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Thunder

10-19-2007, 02:51 AM Post subject: Is The Maths Correct?

#1 (permalink)

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Hi all,

It's been a while since my last post so.....

There I was, noodling around, sucking up as much poker info as I could when I chanced upon a pot odds article by Phil Gordon with a quiz at the end. I was glad to note I got every single one right except for the very last question. And it still doesn't make sense and so I'm hoping for vindication

The scenario: The pot has $1,000 in it after the turn. You've got a straight and the best hand possible. You are absolutely certain that your opponent, Howard Lederer, has a flush draw. Only the river card is yet to come and both of you have very deep stacks. How much is the minimum you can bet to give Howard insufficient odds to chase his flush?

I know that with 9 cards to improve his hands, the odds are 19% and that Phil likes to work on an 'outs multiplied by 2' ratio to get 18%. So I round it up to 20% (or 1 in 5) and conclude that a bet of $200 would be enough. The actual answer came out as being any bet greater than $282 and not $180 - $200 that the maths dictated.

On reflection, I realised that my bet of 1/5 actually increased Howard's odds but still, $282 seemed too low a bet. Howard needs better odds than 1 in 5 (which is 4 to 1; which is 20%) in order to call. **Anything numerically higher than 1 in 5 (4 to 1) or less than 20% is providing odds to call. A bet of $282 makes the pot $1282 thus giving Howard odds of 1 in 5.5. Or 4.5 to 1. Or 18%. This should be enough for Howard to call, not fold.

Even a bet of $300 fails to provide insufficient odds as it provides odds of 1 in 5.3 (4.3 to 1) or 23%. As Howard needs to get less than 1 in 5 - which is less than 4 to 1 but greater than 20% - the required bet would need to be nearer $400 as this equates to 1 in 4.5 = 3.5 to 1 = 22%.

And even then, it's still close enough for Howard to call so I don't see how the supplied answer of $282 can possibly be correct.

I know it's not real life. I just like to make sure my calculations tally.

Thx

** I now 1 in 5 is higher than 1 in 7 and that 4 to 1 is higher than 9 to 1. That is why I mentioned "numerically higher" as inthis respect, 1 in 7 is higher than 1 in 5 and 9 to 1 is higher than 4 to 1.

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taipan168

10-19-2007, 05:42 AM

#2 (permalink)

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I actually disagree with both of you, I think the answer is $347

Running this through Pokerstove, Howard would be 20.455% to win on the river since he has exactly 9 outs:

Code:

Board: 9c Tc Jd 4h
Dead:  

	equity 	win 	tie 	      pots won 	pots tied	
Hand 0: 	79.545%  	79.55% 	00.00% 	            35 	        0.00   { KsQd }
Hand 1: 	20.455%  	20.45% 	00.00% 	             9 	        0.00   { Ac2c }

20.455% to win means that Howard is (1/20.455%)-1 = 3.89 to 1 against to win, so the minimum amount you need to bet is such that he is getting worse than 3.89 to 1 pot odds. This break even amount is $346 ($1346/$346) = 3.89.

Therefore, the minimum amount you need to bet to give Howard incorrect odds is $347.

jyms

10-19-2007, 06:47 AM

#3 (permalink)

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I will let you figure this out yourself in hopes you "get it".

When you bet, what is the pot?
When he calls, how much is the pot?
What are his odds to "call" not your odds to bet.

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Jibalob

10-19-2007, 08:22 AM

#4 (permalink)

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Quote:

Originally Posted by Trainer_jyms

I will let you figure this out yourself in hopes you "get it".

When you bet, what is the pot?
When he calls, how much is the pot?
What are his odds to "call" not your odds to bet.

He's still laying 4.55:1 odds though???
Or maybe I need more sleep?

