I just sat back and listened last night. A fellow, P1, was complaining because he always runs bad and that led to a discussion of odds.
There was lots said back and forth that I didn't really pay attention to, but then I hear P2 say something approximating...
>>If you have AA v KK you're an 80/20 favorite. That' means if you play that out five times, you have a 100% chance of losing once.<<

I may not be all that bright, but I do know two things, P2 was wrong and that there is nothing to be gained from getting involved in that sort of conversation.

It's been a loooooooooong time since I took a stats class, and I've pretty much forgotten all of it. But their convo got me to thinking, what are the odds of losing multiple hands if you are (let's use their numbers) an 80/20 fave?
I think I have to multiply .8 ^ 2 = .64 * 100 = 64% to give me the chance I win an 80/20 bet twice. If that's correct then the chance of winning all of 5 repetitions of that bet are .8 ^ 5 = a 33% chance. Is that correct?
And the reverse side of that is the chance of losing twice when you are the 80/20 fave is .20 ^ 2 = .04 or 4%?


Do I have that right or was their a second idiot at the table last night?

Your probability of winning five hands in a row as an 80/20 favorite is .8^5= .328 or just under 1/3. The probability that you will lose at least one of the five is therefore 67.2%, so the player who stated you will certainly lose one of the five is wrong. You are more likely to lose at least one hand than you are to win all five, but it is by no means a certainty.

"there" dagnabit

Btw the calculation above is your probability of losing at least one hand. The probability of losing exactly p hands is a bit more complex. For example if p=2, the probability is a bit over 20%. Your calculation would be the probability of losing 2 out of 2 hands, not 2 out of 5.

First, the probability of any sequence of two losses and three wins is .8^3 * .2^2 = 2.04%. The question then is how many sequences of two losses and three wins are there? Fortunately there’s a standard math formula for this. It’s called the binomial coefficient. If there are n hands and p losses the binomial coefficient is equal to n!/[p!(n-p)!]. In this case the binomial coefficient is 5! / (3! 2!) = 10. We therefore multiply the probability of each sequence by the binomial coefficient to find that the probability of losing two hands is 20.4%.

If I am playing it’s 100%.

came to say same

my pony is also slow

tree-fiddy

If the question is 'What are the odds that someone at the poker table does not really understand probability and odds?' the answer is 4 out of 3.

Just because someone is wrong doesn't make them an idiot. Absolutely everyone is wrong about something, well not me obviously, but everyone else.

Also, and I'm not saying this is the case here, but sometimes decent players will say things that are absolute nonsense so that they sound as though they don't understand poker e.g. last table I sat, two of the reggiest players were commenting about 'hearts are hitting' or 'lot of paired boards today'. I'm sure that this wasn't going to alter their play in the slightest, but they know it's the type of thing that bad players say.

My friend told me that if there are 100 idiot poker players and 100 excellent players, then given a random opponent, the probability you are playing in a heads-up game against an idiot player is 50%.

Well, my friend is wrong, because almost all the time I’m playing against an excellent player.

For a lot of people, the odds that there is an idiot at their table is 100%.