is completely irrelevant. Others said he should fold because he “only” had K-Q. Others said it was an obvious call. What do you think?

Take a minute to actually think it over.

You can actually figure out the optimal pushing and calling ranges for each player involved in this hand using an Individual Chip Model calculator. Of course, these programs only look at push or fold scenarios. If McKeehen implements a min-raise or limping strategy, all of this changes. On the second hand of the final table, you have to guess as to whether he will have a min-raise and pushing strategy or only a pushing strategy. While I would likely use only a pushing strategy in this exact situation due to being able to apply maximum pressure to the two short stacks, min-raising is certainly an option, especially if you assume your opponents have studied the push or fold scenario.

Using the ICM calculator, you will find the optimal all-in or fold strategy for McKeehen is to push the top 53% of hands, including hands as weak as J-2s, T-4s, 9-5s, 6-4s, 4-3s, K-7o, T-8o, 8-7o, and 7-6o. Clearly this is a very wide range. If the payout jump was “normal”, he could push an even wider range.

Assuming McKeehen is pushing optimally, Chan can call with 15% of hands, including 6-6, A-5s, K-Ts, A-8o, and K-Jo. Since K-Q is in that range, Chan has an easy call. The main time Chan should fold to an all-in from McKeehen is when he is pushing with only reasonably strong hands that flop poorly, such as small pairs, A-x, and decent K-x, which implies he is min-raising everything else, because K-Q is roughly flipping or only slightly ahead against that range.

This analysis results in Chan having a somewhat easy call. Had the payout structure been “normal”, Chan would have a very close decision. If you want to succeed in poker tournaments, be sure to study the math away from the table so you know what to do when any tough situation arises. You can check out ICMizer here. Now that there is no break before the final table, you better study up!

Thanks for taking the time to read read this blog post. Be sure to check back next week for another educational poker hand. Good luck in your games!