River card syndrom (cross-posted in psychology)

[PrayingMantis]

I'm revisiting this old thread (from the psychology forum), to share some new thoughts in the subject. (The subject was why people tend to see the river card as having more "powers" than any other board card in HE, and why they feel they're often get outdrawn on it. There were some very interesting replies on the psychology forum, specifically by PDosterM)

Assuming player A holds AA, and player B holds 22. They are both all-in pre-flop. 22 will outdraw AA a little less than 20% of the time.

I'm doing here a really rough calculation, *without redraws*, not to complicate it. So, I assume 22 beat AA by hitting his set card.

Now, looking backwards, on what card of the board was it more probable that player B would hit his set? (I'm calculating the outs from a 48-cards-deck to begin with, since we are accounted for 4 known cards)

First card: 2/48 = ~0.04167

Second card: 2/47 = ~0.04255

Third card: 2/46 = ~0.04347

Turn card: 2/45 = ~0.04444

River Card: 2/44 = ~0.04545

It is clear that the chances player B outdrew player A by hitting his set *on the river*, are bigger than the chances he hit it on any other specific board card. Not by much, but the difference is there. I assume this kind of small difference will be there for any draw, against any made hand.

This is a very simplistic calculation, but I believe it sheds some interesting light on this "river card syndrom".

Any thoughts? Especially regarding the mathematic side of this problem?

[Gonzoman]

[ QUOTE ]
I'm revisiting this old thread (from the psychology forum), to share some new thoughts in the subject. (The subject was why people tend to see the river card as having more "powers" than any other board card in HE, and why they feel they're often get outdrawn on it. There were some very interesting replies on the psychology forum, specifically by PDosterM)

Assuming player A holds AA, and player B holds 22. They are both all-in pre-flop. 22 will outdraw AA a little less than 20% of the time.

I'm doing here a really rough calculation, *without redraws*, not to complicate it. So, I assume 22 beat AA by hitting his set card.

Now, looking backwards, on what card of the board was it more probable that player B would hit his set? (I'm calculating the outs from a 48-cards-deck to begin with, since we are accounted for 4 known cards)

First card: 2/48 = ~0.04167

Second card: 2/47 = ~0.04255

Third card: 2/46 = ~0.04347

Turn card: 2/45 = ~0.04444

River Card: 2/44 = ~0.04545

It is clear that the chances player B outdrew player A by hitting his set *on the river*, are bigger than the chances he hit it on any other specific board card. Not by much, but the difference is there. I assume this kind of small difference will be there for any draw, against any made hand.

This is a very simplistic calculation, but I believe it sheds some interesting light on this "river card syndrom".

Any thoughts? Especially regarding the mathematic side of this problem?


[/ QUOTE ]

The math here is off. The probability of 22 hitting a set on the 2nd card (which implies that it didn't hit the set on the 1st card) is
( (1 - p(hitting set on 1st card) ) * (2/47)), or ( (46/48) * (2/47) = about .04078

3rd card = (46/48) * (45/47) * (2/46) = .039989

4th card = (46/48) * (45/47) * (44/46) * (2/45) = .03900

River card = (46/48) * (45/47) * (44/46) * (43/45) * (2/44) = .03812

What you are really asking is: Given that one player has AA and the other has 22, what is the probability that the first of the remaining 2's comes at the 1st, 2nd... card.

[PrayingMantis]

[ QUOTE ]
The math here is off. The probability of 22 hitting a set on the 2nd card (which implies that it didn't hit the set on the 1st card) is
( (1 - p(hitting set on 1st card) ) * (2/47)), or ( (46/48) * (2/47) = about .04078

3rd card = (46/48) * (45/47) * (2/46) = .039989

4th card = (46/48) * (45/47) * (44/46) * (2/45) = .03900

River card = (46/48) * (45/47) * (44/46) * (43/45) * (2/44) = .03812

What you are really asking is: Given that one player has AA and the other has 22, what is the probability that the first of the remaining 2's comes at the 1st, 2nd... card.



[/ QUOTE ]

Thanks for the reply. I know that my math was off, if you look at it in the way you took it. But I think I'm asking a different question than what you suggest. That's why I didn't took in calculation the probablity of the occurance happening (or not happening) on the former streets. I was trying to look at each street for its own, only I'm not sure I did it right, and how one should actually do it. I didn't try to solve it, or to get the right probablity, but only to see, in a way, what is "more probable".

I will rephrase my question (or actually phrase it for the first time):


Player A had AA, player B had 22. They were both all-in pre-flop. You *know* player B outdrew player A by hitting his set on one of the streets. What is the probablity he hit his set on any specific street?

[Gonzoman]

[ QUOTE ]

Thanks for the reply. I know that my math was off, if you look at it in the way you took it. But I think I'm asking a different question than what you suggest. That's why I didn't took in calculation the probablity of the occurance happening on the former streets. I was trying to look at each street for its own, only I'm not sure I did it right, and how one should actually do it. I didn't try to solve it, or to get the right probablity, but only to see, in a way, what is "more probable".

I will rephrase my question (or actually phrase it for the first time):


Player A had AA, player B had 22. They were both all-in pre-flop. You *know* player B outdrew player A by hitting his set on one of the streets. What is the probablity he hit his set on any specific street?



[/ QUOTE ]

Each street is equally likely. You assume one and only one of the 2's will be on the board of 5 cards. This problem is equivalent to having 5 cards in your hand 1 of which is a 2. Shuffle them out and turn them over face up. The 2 has an equal probability of being in each of the 5 streets.

