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Poker Video: Misc/Other by bellatrix (Micro/Small Stakes)

Bellatrix,

PLEASE don't drop the additions like G-bucks and Sklansky bucks. The fact that some may struggle with them doesn't work for me: the fact is, under that argument, you simply wouldn't do the course.

It's unreasonable to expect Bella to answer every single question that comes up. If that were the case, she is likely to find this series equivalent to a fulltime job, and unlikely to complete it. I would suggest that people step forward act as teaching assistants. In fact, we should probably start a SecretHQ group.

I won't start it, since I don't have questions yet, but I'll join and assist if someone else does.

But it might be worthwhile if Bella could do a Tooltime video of installing Auctex and producing a simple Latex file. The average DC member probably won't have heard about Latex.

I won't start it, since I don't have questions yet, but I'll join and assist if someone else does.

Scratch that. I had a question, so the SecretHQ group is at http://www.deucescracked.com/forums/28-Secret-HQ/topics/365481-Maths-Attacks-Study-Group

Don't know what to say about someone that doesn't feel comfortable talking probabilities. Sort of shocking, but hey we all have our strengths and weaknesses. Sounds like the use of EXAMPLES was a big winner and deviating from the book or not was less of the issue.

Statistics is pretty simple once you're able to relate the concepts to something meaningful to you. I would definitely start youtubing some statistics stuff or googling terminology if I didn't have a grasp of basic stuff already... I can't fathom playing poker seriously without having this understanding. I'm sure there's even full courses online available for free about statistics.

For question 2b... how many green spaces are on the wheel? I've heard this isn't always standard

For question 2b... how many green spaces are on the wheel? I've heard this isn't always standard

European roulette has one zero, American roulette has two (zero and double zero).

You should probably assume American.

I'll send you solutions for both

Time Link to 00:22:42

It should be 8/44 because we can exclude two cards which the villain holds

It should be 8/44 because we can exclude two cards which the villain holds

Argh, I don't remember exactly what I said now.

I think at some point I said "you knew" he had an overpair. In that case - sure. But if you just "know" that hitting your draw will make you win the pot, then you can't do that, since one of the 8 cards could well be in opponent's hand.

Bellatrix.. I have a question about the example of pot odds where we had an open ended str8 draw.

We are on the turn obviously,(46 cards left unseen) Opponent goes all in for $20 into an all ready $100 pot. Wouldn't this actually make out pot odds 6 to 1 ? 120 into 20 = 6/1

Then you go on to say something about 38/8. This is where I'm confused.

there are 8 cards in the deck that can make our str8 and there are 46 cards unseen. This would make it 46/8 or 5.75% to 1. Is this not correct? So where does the 38/8 come into play. We cannot subtract our 8 outs from the 46 to figure are correct chances of hitting our draw can we?

So if my calculations are correct. We are being offered 6 to 1 pot odds. Our odds of hitting our str8 are 5.75 to 1. So this makes a call profitable. Correct ?

Yes, as Soepgroente pointed out, I didn't word myself correctly (even if he was wrong on his odds ;-) )

We are getting 6 to 1 odds. 120 to 20. Those are the pot odds.
For the call to be profitable we need better than 38 to 8 or 4.75 to 1 odds (not 5.75).

If the pot before was 80$ instead of 100$ in our example, the pot odds would be 5:1 and a call would still be profitable. The expected value of that call is still positive, even though it's much smaller than in the example (~87cents).

Bellatrix.. I have a question about the example of pot odds where we had an open ended str8 draw.

We are on the turn obviously,(46 cards left unseen) Opponent goes all in for $20 into an all ready $100 pot. Wouldn't this actually make out pot odds 6 to 1 ? 120 into 20 = 6/1

Then you go on to say something about 38/8. This is where I'm confused.

there are 8 cards in the deck that can make our str8 and there are 46 cards unseen. This would make it 46/8 or 5.75% to 1. Is this not correct? So where does the 38/8 come into play. We cannot subtract our 8 outs from the 46 to figure are correct chances of hitting our draw can we?

So if my calculations are correct. We are being offered 6 to 1 pot odds. Our odds of hitting our str8 are 5.75 to 1. So this makes a call profitable. Correct ?

