a razz odds question

appears · #1 10-04-2007, 03:56 PM

I've read in more than one place that the odds of outdrawing a single opponent on 4th st. is 4-1 . Can someone show me the math? I can't seem to come up with that number. Also, if this number is correct, does this mean that on any given street, if you are getting 4-1 pot odds, you need to be calling if you feel you are at most only one card behind? Two examples : 1) defending the bring-in against a single opponent showing a baby upcard when you have a hand like A2/K ; 2) you have caught bad on fourth while your opponent appears to have caught good ... say A2/72 v xx/A2 . (Assume in both examples that the cards that are out don't have any particular sway.)

I also guess it's worth mentioning that I read through this whole lengthy post on defending the bring-in but wasn't exactly sure what to draw from its conclusions. I assume that the odds of outdrawing someone on 4th st. are pretty relevant to answering that question. But I'll hold off on any follow-up questions until someone responds and sets me on the right path.

RustyBrooks · #2 10-04-2007, 04:07 PM

Say you have AT8 and your opponent has A68. You know your 3 cards, your opponent's up-card and 6 other up cards, for a total of 10, so there are 42 unknown cards.

You have 24 cards to improve - so you'll improve 57% of the time.
Your opponent needs to catch J K Q or 8 - we'll ignore his pair cards for a moment. So there are 14 bad cards for him, he'll catch bad 33% of the time.

So for him to catch bad, and you to catch good is 33% * 57% = 20% or in this case 4:1.

This ignore subtleties like, is a T a bad card for him if you catch good? Your board will be better but your hands could be about equal so that's a slight advantage. Also, there are more cards for him that are bad but you don't know what they are (his pair cards) and he doesn't know what cards pair you, but I tend to think that sort of evens out.

As the streets progess the numbers change a bit because a) there are fewer cards to come and b) there are more known cards. Being "one card ahead" is worth more the later the street it is, and also, how low your current hand is matters a lot. On earlier streets a primary concern is not only what you and your opponent's current hand is, but what the next best hand you're drawing to is. So there's a big difference between A2T9 and A2T6 and a tremendous difference between A2T69 and A2T65 etc. The more likely that you can not only outdraw your opponent's current hand, but ALSO outdraw what he's drawing to, the better.

RustyBrooks · #3 10-04-2007, 04:13 PM

Note that on 3rd your chances are worse if he has a much lower 3 card hand than your 2nd-lowest card. This is a close example because he has a 3 card 8 and you're drawing to a 3 card 8. If he has a 3 card 6, like A56 and you have the same AT8 then you will still "outdraw" him 1 in 5 times but a lot of those times he'll call anyway, like if you have AT8 and he has A56 and he draws a T and you draw a 7, he'll have A56T vs your xx87, not only are you not *actually* ahead because of your T, but he might call you.

Realistically in the 3rd st case you are making a call as an underdog with the hopes of picking it up outright or becoming a favorite, in this example you have fewer chances of becoming an outright favorite and fewer chances of winning the pot outright. The better his 3 card hand appears to be, and the worse your 2nd low card is, the trickier it's going to be from 4th on.

Praxising · #4 10-04-2007, 04:23 PM

[ QUOTE ]
So for him to catch bad, and you to catch good is 33% * 57% = 20% or in this case 4:1.

[/ QUOTE ]
Rusty, isn't 80/20 4 to 1? Worst case here, rounding the 33 down to 30 and the 57 up to 60, is 60/30 or 2 to 1.

RustyBrooks · #5 10-04-2007, 04:25 PM

When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

so 20% of the time it does happen and 80% of the time it doesn't, so 80:20 or 4:1

Praxising · #6 10-04-2007, 04:42 PM

[ QUOTE ]
When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

[/ QUOTE ] To outdraw him, I only need him to catch worse, as in he can stay the same, get a brick, or a worse card than my draw. No? And, I'm not sure these are considered independent events, as one event affects the outcome of the other. (He gets my good card, so I have one less good card to get or vice versa depending on who goes first.)

RustyBrooks · #7 10-04-2007, 04:42 PM

The tendencies of your opponent are SUPER important here. If your opponent will rigorously fold if he bricks, then you are correct to call with ANY baby face up if you're getting better than 4:1 immediate odds from the pot.

However, also consider that the pot is small, and you are unlikely to be getting MUCH better than 4:1 from the pot, and if you are then he'll call more liberally on 4th. So this is realistically worth only a fraction of a small bet every time you try it.

Certainly it makes you look looser than you are though.

RustyBrooks · #8 10-04-2007, 04:50 PM

[ QUOTE ]
[ QUOTE ]
When there are two independent events, you multiply the probabilities together. You need to catch good, .57, and he needs to catch bad, .33, so BOTH of those happening happens .33*.57=.20

[/ QUOTE ] To outdraw him, I only need him to catch worse, as in he can stay the same, get a brick, or a worse card than my draw. No? And, I'm not sure these are considered independent events, as one event affects the outcome of the other. (He gets my good card, so I have one less good card to get or vice versa depending on who goes first.)

[/ QUOTE ]

Yes of course, if you catch a good one, he has one less good one to catch, the effect is minimal enough to consider independent (about 1/42 or a little more than 2%)

What cards is he going to draw to "stay the same" in your opinion?

In order for you to have a better hand than him on 4th, he has to get a T or worse, period, and you have to get a 9 or better - except that a 9 is not really that good so I'm not going to include it. If you draw a 9 and have T98A vs his A68T, you are not winning, and he is not folding. A 9 is only good for you if he catches really bad and maybe even then it's not great.

My math is correct - we're arguing really over what constitutes a situation good enough to warrant calling 3rd, and it's well established that we disagree on that.

RustyBrooks · #9 10-04-2007, 04:51 PM

I'm at the tables at the moment but I'll post some graphs later about this.

appears · #10 10-04-2007, 05:31 PM

1.) Why didn't you count 9s and Ts as bad cards? On 4th st. if you have 26/A8 and your opponent shows xx/A9, you've technically outdrawn your opponent but he probably isn't going anywhere in the hand ; but if you omit 9s and Ts as bad cards, then doesn't that also completely omit from the equation those instances where you might connect with xx/A2 v xx/AT and induce a fold there?

I suppose part of the problem is that I'm not entirely sure what the odds need to be on 4th to continue if I catch a 9 or a T and my opponent catches a baby card. If you can enlighten me on this subject also, I would be much obliged.

2.)The way you did the math, all six upcards must be 9s or Ts because you didn't factor them into the equation. But how should six random upcards be factored in?