Calculating Poker Odds Charts

[rhughes1138]

I've been looking at Sklansky's odds chart and trying to derive the numbers, but I can't seem to do it.

Say I'm at the flop and I need to draw a six to win. That means I've got four outs with 47 cards in the deck. The probability of getting it on the turn and river would therefore be 4/47 and 4/46 respectively, right? This should give a probability of 17.2% (4/47 + 4/46). I even went back and checked my stats book and the math seens right. However, Sklansky's odds chart says that four outs have a 16.5% chance of drawing on both the turn and river. From where comes my discrepancy? Any help is greatly appreciated.

Thanks in Advance,
Ryan

[Koller]

1- 43/47 * 42/46 = 16.5%

[DeucesUp]

Your calculation double counts the times you hit on both the turn and river, you only need one or the other.

Koller shows one way to correct for this (1 minus the odds of NOT hitting on either street). The other alternate form is:

(odds of hitting on the turn) + (odds of hitting on the river) - (odds of hitting on both turn & river).

or:

4/47 + 4/46 - (4/47)*(4/46)

[TStoneMBD]

use the 4 and 2 rule. if youre on the flop with 9 outs do 9x4 for 36%. if you have 9 outs on the turn 9x2 for 18%. the 4 and 2 rule is accurate but off by a few % points. once you get up to like 13 outs tho the rule doesnt work well.

[JoshuaD]

[ QUOTE ]
use the 4 and 2 rule. if youre on the flop with 9 outs do 9x4 for 36%. if you have 9 outs on the turn 9x2 for 18%. the 4 and 2 rule is accurate but off by a few % points. once you get up to like 13 outs tho the rule doesnt work well.

[/ QUOTE ]

The 4 and 2 rule is almost exactly accurate at 5 and 6 outs. Anything below that the actual chance is a little higher, anything above that the actual chance is a little lower. The farther you get from that "balance point", the larger the gap between the true value and the 2/4 value.

So if you've got 20 outs, you're probably about 75% to hit by the river. (the 2/4 method would suggest 80%)

[BruceZ]

[ QUOTE ]
Your calculation double counts the times you hit on both the turn and river, you only need one or the other.

Koller shows one way to correct for this (1 minus the odds of NOT hitting on either street). The other alternate form is:

(odds of hitting on the turn) + (odds of hitting on the river) - (odds of hitting on both turn & river).

or:

4/47 + 4/46 - (4/47)*(4/46)

[/ QUOTE ]

You are making two errors in evaluating these terms, and these two errors exactly cancel, so that your final equation is correct, but a different form than you intended. Your last term for the probability of hitting on both the turn and river should be (4/47)*(3/46). Then, the first two terms should be 4/47 + 4/47, since the probability for the turn and river are each 4/47 before either card is dealt. So all together you want 4/47 + 4/47 - (4/47)*(3/46). These cancelling errors were recently discussed on the probability forum, except in this case they are more clear.

[rhughes1138]

This makes a lot of sense. I knew I was missing some mutual combination. Thanks, everybody.

One other thing for BruceZ, though. Why do the first and second terms, 4/47 + 4/47, both assume 47 possibilities while the third term, (4/47)*(3/46), assume 47 and then 46 possibilities? Your method is precise with Sklansky's chart, but I just don't understand how the river has 47 outcomes when a card on the flop would bring the number of possibilities down to 46.

Thanks Again,
Ryan

[BruceZ]

[ QUOTE ]
This makes a lot of sense. I knew I was missing some mutual combination. Thanks, everybody.

One other thing for BruceZ, though. Why do the first and second terms, 4/47 + 4/47, both assume 47 possibilities while the third term, (4/47)*(3/46), assume 47 and then 46 possibilities? Your method is precise with Sklansky's chart, but I just don't understand how the river has 47 outcomes when a card on the flop would bring the number of possibilities down to 46.

Thanks Again,
Ryan

[/ QUOTE ]

P(A or B) = P(A) + P(B) - P(A and B).

In our case, let A be the event that the turn card is one of the 4 outs, and B be the event that the river card is one of the 4 outs. The probability for each card is 4/47 before either card is dealt. If we simply add these probabilities, P(A) + P(B) = 4/47 + 4/47, then we double count the times that it happens on both, so we must subtract off the probability that it happens on both, which is P(A and B). Now P(hitting on both) = P(hitting on turn)*P(hitting on river GIVEN that we hit on turn) = (4/47) * (3/46). Note that we cannot simply multiply P(hitting on the turn)*P(hitting on the river) = (4/47)*(4/47) since these events are not independent. That is, it is not true that P(A and B) = P(A)*P(B), but it is generally true that P(A and B) = P(A)*P(B|A) ,where P(B|A) means the probability of B GIVEN that A has occurred. The probability of hitting on the river must be evaluated under the condition that we hit on the turn, so we need to multiply by the conditional probability P(B|A) = 3/46. So in general, we are doing:

P(A or B) = P(A) + P(B) - P(A)*P(B|A).

