Probability Skewing in Drawing Hands

Poker Strategy
1

I will start by saying this isn’t a game changer either way. It’s a minor statistical anomaly that I’ve been thinking about lately, and yes, I do have way too much time on my hands.

Let’s say your hole cards are both spades, and 2 more spades come on the flop. Conventional thinking goes something like this…

There are 13 spades, I know the location of 4 of them. This means 9 spades are still out. A deck contains 52 cards, and I can see the 3 in the flop, plus my 2 hole cards. This leaves 47 cards unaccounted for.

Since there are 47 cards remaining and 9 of them are spades, the next card will be a spade 19% of the time. ((9/47)X100)

But there’s another way to look at it…

There are 9 people at this table, so 18 hole cards have been dealt. Of these, 4.5 will be spades, on average. I have 2 spades, so 2.5 spades additional are in the initial deal. The flop contained 2 spades, so I know the location of 4 of them, and statistically, 2.5 more are in other people’s hole cards. This leaves 6.5 unaccounted for.

I know my hole cards and the 3 in the flop. Although I can’t see the other 16 hole cards, I can discount them because statistically, they will contain 2.5 spades. Thus there are 31 cards unaccounted for, and 6.5 of these are spades. The odds of the next card being a spade is 21%.

The difference is only 2%, but which method is more correct, and why?

Sklanski’s fundamental theorem of poker basically states, “If seeing your opponent’s hole cards would make you play differently, your initial play would be incorrect.”

If you knew for a fact that 2 or 3 of your flush cards were missing from the remaining deck, would you still take that flush draw? Does Sklanski’s theorem even apply here?

I dunno. What do you think?

2

Hi SPG,

I have always considered the opponents hands to be part of the pool of unknown cards. Their composition is the same as the remainder of the deck, barring outrageous reads on players who only play 100% suited cards or pairs, for instance.I think the flaw may be assuming there are 2.5 spades in 16 cards after 4 spades are already removed from a deck of 52.

We have 2 spades and the board has 3 cards, of which 2 are also spades meaning another random card anywhere is 9/47 = 19.1489% to be a spade.

Are you wondering if removing the 16 cards before or after we are given ours could make a difference, by the way?

I found two simplistic ways to calculate the number of spades missing if we remove 16 other random cards.Someone might know of a source to find the correct number but I ran some figures like this:

16 random cards taken from the 47 and nominated as opponents hole cards comes to to 16*19.1489% = 3.0638 spades we cannot draw to complete our flush.
16/47 = 34.042 % of the deck is gone and 34.042% of 9 is also 3.0638
Therefore,
3.06382 unobtainable spades + 2 hole cards + 2 on flop means 7.06 spades removed and 5.936 available in 31 cards still in the deck
5.936/31 = 19.1489% chance of the next card being a spade, or any card of the deck

Does that work?

Rob

3

I have no idea! My basic assumption was that, on average, 16 cards would contain 4 of each suit.

I guess what I’m wondering is why there is a 2% difference between methods.

By the way, this also works (doesn’t work?) with straight draws. Those 16 cards others have as hole cards will contain, on average, slightly more than one of each denomination, reducing your outs to an open ender from 8 to 6.

Oddy enough, this also increases your odds of hitting the straight, since 8 outs in 47 cards (after the flop) is about 17% and 6 outs in 31 cards is 19%.

If this “new” way of looking at it was correct, I’m sure it would have been discovered long ago. So I’m sure it’s wrong, but can’t for the life of me see why it’s wrong.

4

I think that the 16 cards do not contain 4 of each suit because there are already a number of spades accounted for. The 16 cards will have less than 4 spades on average. because we see 4 which cannot be among them
Going back to the ordering of the dealing, if the hands are all dealt at the same time it would appear the 16 cards would contain 4 spades. But I think that even if all the cards were dealt simultaneously and 4 of the 5 we see are spades, the average number of spades in the 16 would not be 4 because they would be 16 random cards of the 47 we cannot see and there are just 9 spades among them.
I, too seem to have more time on my hands today, lol.

5

Yes you do! Maybe we need more collusion to keep ya busy? :slight_smile:

My original post stated that there are 18 hole cards dealt, and that 4 1/2 of these will, on average, be spades. Knowing that 2 of these are in our hole speaks to distribution, not frequency. The remaining 16 cards should still contain 2.5 spades.

After the flop, there are 21 cards out, and 5.25 of these should be spades. Since we know the location of 4 of them, there is less of a chance someone else has spades, since only 1.25 additional spades should be out.

This would leave 31 cards to go, and 7.75 of them will be spades, on average.

Unfortunately, this now means that ((7.75/31)X100)=25%, or that the next card will make your flush 25% of the time!

All this is actually showing is that, when we see a statistically average number of spades in the first 21 cards, we will see a statistically average number of spades in the remaining cards as well. This makes both logical and intuative sense.

However, our bothersome 2% difference has now grown to an irritating 6% difference.

6

The 16 cards we put to one side still have the same chance of being spades as the 34 others remaining in the deck before the flop, unless we know something about them which does not apply to the 34.