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Thunder

10-19-2007, 09:53 AM

#5 (permalink)

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Trainer,

Q: When you bet, what is the pot?
A: 1282
Q: When he calls, how much is the pot?
A: 1564
Q: What are his odds to "call" not your odds to bet.
A: He has to call 282 to win 1282 = 4.5 to 1. He has to call 282 to win a total of 1564 = 1 in 5.5. Circa 18%.

These are the sums I ran through before posting and I keep getting the same answers. Gordon uses the rule of 2 and 4 to reach an 18% chance of success and thus this formula leads to odds 1 in 5.5 or 4.5 to 1 and these are higher than 1 in 5 (4 to 1).

So in this scenario, I need Howard to get unfavourable odds to call and so anything around 4.5 to 1 is going to be in his range. He needs worse odds than 1 in 5.5 (4.5 to 1) such as 1 in 4 (3:1 or 25%).

Maybe I just don't get it?

Taipan, as this was a "work it out in your head" quiz and because I am following Gordon's own formula for accuracy I need the answer to correlate to his line of thinking.

Jibalob

10-19-2007, 10:24 AM

#6 (permalink)

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I agree with thunder on this one.

The answer in the book must be a mis-print

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bigslikk

10-19-2007, 05:51 PM

#7 (permalink)

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[quote="taipan168"]I actually disagree with both of you, I think the answer is $347

/quote]

I got the same exact answer, working it out on paper before reading any responses.

I like to think of pot odds as thus: A flush draw comes once every three times, thus you need to be getting three to one (the *final* pot, which includes the last street's pot, plus your opponent's bet, and your eventual call, must be three times larger than the price to call in order to break even).

Example: If an opponent puts you allin by betting the pot and you're on a flush draw, you call.

The pot is x. Opponent's bet is x. Thus you're getting:

x (the original pot) + x (opponent's current wager which increases the pot) + x (your eventual addition to the pot by calling) for the price of x (your addiction to the pot by calling. 3x:1x or 3:1. My mindet may be weird /confusing, but it's the easiest for me, because all I need to do is use the rule of two / rule of four (muliply the outs by these magic numbers) to get percentages. Then, find the percent that the call price is to the pot. You know the rest now I'll just stop my setup. Anyway, my thinking boiled this problem down to this equation:

(x) / (2x+1000) > 9/44

You want Lederer's price / pot ratio to be greater then the chance he'll will it. X is Lederer's price to call. 2x + 1000 is what he would rake in if he won (the 1000 in the middle, your bet, and Lederer's call).

9/44 is the odds that he'll catch the draw. Do the math (make the greater sign an equal sign) and the break-even price is $346 and change. Since $346 would barely price him in, $347 is the minimum bet needed to price him out. However, it's a pretty bad bet since its +EV is probably pennies. If it's the biggest that he'd call then I guess it's a good move.

Thunder

10-19-2007, 07:27 PM

#8 (permalink)

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BigSlikk,

You're right, your way is confusing and turns simple equasions into long winded ones

The most common way of calculating odds that I have seen is the manner as I illustrated and, most importantly, it is the same way that Gordon uses once he has number of outs.

Remember, the article contains his methodology and then the quiz to showcase how it works. So, if using Gordon's own method indicates that $282 is too low then how did he reach it?

Quote:

I like to think of pot odds as thus: A flush draw comes once every three times

With one card to come, the odds of making the flush are not 1/3 . However you do it, be it 9/44 or (9 x 2) +1, his odds should still be as indicated above: 1 in 5.5 - which is 4.5: 1 - which is 18%.

Quote:

which includes the last street's pot, plus your opponent's bet, and your eventual call

There is no bet to call. Your opponent has not made any bet.

Quote:

working it out on paper before reading any responses.....all I need to do is use the rule of two / rule of four to get percentages.

It is not my intention to be rude but if you had read my responses you would see that said rule has already been implemented, as well as breaking down why it shows that $282 doesn't fit.

