I did notice that flaw in my logic before I posted. I was hoping our BS in Math or a probability expert would enlighten. At what point does it break down? Because it seems like your chances of being dealt AA over a two hand sample instead of 1 would be about double.

This seems to be the answer....too tired to digest it tonight:

https://math.stackexchange.com/quest...ultiple-trials

In layman's terms, the probability of something happening equals "1 minus the probability of it NOT happening". So, the odds of getting AA at least once in a specific two-hand sample equals 1 minus the odds of NOT getting aces in either hand. Or, 1 - ((220/221)^2) [with the '^' meaning you raise the 220/221 to the second power, i.e. you square it].

This can be expanded for a greater number of hands by substituting the '^2' with '^(number of hands)', but this is valid ONLY for a specific trial. In other words, we can compute the probability of whether we will get aces at least once over the next 10 hands or 221 hands, or however many hands we like. For 10 hands, the probability that we will get AA at least once is 0.0443385, or roughly 4.4%.

As Illiterat noted, this doesn't apply across a session, which will include multiple ten-handed sequences (hands 1 through 10, 2 through 11, 3 through 12, etc.), which makes the OP's question more dynamic and problematic to address. In some ways, it reminds me of a conversation had between Joe Hachem and Dutch Boyd years ago when they were heads-up for a WSOP bracelet. There were four hands where Hachem was all-in and ahead, and he won all four. He was probably a 60-70% favorite in each hand. Boyd lamented "I can't believe you won all four of those!" Hachem countered with "I was a favorite in each one" and Boyd responded with something along the lines of "Yeah, so you had a 70% to win each one, but less than 25% chance to win all four."

One other math-nit item to point out. If the probability that something will happen is 1 in 221, then the odds against it happening are 220:1, NOT 221:1. A relatively easy way to remember this is to think of the payouts on roulette. "Even money bets" have probabilities of occurrence just less than 1 in 2 (to account for the 0 and 00), and pay at 1:1. Similarly, the First 12, Second 12, Third 12 have rough probabilities of 1 in 3 (again, slightly less due to the 0 and 00), and pay 2:1.

Hoping this helps, and hasn't confused further.

Super helpful. Thanks!

The probability for an event to occur at least one time in N independent trials is exactly 1-(1-P)^N where P is the probability the event occurs in an individual trial.

It turns out that if P is quite small and N is also not too big then
1-(1-P)^N is very close to P*N.

By the way:
This is a consequence of the binomial theorem which gives a formula for (1-P)^N in terms of the powers of P and binomial coefficients (see “binomial theorem on Wikipedia)

The approximation “1-(1-P)^N is very close to P*N” is valid only when P*N is quite small (much less than 1). If P*N is around 0.5 for example then the error in the approximation will be quite bad.

For example consider the odds of being dealt aces at least once in 110 hands.

You would guess this to be P*N = (1/221)*110 = 0.4977 = 49.77% … using the approximation.

But the true probability is 1-(1-P)^N = 1-(220/221)^110 = 0.3297 = 32.97%

You see the error is quite substantial here as P*N is around 0.5 and not much smaller than 1.

Contrast this with the same calculation over N=10 trials and you will see the error is much smaller, for then the approximate probability P*N = 0.045.. = 4.5% is very small. The true probability is 1-(220/221)^10 = 0.044.. = 4.4%, close to the approximation


Tl;dr the approximation works well only when the estimated probability P*N is small.

Edit: and yeah, if N=2 then the probability is about double…but not exactly double. The true probability is 1-(1-P)^2 and if you expand that out using FOIL you get 1-(1-2P+P^2)=2P-P^2. Your guess is 2P, which is very close to the true answer if P is small (because P^2 is very very small if P is small).

Yes and no. Yes there are multiple sequences, but OP "froze" (or chose) one and for that specific sequence of N hands, given the known all-in preflop percentages, there's a resulting combined probability for these N hands and this is that combined probability that OP wanted to know.

I believe that's what the OP thought he wanted to know, but that the actual answer is what has been pointed out. This isn't 9th grade math class. Part of solving problems is to figure out what is being asked. Yes, things can be rare if you only have one trial, but I guarantee someone is going to win MegaMillions sooner or later.

A tongue-in-cheek argument could also be made that the probability of something happening that has already happened is 1. Another tongue-in cheek argument could be made that it's 50/50. Either it happens or it doesn't. (I jokingly use this alot at the poker table, when someone asks "Wow...what were the odds?")

Also, did anyone ever clarify with OP whether he meant the SAME opponent for all five all-ins, or rather that each all-in was with just one villain, and not multi-way? If OP was a favorite with the percentages given, it wouldn't matter whether it was multiway or heads-up if all action has been closed (i.e. all remaining players are all in except the biggest stack). However, if there was a possibility that future action could fold out one of the Vs, then OP's win percentage would need to be adjusted (likely upwards, unless the remaining V gained more equity from the subsequent card than the folding V had).

Well, we don't know for sure because OP hasn't come back yet to clarify, but I still believe that's what he truly wanted to know unless specified otherwise.

Answer is "who cares"? Poker can be brutal. Every player experiences some sort of "how can this even be real" doomswitch event. Toughen up, try to learn from it and move on is the only sensible approach.

Yes, that is what he wanted to know. It's just not what he needed to know.

I agree with the last 2 posts.

That is what I wanted to know. There were, of course, many hands in between these all ins. I wanted to know the odds of losing 5 all-ins as huge favorite each time.

Thx for the detailed responses.