Yet another form is 4/47 + (43/47)*(4/46). That is P(hitting on the turn) + P(missing the turn)*P(hitting the river).

[PE101]

Good post.

Anybody that would puzzle over this problem this much will enjoy this book:
'Hold'em Odds Book' at Amazon (http://www.amazon.com/exec/obidos/tg/detail/-/0968122302/qid=1098460165/sr=8-2/ref=sr_8_xs_ap_i2_xgl14/102-1850103-3060121?v=glance&s=books&n=507846)
by Mike Petriv

You won't find this at your local book store (unless you live in Vegas).

You can also get it at Gambler's Book Store (http://www.gamblersbook.com/)
Enjoy

[DeucesUp]

[ QUOTE ]
You are making two errors in evaluating these terms, and these two errors exactly cancel, so that your final equation is correct, but a different form than you intended. Your last term for the probability of hitting on both the turn and river should be (4/47)*(3/46). Then, the first two terms should be 4/47 + 4/47, since the probability for the turn and river are each 4/47 before either card is dealt. So all together you want 4/47 + 4/47 - (4/47)*(3/46). These cancelling errors were recently discussed on the probability forum, except in this case they are more clear.

[/ QUOTE ]

I don't believe this is a case of canceling errors, this is simply a different method of evaluating the odds which looks at the problem from a slightly different temporal perspective.

I think you agree that the only way to make the hand is to hit a 4-outer on either the turn or river. And the general odds of hitting a 4-outer on the turn are 4/47 and the general odds of hitting a 4-outer on the river are 4/46. Thus the odds of one of these two events occuring are:

4/47 + 4/46 - (4/47)*(4/46)

I chose my terms from the perspective of having 1 of 2 independent events occuring (either hitting a 4-outer on the turn or hitting a 4-outer on the river). You choose to look from the perspective that these non-independent events will occur before either card is dealt which, as you've said, can be written as:

4/47 + 4/47 - (4/47)*(3/46)
or
4/47 + 4/46 * 43/47
or
1 - (43/47)*(42/46)
representing 1 - (odds of missing twice)

Any of these are fine, but your assertion that my equation is wrong because of "self-canceling errors" and that it is just some sort of cosmic coincidence that it gives the correct answer is wrong. This is legitimate and logical way to compute the odds. Though it scares me to death to question the guru, I believe that my equation is correct and my description of the terms is also correct with the caveat that I should have been more specific that when I say


[ QUOTE ]
(odds of hitting on the turn) + (odds of hitting on the river) - (odds of hitting on both turn & river)

[/ QUOTE ]

I am looking at the two different ways to make the hand, not the odds of these events occuring in succession before the turn card is dealt. I probably should have said:

(odds of hitting a 4-outer on the turn) + (odds of hitting a 4-outer on the river) - (odds of hitting a 4-outer on both the turn & river)

[gaming_mouse]

[ QUOTE ]
(odds of hitting a 4-outer on the turn) + (odds of hitting a 4-outer on the river) - (odds of hitting a 4-outer on both the turn & river)


[/ QUOTE ]

But the original question is the chance that you hit one of 4 cards on the turn or river. You are not guaranteed to have a 4-outer on the river, so this this conception doesn't make sense.

I must admit, though, that I find it strange that you just happened to get the right answer. Barring a better explanation, though, I think Bruce is right and that this method is not justified.

I'd be curious to hear more arguments tho.

gm

[jason_t]

This principle is called the "principle of inclusion/exclusion" and that is what the OP should look up in his stats book if he wants to understand it further.

[DeucesUp]

[ QUOTE ]
But the original question is the chance that you hit one of 4 cards on the turn or river. You are not guaranteed to have a 4-outer on the river, so this this conception doesn't make sense.

I must admit, though, that I find it strange that you just happened to get the right answer. Barring a better explanation, though, I think Bruce is right and that this method is not justified.

[/ QUOTE ]

Well, admittedly I hate to disagree with Bruce because I forsee having to eat crow in the near future, but I believe I am correct.

You only care about the times when you DO have a 4-outer on the river, because that's the only way to make your hand on the river. If you've only got 3 outs on the river, you already made your hand on the turn. This is not an "event" that we care about, it is just an irrelevant "branch" of the case where we hit on the turn and is already incorporated into the equation.