As a side issue, If you use the rule of 2 and 4 to get your percentage I can't see why you overcomplicate the process. Simply convert that percentage into your preferred ratio format.

euphoricism

10-19-2007, 10:20 PM

#9 (permalink)

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Just memorize 4.5:1 sheesh ;p

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taipan168

10-19-2007, 11:11 PM

#10 (permalink)

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Quote:

Originally Posted by bigslikk

Anyway, my thinking boiled this problem down to this equation:

(x) / (2x+1000) > 9/44

OMG I need to brush up on my maths, it's been 20 years since I've done any of this stuff...

bigslikk

10-21-2007, 11:01 PM

#11 (permalink)

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The problem for me is this: I've learned to think in these terms: If your draw is 1 in 3 to come, you need 3 for 1. (a final pot three times as large as the price). For a 1/4 you need 4 for 1. For 1 in x you need x for every 1. See the easy pattern? Unfortunately, this makes calculations a bitch.

Okay, I will *attempt* to use the easier ratio / horsetrack method.

bigslikk

10-22-2007, 12:04 AM

#12 (permalink)

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Quote:

Originally Posted by Thunder

BigSlikk,

You're right, your way is confusing and turns simple equasions into long winded ones

My method wasn't so long winded as it was very heavily explained. My post was needlessly long, but the method wasn't really.

My method, which provided a solvable equation of

(x) / (2x+1000) > 9/44

in which x is the bet necesssary to deny Lederer proper odds, was merely another version of taipan's:

Quote:

Originally Posted by taipan168

This break even amount is $346. ($1346/$346) = 3.89.

taipan's math can be rewritten as:

(1000 +x) / (x) < (44/9 - 1), or (1000 + 346) / (346) < (4.89 - 1)

Which, if you add one to both sides, and invert (this results in the arrow pointing the opposite way) is the same equation that I used.

I'm hesistant to see how using 3.89 to 1 is easier than (4.89 total, per $1 in price paid). However, since everyone in the poker world seems to use it, it must be.

Also, Gordon's answer of 282 is correct, if you use his dumbed down rule of 2. I like those rules, they are what I use at the table. However, his shortcut figured Lederer's chances too low; Gordon believed Lederer had an 18% to chance to catch whereas really he had a 20.5% chance. He underestimated Lederer's chances and consequently bet lower than he should have. The 2.5% oversight is signficant when there's a thousand bucks in the middle.

But according to the rule of 2, Gordon's correct. According to the laws of the universe, taipan and I are correct. If I had unlimited time at the table, and if it were socially acceptable to actually work out math like this at the table, I would choose my (our) method over Gordon's. But since I care what people think of me, I will continue to follow the advice from his Little Green Book. I recommend that you do too. He knows his shit.

Robb

10-22-2007, 08:55 PM Post subject: Re: Is The Maths Correct?

#13 (permalink)

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Quote:

Originally Posted by Thunder

I know that with 9 cards to improve his hands, the odds are 19% and that Phil likes to work on an 'outs multiplied by 2' ratio to get 18%. So I round it up to 20% (or 1 in 5) and conclude that a bet of $200 would be enough.

This is one of the most misunderstand points of probability vs. odds. If your probability is EXACTLY 20%, you need to bet $333 to give him break even odds. If you're 1 in 5, your odds (against) are 4:1.

You play this 5 times. Each time, you bet $333. Each time, he calls. He's getting 4:1 on his bet. Four times, he catches air, and you win 4 x $333 = $1,333. The one time he hits his draw, he makes $1,333, breaking even.

This is easier from hero's perspective. If hero has a fractional probability of hitting the hand that is 1/3, hero needs to get 2:1 pot odds.

In practice, I like to bet at least half the pot (in the example, $500). Why? This goes back to your suited connector thread. He might have a combo draw. And I play fish, not Lederer. So I want to make sure that they don't have anywhere near the correct pot odds to call, even if they have a few extra outs. Note that even with a Flush/Inside Straight Draw and 12 outs, he would still be only even money. I try to bet 50% of the pot miminum against draws. At my levels, I get assinine calls ALL the time.