[PrayingMantis]

[ QUOTE ]

Each street is equally likely. You assume one and only one of the 2's will be on the board of 5 cards. This problem is equivalent to having 5 cards in your hand 1 of which is a 2. Shuffle them out and turn them over face up. The 2 has an equal probability of being in each of the 5 streets.

[/ QUOTE ]

Yes, this looks quite obvious and simple. However, It clearly contradicts your first answer, in which you showed (and rigtly so) that the probability of the 2 falling on the first card is greater than on the second, and on the second it's greater than on the third , and so on (this is if we don't KNOW the 2 already hit and player B outrew player A...).

How do you settle this?

[Gonzoman]

[ QUOTE ]

Yes, this looks quite obvious and simple. However, It clearly contradicts your first answer, in which you showed (and rigtly so) that the probability of the 2 falling on the first card is greater than on the second, and on the second it's greater than on the third , and so on (this is if we don't KNOW the 2 already hit and player B outrew player A...).

How do you settle this?

[/ QUOTE ]

Of course, the two questions you've posed are different. A question more related to the first situation goes like this:

Assume at least one 2 is on the board. What is the probability that a 2 exists at slot i, but not any j < i, 0 < j <= 5.

Notice here that the probabilites here will decrease from 1 to 5 since any time a 2 comes at the first card, it will always be the first 2 on the board. However, if a 2 comes as the 2nd card, there is a slight chance that the other 2 already came at the first card. So the probability that you will be beaten by the 2nd card is slightly smaller than the first. This is the reason the probabilities decrease in my first answer.

However, suppose there is only 1 card in the deck which can make the underdog lose.

P(hitting the card in the 1st card) = (1/48)
P(2nd card) = (47/48) * (1/47) = (1/48)
P(3rd card) = (47/48) * (46/47) * (1/46) = (1/48).
.
.
.

So we come to the riveting result that a particular card is equally likely to show up on each street. [img]/images/graemlins/smirk.gif[/img]

[PrayingMantis]

After the short discussion with Gonzoman (above), I would like to restate what I said in the original post, and clarify it. This has to do with psychology and probability alike.

The situation is: Player A holds AA, Player B holds 22, they are both all-in PF. (I'm discounting redraws and other draws).

Player B has 2/48 (0.04166) prob. to hit his set on the first card.

If he *did not* hit it, he will have now a 2/47 (0.04255) probablity to hit on the 2nd card. *In his view*, and rightly so, this is a greater probablity than he had to hit it on the first card, *alone*.

If he does not hit it on th 2nd, he *now* has 2/46 (0.0435) prob. to do it on the 3rd card and so on.

When he'll be facing the river card (if he didn't make he's set until then), he'll have a specific 2/44 (0.04545) probability of hitting the set NOW. This is a greater *specific* probability to hit his set than on any other specific street.


We are very used to see these problems, in a way, from an "objective" point of view. In this view, as Gonzoman stated, the chances to hit the set are getting *smaller* as more cards fall. But this is an "external" view of probability. It is improtant to notice, IMO, that from an "internal" view, things can look quite different (if not opposite).

I find this concept of "internal probabilty" (that's the best name I can think of it now), very intriguing, since it has much to do with how we deal with chance in general. It is obviously very relevant to poker.

I believe that behaviours like the "river card syndrom" (I apologize for this fancy title, but can't think of any better option), can be explained very thoroughly with such a tool.

Any more thoughts will be greatly appriciated.

[PrayingMantis]

More discussion and clarification of this idea, in the psychology forum:

internal probability

[Lexander]

The results you posted in my mind are pretty trivial in general (but useful to keep in mind). The chances that the river cards will hit the opponent are higher at that point if we assume that none of the previous cards have done so (which is required for your calculation).

I think the more interesting thing is that the river has three properties that make losing at the river more frustrating:

#1) You can't catch to win if you are behind at the river.
#2) You have generally put a lot of money in the pot that you have now lost.
#3) Many times your opponent made a poor call on the turn and got lucky.

[PrayingMantis]

[ QUOTE ]
The results you posted in my mind are pretty trivial in general (but useful to keep in mind). The chances that the river cards will hit the opponent are higher at that point if we assume that none of the previous cards have done so (which is required for your calculation).

I think the more interesting thing is that the river has three properties that make losing at the river more frustrating:

#1) You can't catch to win if you are behind at the river.
#2) You have generally put a lot of money in the pot that you have now lost.
#3) Many times your opponent made a poor call on the turn and got lucky.


[/ QUOTE ]


I do agree my calculaions and results are trivial. However, these by themselves were not the point of this thread, here and in the psychology forum. I wanted to show how we can solve a seemingly conflict between the mathematical ("chances are equal for any card to bring the suckout, or even: chances are smallest for the river to bring it"), and the very common psychological point of view, which many posts here, on various forums, are an evidence for its existance ("river suckouts are more frequent than they sould be on XYZ poker site").


I believe the 3 points you give do not really solve it.

1. The fact that you can't catch to improve after you're behind on the river is correct, but I think that the "river syndrom" will appear regardless, even if you are not able to improve (redraw) at all, no matter on what street, i.e., any situation where you'll have 0 outs for a redraw. Conditional probability calculations ("internal prob.") will be much more helpful in explaining the river suckout feeling, IMO.

2. The fact that you have put lots of money and you've now lost, is a variation on your first point. You have now lost, i.e, you cannot catch and improve any more.

3. That's true, but is not really relevant to most aspects of the syndrom, as I described it. I'm refering basically to a scenario where both opponents are all-in PF (or at least on the flop) and are dealt few community cards for them to make hands with. In the case where a bad call was made on the turn, there's no difference between conditional and non-conditional probabilities (interior vs. exterior), because no matter how you look at it, there's only one more card to come.