Why an Ace always comes...

[Masquerade]

The scenario. Late in a tourney. It's folded to the button who allins. The small blind folds and you have 88. Correctly putting him on overcards, probably a big Ace, you call and he flips up. AJo. You know you're a 55-45 fav but an Ace duly shows up and you're out. But you got your money in as the favourite ... or did you?

The odds of at least one Ace showing up in the next five cards, given that you've seen four cards, is easy to calculate: 28.6%. (The same for a Jack of course). But the odds of at least an Ace or a Jack are almost exactly 50%. [The probabilities arent linearly additive because there is some overlap.] Of course not all these cases win as the 8s can improve as well and you need to use an odds calculator but it gives the rough picture.

Now we are going to make an extreme assumption. We are going to model a weak online NL player as one who plays ANY ace. Obviously this isnt totally realistic but let's see show that affects the analysis. Let's also assume a 10 player table.

As none of the other 8 players entered the pot we can assume that the three remaining Aces must all be in the 32 remaining cards. This raises the probability of at least one Ace showing up by river to an incredible 41%. When we correctly account for the Jack as well, the probability of AJ catching goes from 50% to 60%, and overall becomes approximately a 55-45 favourite over 88!

These figures of course reflect the extreme assumption that as noone else played, no-one had an Ace. But that's not SO far off the mark, particularly in the later stages of a tournament when any ace is likely to be played into an unopened pot - particularly at lower limits. Do we need to rethink tournament strategy accordingly? And is this why an Ace always seems to come, busting our pair?