Rule of thumb on the river: always bet AT LEAST one third of the pot against draws, and AT LEAST 1/2 the pot if you think it might be a combo draw. And bet the whole pot if you're against a fish who will call it.

Robb

10-22-2007, 09:10 PM

#14 (permalink)

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Quote:

Originally Posted by bigslikk

Also, Gordon's answer of 282 is correct, if you use his dumbed down rule of 2. I like those rules, they are what I use at the table. However, his shortcut figured Lederer's chances too low.

This is why I would bet MORE than $250, or $282, or whatever I thought his break even point was, even just rounding up to $300 or $350. The point isn't get into a coin flip. The point is to get more money in the middle ONLY if I'm ahead or have better odds than my opponent.

And if you overbet, he folds. You win $1,000. Also, the Rules of 2 and 4 slightly overestimate YOUR pot odds, when you're playing from Lederer's perspective. So they won't get you in trouble when used for their intended purpose. When you're playing against an unknown draw, big a bit bigger than they suggest.

Thunder

10-22-2007, 09:30 PM

#15 (permalink)

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I too would bet big in this situ but (at the risk of soundign numb) I cannot see how the $282 or anything much less than $400 would be the minimum.

Hoever we each work out the odds (and I too love Gordon's rule of 2 and 4) I cannot see how it adds up.

My point being, with just 9 cards left (18%chance) and with $1282 in the pot, calling $282 to win $1564 is providing him with odds of 1 in 5.5 which is 4.5 to 1.

If you go over my posts, and if you could *please* use the formula/logic there, you'll have a much better chance of getting through to me. When I see your stream of equasions I slip into a coma.

taipan168

10-23-2007, 12:54 AM

#16 (permalink)

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Quote:

Originally Posted by Thunder

My point being, with just 9 cards left (18%chance) and with $1282 in the pot, calling $282 to win $1564 is providing him with odds of 1 in 5.5 which is 4.5 to 1.

If you go over my posts, and if you could *please* use the formula/logic there, you'll have a much better chance of getting through to me. When I see your stream of equasions I slip into a coma.

OK, if you use the rule of 2 and 4:

- If you estimate that Howard is 18% to win then he is 4.56 to 1 against to hit on the river, so any bet has to be large enough to offer him worse than 4.56 to 1 on the call. This amount is $281 ($1281/$281) = 4.56 to 1. So if you use the rule of 4 and 2, Phil Gordon's calculation of $282 is correct.

- If you round up to 20% then Howard is 4 to 1 against to hit on the river, so the corresponding bet is $333 ($1333/$333) = 4 to 1, so the minimum bet would need to be $334.

Robb

10-23-2007, 03:11 AM

#17 (permalink)

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Quote:

Originally Posted by Thunder

My point being, with just 9 cards left (18%chance) and with $1282 in the pot, calling $282 to win $1564 is providing him with odds of 1 in 5.5 which is 4.5 to 1.

He's not winning $1,564. He's betting $282 to win $1,282. That's 4.5 (in the pot) to 1 (his bet). You're counting his wager in his winnings.

Again, his probability is 1 in 5.5, or 1/5.5. But his odds are 4.5 to 1.

Suppose there's $100 in the pot, villain bets the pot, and I have EXACTLY 1/3 chance of making the best hand with one card to come. Play it out three times.
1. I call $100, making the pot $300. Lose my $100 bet.
2. I call $100, making the pot $300. Lose my $100 bet.
3. I call $100, making the pot $300. Win $200 on my bet.

I've made a $100 bet three times, lost it twice, but won $200 once. Breakeven. This is 2 to 1 odds against me making my hand, since I win once and lose twice in every three rounds, on average.

