Now let’s consider a real life example. Say that our opponent raises to $30 and I reraise to $90 with 5‐5 and he goes all in for $1,000. This is a classic “way behind or even” situation and it’s clear a fold in most situations. Here then is a simple math concept put into words that shows what the math obviously proves. But let’s say he would do this only with A‐K, K‐K, or A‐A and consider the EV calculations involved. A‐A and K‐K are effectively the same if we go all in vs. 5‐5 so instead of analyzing them individually we can put the hands into different groups – A‐K and the overpair groups. If we think he has A‐K 50% of the time and a high pocket pair 50% of the time the EV equation is:
(1/2)(1/2($1,090) – (1/2)($910)) + (1/2)((4/5)(–$910) + (1/5)($1,090)) = –$210
This equation is slightly more complicated than the previous one as it’s split up into two sections. The first side is a representation of what happens the 50% of the time the opponent has A‐K, and the second side a representation of when he has A‐A or K‐K. When he does have A‐K vs. our 5‐5 we will win roughly 50% of the time, so we win $1,090 and the other 50% of the time we lose so that means we lose $910.
Then on the other side of the equation when the 5‐5 goes all in vs. a higher pocket pair we will lose roughly 4/5 of the time, and lose the $910 we put into the pot to call. And the other 1/5 the time we win the $1,090 in the pot. Add all of those together and the result is an expected value of around –$210 which is very significant (although still surprisingly low). The fact that we are risking $910 to win $1,090 makes a little difference, and also the fact that we can outdraw a big pair with the 5‐5 helps. Also obviously the fact that half the time we are even money makes a huge difference. So, if we compare calling with the 5‐5 to our alternative plays, here there is only one more which is folding. The EV of that play is 0, and the EV of calling is $210, so we save ourselves $210 by exercising discipline and folding.
Now let’s make this representation even more accurate. In a real situation he isn’t going to have A‐A or K‐K half the time and A‐K the other half the time. There are more ways to make A‐K than there are A‐A and K‐K combined – there are 16 combinations of A‐K and 12 of A‐A and K‐K in total. So the odds he has A‐K will be 16/28 vs. the chances he has A‐A or K‐K, which will be 12/28. So the equation becomes:
(16/28)(1/2($1,090)-(1/2)($910)) + (12/28)((4/5)(-$910)+(1/5)($1,090)) = -$167
So here things start to come back in our favour a little. But now let’s say that although he will get A‐A and K‐K 42% of the time versus the 58% of the time he gets A‐K, he might not always play this way with A‐A or K‐K and always play this way with A‐K. An opponent might make this play 7/8 times with A‐K because he just wants to get all‐in instead of playing post‐flop, whereas playing A‐A or K‐K is easier post‐flop because they are already strong hands, so he might want to trap as much as 50% of the time. In this case the equa‐ tion changes dramatically and now we must weight the hand combinations and then put them in the EV equation:
0.5*12=6wayshe’llplayA-AorK-Kthisway 0.875 * 16 = 14 ways he’ll play A-K this way.
So now there is a 6/20 chance he has A‐A or K‐K and a 14/20 chance he has A‐K, which makes the equation:
(14/20)(1/2($1,090) – (1/2)($910)) + (6/20)((4/5)(–$910) + (1/5)($1,090)) = –$90
So clearly unless the circumstances change dramatically this will al‐ ways be a ‐$EV play. However, there are still a lot of factors that can be changed in this or other models. For instance, maybe the oppo‐ nents could have an underpair and then there will be three groups of hands. Again these would need weighting as it’s unlikely each group would have a probability of 1/3 – it’s probably a 50% chance he has overcards, 10% chance underpair, and 40% overpair.
Or as we have already seen you can change his stack size (and hence his bet size), or add in more hand combinations like Q‐Q or J‐J in which case an overpair becomes a lot more likely and a call a lot worse. Or maybe he does this with just A‐K or A‐Q and J‐J because he likes to trap with his really big hands, and just goes all‐in pre‐flop with hands he is afraid to play post‐flop. A‐K and A‐Q have a total of 32 combinations vs. 6 possibilities for J‐J so if he always plays those three hands in exactly this manner it weights heavily towards a call.
This is the underlying math involved in the game and it takes place in every single hand. This is a relatively simple problem to figure out because the opponent has gone all‐in, there are no future streets and no more action – just try doing an EV calculation on the flop with middle pair and adding all the possible scenarios such as the possibility you are ahead, behind, the times he bluffs you when you are ahead, the times he gives you a free card when you are behind and you have implied odds and so on. It’s impossible.
That’s why – instead of doing the math everytime – we use concepts to guide us through these situations. For instance if we are either way behind or even money that is an excellent reason to fold. But there are other concepts too that can be important – so here very good pot odds would counter the first concept. It’s useful to do the equations yourself, simplified down to the basic concepts involved in a hand, and then to play around with the numbers to see which concepts outweigh others and so on.