In this case, I simply don't see how it can be wrong to look at the only two events which have any relevance, hitting a 4-outer on the turn or a 4-outer on the river, and calculate the odds of one or the other occuring. It doesn't matter that the 2nd event only occurs when the 1st misses (because its an either situation), as long as the events we consider represent every relevant possibility. And 100% of the time, you can only make your hand by hitting a 4-outer on one street or the other.


Boy, this is WAY too long a post for what is really a simple concept. But the difference in the equations really comes down to semantics. Bruce's equation can most accurately be described as the odds that the current order of the cards in the deck will make your hand (odds that the 1st card is an out + odds second card is an out - odds that both the 1st and 2nd are outs). While mine describes the odds of the different situations where you make your hand.

[gaming_mouse]

[ QUOTE ]
You only care about the times when you DO have a 4-outer on the river, because that's the only way to make your hand on the river. If you've only got 3 outs on the river, you already made your hand on the turn. This is not an "event" that we care about, it is just an irrelevant "branch" of the case where we hit on the turn and is already incorporated into the equation.


[/ QUOTE ]

But if you are ignoring the case when you have a 3-outer -- that is, the case when you hit the turn first -- then what are you subtracting off? the " - " term of the equation is to subtract off the event of hitting both the turn and the river. But you are saying that you are already accounting for that. I still think your logic is not adding up....

gm

[DeucesUp]

[ QUOTE ]
But if you are ignoring the case when you have a 3-outer -- that is, the case when you hit the turn first -- then what are you subtracting off? the " - " term of the equation is to subtract off the event of hitting both the turn and the river. But you are saying that you are already accounting for that. I still think your logic is not adding up....

[/ QUOTE ]

OK, as simple as I can make it: I am simply calculating, using the most basic form, the odds of either event A occuring or event B occuring. Odds of Event A occurring are 4/47, odds of event B occurring are 4/46.


I will leave it up to you to decide whether this applies to this situation. I contend it does, because event A and event B are the ONLY way you can make your hand.

[gaming_mouse]

[ QUOTE ]
odds of event B occurring are 4/46.

[/ QUOTE ]

No. Event B is the chance of one of your 4 outs falling on the river. Given that you have seen only the flop and not the turn, the odds of this event are 4/47.

[DeucesUp]

[ QUOTE ]

No. Event B is the chance of one of your 4 outs falling on the river. Given that you have seen only the flop and not the turn, the odds of this event are 4/47.

[/ QUOTE ]

No. Ignore the original question for a moment. Consider another example of John and Jane sitting at two different tables, John trying to catch a 4-outer on the turn (he will fold before the river card) and Jane trying to catch a 4-outer on the river (the turn has already been dealt). Clearly my equation is correct for the odds that either John or Jane will make their hand because these are completely seperate and independent events, indeed it's the standard and most general form of the equation for the probability of either of two independent events occuring. The only reasonable way to write John's odds is 4/47 and Jane's 4/46. Thus either John or Jane will catch:
4/47 + 4/46 - (4/47)*(4/46)

Now let's make the events conditionally non-independent. Assume that if John catches on the turn, some magical force makes Jane's odds a bizillion-to-1 of catching on the river, what are the odds that either John or Jane make their hand? Exactly the same of course, if the odds only change when John catches, we don't care that Jane's odds changed.

I contend that this is exactly the same for a single player in a single hand. Sure when he catches on the turn his odds change as he's only got 3 "outs" on the river (analogous to the magical force /images/graemlins/wink.gif ), but we don't care. His odds could be 3/46, 1/1000 or 1/2, doesn't matter he made it on the turn. If he misses the turn, his odds on the river are EXACTLY 4/46, there is no other possibility. Thus there are two possible ways to make the hand and they occur 4/47 and 4/46.


Again, different perspective/methodology. I'm not calculating the odds of the cards coming a certain way starting from the flop. I contend that the odds of success in this situation can be completely characterized by the two events I have defined - hitting on the turn or hitting on the river when the turn misses. These are the only two events which make the hand. If you are to find a flaw in my argument, you must explain why these events do not fully characterize the odds of making the hand or why these events can not be treated independently.

[gaming_mouse]

I contend that the odds of success in this situation can be completely characterized by the two events I have defined - hitting on the turn or hitting on the river when the turn misses.

If this is how you define the two events, then your subtraction term is incorrect. The term you subtract off is the chance that events A & B both happen. In this case, it is impossible for them to both happen at the same time. Or are you somehow claiming that you are not using inclusion-exclusion here? If so, I don't follow. If you are using it, however, you are applying it incorrectly.

On a slightly different, I want to clarify, once more, my previous statement, because I'm not positive you understood it:

Event B is the chance of one of your 4 outs falling on the river. Given that you have seen only the flop and not the turn, the odds of this event are 4/47.

Be assured that this is the correct probability of hitting one of your outs on the river, when we have not yet seen what the turn card is.

HTH,
gm