Pots Odds Question!?

First and foremost, I like to say that I'm a new member of the CardsChat community. I've only began playing poker with real money for almost a year. I've lost almost 2000 of my own money so I decided to take a break and learn poker theoretically. I subscribed to DeucesCracked only shorty and I've been watching wiltontilt math videos. On episode 2, he gave an example on pot odds.

2:1 dog

The pot is a 1000 $ and you have 500 $ to call to win. If you win once, that's plus 1000, but if you lose twice -500 x 2 = -1000.

Assuming this is a heads up match and when you make the call for 500, the total pot will be 1500. Altogether I've put forth 750. If I called the 500 and lost twice, wouldn't I of lost 250 in the 3 games?

500 of the pot before, 500 of his raise and my 500 call (1500).
-500 twice (-1000) added to his/ her 750 money (the one profitable game/ call)- might of been confusing.
I'm still 250 down.

He said 2:1 dog is break-even and any better and I'll win money, any lower and I'll lose money.

But I see 2:1 dog a still losing act.

Can someone please explain this to me. I don't know if he made a mistake or if I'm not just understanding it. Thanks.

Man your explanation confused me even more so than I normally am.

This should explain your pot odds question to you in a fashion in which anyone can easily understand


https://www.cardschat.com/odds-for-dummies.php

I think what you're missing is that whenever you calculate pot odds, you don't include what you've already put in the pot. Once money is in the pot, it's not yours anymore. You either win the pot or you don't, but the chips in the middle are gone for all extents and purposes. If you fold at this point, you've already lost the 250, so you will be -250 either way. Therefore it's removed from the pot odds calculation.

wanderingthehall said:

I think what you're missing is that whenever you calculate pot odds, you don't include what you've already put in the pot. Once money is in the pot, it's not yours anymore. You either win the pot or you don't, but the chips in the middle are gone for all extents and purposes. If you fold at this point, you've already lost the 250, so you will be -250 either way. Therefore it's removed from the pot odds calculation.

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Yes, This is where you are going wrong. The money already in the pot is not yours to lose.
You are calculating what is in the pot compared with what you have to further contribute 1000:500

wanderingthehall said:

I think what you're missing is that whenever you calculate pot odds, you don't include what you've already put in the pot. Once money is in the pot, it's not yours anymore. You either win the pot or you don't, but the chips in the middle are gone for all extents and purposes. If you fold at this point, you've already lost the 250, so you will be -250 either way. Therefore it's removed from the pot odds calculation.

Click to expand...

Thanks for clearing that up, but then why would he say it's a break even act if your going negative in the end result. Like if you call these however so many times, your bankroll would decrease substantially.

You're a 2:1 dog (based on odds of the cards), it'll cost you $500 to call, with a chance to win $1,000. So if you lose this twice, you lose $1,000 ($500x2, your two calling bets), and if you win once you get $1,000 (the money already in the pot). So basically if the money (the amount you have to put in vs. the amount you can win) is also 2:1, and those are your odds of winning the hand 2:1 (33% chance), you will break even with this play (You will lose twice and win once). If your chances are even better than 33% to win with this amount of money, then you should call and you will win money with this play in the long run. If your chance of winning is lower than 33% then you shouldn't call, because you won't win often enough and lose money in the long run (You will lose $500 more than twice compared to the $1,000 you will win once).

OpenArms said:

Thanks for clearing that up, but then why would he say it's a break even act if your going negative in the end result. Like if you call these however so many times, your bankroll would decrease substantially.

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If you fold, you end up -250. If you call and you're exactly a 2:1 dog, the EV on your call is 0, so on average, you end up -250.

But if you're even a bit less than a 2:1 dog, the EV on your call is +ve. Let's say it's +10. So if you fold, you end up -250, but if you call, you end up -240 on average.

Which is preferable, -250 or -240? Part of winning poker is losing less. Money you don't lose counts exactly the same in your BR as money you win.