The expression of odds is dealt with in first couple of episodes of WiltOnTilt's Maths of No Limit Holdem series. It's clear that you aren't understanding odds, and I'd strongly recommend watching at least the early episodes of that series as background for this. It's also the case that the whole of that series is some of the best video on the site, but not all of it is directly helpful for this video.

Your question is about a notation difference. "38 losses to 8 wins" or "38 to 8" expresses the same idea as 8/46. I'm sure you simply made a typo with "5.75% to 1", but when expressing odds, you set losses against wins, so 38/8 calculates the number of losses if we want to express them "x:1".

You are correct about the 6:1 pot odds, but our odds of hitting the straight are (38/8):1 = 4.75:1.

Funny you mention the series of Mathematics of NLHE by WoT. This is a great series and I was just watching it when this same situation came up.

I now understand that when we have 8/46, we must subtract our 8 outs from the 46 remaining cards in order to make the correct ratio. Which is 8/38 or 4.75/1

Thanks to Bella and SlowJoe for your very fast responses.

Also thanks to everyone at Deuces Cracked. This is an amazing site and because of it I pretty much have a lock on every game in my home town and some casinos near by as well.

Keep the good stuff coming. Ill be a member for life!

4 because of 4 suits
13 choose 2. The number of combinations within 13 cards (all of one suit).
Now we have to divide by the number of all probabilities of having 2 cards (52 choose 2).

I agree that for just one suited cards it seems like much easier to write 12/51, but it might be easier to get youself used to the combinatorics. Especially, when you start counting combinations later to put people on a range (e.g. is it more likely that he has a flush draw or that he has a TP+ on the turn...).

I was thinking about other ways to compute this probability on my way to work today. 12 / 51 is obviously the simplest way. The 4 * (13 choose 2) / (52 choose 2) is straight by definition of the probability. How about the following (admittedly useless) way:
P(not a pocket pair) * 0.25 = (16/17) * (1/4) = 4/17, as expected.
The 0.25 is the ratio of suited to all the combos of 2 card (e.g. KQ: 4 KQs of 16 KQ total). P(not a pocket pair) = 1 - P(pocket pair) = 1 - 3/51 = 1 - 1/17 = 16/17. Alternatively, P(pocket pair) = 13 * (4 choose 2) / (52 choose 2) = 1/17. I find it interesting that for every pair that contains say 2d, there is 16 non-pair combos with 2d. Just to visualize: we could map
2d2h to the set of remaining 12 harts + 4 diamonds;
2d2c to the set of remaining 12 clubs + 4 other diamonds;
2d2s to the set of remaining 12 spades + 4 remaining diamonds

I was thinking about other ways to compute this probability on my way to work today. 12 / 51 is obviously the simplest way. The 4 * (13 choose 2) / (52 choose 2) is straight by definition of the probability. How about the following (admittedly useless) way:
P(not a pocket pair) * 0.25 = (16/17) * (1/4) = 4/17, as expected.
The 0.25 is the ratio of suited to all the combos of 2 card (e.g. KQ: 4 KQs of 16 KQ total). P(not a pocket pair) = 1 - P(pocket pair) = 1 - 3/51 = 1 - 1/17 = 16/17. Alternatively, P(pocket pair) = 13 * (4 choose 2) / (52 choose 2) = 1/17. I find it interesting that for every pair that contains say 2d, there is 16 non-pair combos with 2d. Just to visualize: we could map
2d2h to the set of remaining 12 harts + 4 diamonds;
2d2c to the set of remaining 12 clubs + 4 other diamonds;
2d2s to the set of remaining 12 spades + 4 remaining diamonds

Wow, nice. Took me a while to understand, but now that I do, I like it!
Just be careful not to crash while thinking these things if you drive yourself to work ;-)

Ok, I just re-read Chapter One, and I'm pretty sure that NOWHERE in it, does it bother to explain that monstrousity in your bonus question! Nor do they bother defining/explaining the terms used in other similar math formulas in their own book! This REALLY frustrates me to no end!! So, I take it you have to have a degree in math to be in the know here, or else research all the arcane symbols as best you can, and try to make sense of them! WHY would they not include these definitions in their book? Like a glossary, perhaps? I mean, if you are going to TEACH something, then TEACH it! Grrr....!!

Edit: So, I can't possibly "prove" it, since I can't even decipher what it effing MEANS!