Probability -- Odds against (of hitting my hand)
1/4 -- 3 to 1
1/5 -- 4 to 1
1/6 -- 5 to 1

And so on...

bigslikk

10-23-2007, 05:25 PM

#18 (permalink)

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Quote:

Originally Posted by Thunder

My point being, with just 9 cards left (18%chance) and with $1282 in the pot, calling $282 to win $1564 is providing him with odds of 1 in 5.5 which is 4.5 to 1.

You said it yourself. At this price, if Lederer calls, he has an 18% chance to win 1564, in which he is returned his 282 call and profits a net of 1282. He has an 18% chance to net 1282. He is 82% to lose his 282 call.

18% of 1282 is 231. 82% of 282 is 231. In the long run, he breaks even.

In other words, Lederer nets a profit which is 4.5 times greater than his risk, once every 4.5 trials (or twice every nine trials, for clarity's sake). 1 in 4.5 = 18%. He will break even in the long run if he makes this call.

Thus a bet of $282 is the MINIMUM to give him break-even / bad odds. Of course we want to bet more than 282, so that we may *profit*. How much depends on how often you'll think he calls, given the price.

spoonitnow

10-23-2007, 08:04 PM

#19 (permalink)

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$333 giving him 4:1 i think without thinking hard

45 cards that we don't know (52 - 4 hole cards - 4 board cards) shit that's 44 left nm.

9 outs to 44 cards is 1 to 3.888 with 8 repeating (w/o use of calculator ldo) so wtfever 1000/3.888 is.

Quote:

Originally Posted by Ripptyde

I only have 2 simple rules when I am coaching a new student.

Rule # 1: don't ask questions

Rule # 2: don't ask questions

I have no interest in discussing strategy with a protege'. Your job is to remain quiet and listen. I have a very systematic approach that I will share with the right candidate and I promise that I will turn you into a force of nature and show you elements of the game of poker that you never knew existed.

Thunder

10-29-2007, 01:29 AM

#20 (permalink)

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Hi guys,

Thx for replying. I’ve been away for a few days and not had time to respond.

Taipan, okay, technically, it is correct…….but to a margin of 0.01% and to me, that is break even money. Which is why I kept getting odds that priced Howard in.

Although I fail to see how anyone would quibble over 0.01% (and in this case a mere $1) I do understand that on a pedantic level, it is correct. I thought that at least 5 to 1 would be required to put Lederer off and is why I personally assumed, off the top of my head, it would require a bet closer to $400.

However, just as I conceded to your breakdown, Bigslikk says it’s $231 as opposed to $281 and I’d like to know how there is a $50 discrepancy.

Robb,

Yeah, I understand what you say about 3 to 1 being ¼ and is something I mentioned in my posts. I prefer to look at the overall pot – including your bet – ($282 to win $1564) as this provides the “in” ratio rather than the “to” ratio. In my mind, it’s easier to 1/5 rather than view the more “separate” 4 to 1. ‘Cause overall, there’s 5 portions and you have one of them. And that’s why I rprefer to view it as betting $282 to win $1564 overall instead of calling $282 to win the $1282 already in the pot.

Horses for courses, I guess.

Robb

10-29-2007, 01:38 AM

#21 (permalink)

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Quote:

Originally Posted by Thunder

Robb,

Yeah, I understand what you say about 3 to 1 being ¼ and is something I mentioned in my posts. I prefer to look at the overall pot – including your bet – ($282 to win $1564) as this provides the “in” ratio rather than the “to” ratio. In my mind, it’s easier to 1/5 rather than view the more “separate” 4 to 1. ‘Cause overall, there’s 5 portions and you have one of them. And that’s why I rprefer to view it as betting $282 to win $1564 overall instead of calling $282 to win the $1282 already in the pot.

Horses for courses, I guess.

Just FYI, very few folks write this way. There is a common lingo regarding pot odds using the ratio method rather the fraction method you prefer. Folks won't understand your posts when you use it.

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