If we expose 21 cards and choose two spades for our hole cards and two more for the flop, then yes, there will surely be a dearth of spades in the 16 other cards, or possibly not even enough to take 4 in the first place. Once we look at the other 16 cards, the probability of the top card of the 34 being our suit will be affected by the cards we remove from the deck.

If we deal 21 cards over and over until a deal happens where we get two suited cards and exactly two more appear on the flop, the 16 unknown cards are still calculated as part of the 47 unknown cards and are not drained of our suit.

7

Right, if we deal 21 random cards, 1/4 of them should be spades, at least on average. This works out to 5.25 spades, and with 2 in our hole and 2 on the flop, we should expect 1.25 other spades in the remaining 16 cards.

How often will these other 16 cards contain at least one spade?

I calculated it and came up with 98.5% of the time.

What this means is the idea of having 9 outs from that flop will be incorrect 98.5% of the time, and when the number of outs is wrong, probabilities calculated using 9 outs must also be wrong.

Do as you will, but I’m not going to use a system that’s incorrect 98.5% of the time.

PS: The traditional method says there are 9 spades in the remaining 47 cards, and this is true. The problem here is that 98.5% of the time, one or more of those 9 spades will be dead, and can’t be seen as outs. It’s not an out if it’s impossible to get.

8

I have only skimmed the other responses, but this seems like a case of Schrodinger’s cat or Schrodinger’s spades. Given no additional information, we have to consider that all 9 spades could still be in the deck, or not, based on the 47 cards unaccounted for. I guess I don’t see the difference between whether those spades are in the deck vs the in the hands other players have folded. It is impossible to know (unless another player is acting in a way that indicates that they have spades), so the location of those potential spades does not matter. It is possible that all 9 are in the deck or all 9 are in the hands of your opponents, so it does not provide additional information. Thus the Schrodinger’s cat reference, the spades simultaneously are in the deck and not in the deck until we know for sure where they are.

I completely appreciate your post and the deeper contemplation about probability in poker. I am interested in this idea in terms of how the hand actually plays out (i.e., if there are 8 players going to flop and one of them flats or check raises it is useful to consider the probability that one of the 8 players has two spades as well or is bluffing with the ace of spades or something like that). I haven’t had a chance to examine all of your math, but wouldn’t the proportion of spades per card that is dealt to other players be the same as the proportion of spades per card left in the deck, therefore the 9 outs would essentially apply the same way, but instead of 9 out of 47 it would be 6.07 out of 31 (i.e., the same probability).

9

The difference is that those in folded hands aren’t available to us, so aren’t outs.

And yeah, we can’t see the other hole cards, at least not physically, We can, however, see them statistically. At a full table, with 2 of a suit in my hand and 2 more of that suit on the board, there is a 98.5% chance that at least one of our outs is in the muck or in the opponent’s hand, and thus unavailable.

I was hoping you would respond to this Joe. You are far better at statistics (and poker) than I am.

My take away from this is that the chance of making that flush on the turn depends on how many people are at the table. The more people at the table, the bigger the chance that some of your outs are dead.

10

My first assumption was that, if we deal 18 hole cards and 3 flop cards, these 21 cards would, on average, contain 1/4 of any suit, say spades. This would be 5.25 spades out.

Of these, we know the location of 4, leaving 1.25 in the other 16 unseen cards. These 16 cards should contain at least 1 spade 98.5% of the time, so I don’t think it’s a bad assumption.

This leaves, on average, 7.75 spades in the remaining 31 undealt cards, and this is exactly a 25% probability of the next card being a spade.

11

I think your assumption may be flawed in that you have additional information beyond your original calculation about the number of spades dealt, specifically that two are in your hand and two are on the board, so you would expect more than 1.25 additional spades dealt (although the math to prove that might be complicated).

There are 9 spades unaccounted for, but we are assuming we have no additional info about whether they’re in the deck or dealt to others. So according to my math it’s likely that 3 others were dealt and 6 remain, but since we have no other info, any of the 9 could be dealt on the turn or river, so the probability would still be equivalent to 9/47. I’ll think more about this question. It reminds me of the Monty Hall problem (Google it), but in that case there is additional information given.

12

Well, yeah it must be or else this would be the way it’s done. With millions of players spanning over 100 years, someone thought of this already and had it disproven. I’m not expecting to win the Nobel Prize in poker, I realize I must be wrong.

Yet the math holds for me. There will be, on average, 5.25 spades in those first 21 cards, and the fact that I have 2 and the board has 2 speaks to distribution, not to frequency.

When I watch poker on TV, I calculate the probabilities on the fly and they almost never match the numbers they show on screen. This is because they know what cards have been folded and take this into account, and my “conventional” methods don’t. This “new” way makes a probabilistic guess at how many were folded or are out, and does seem to provide better results overall.

So many people think we see more flushes and straights than we should expect. The assumption is that our expectations are right, so the deal must be wrong, but what if the deal is right, so our assumptions must be wrong? Either would produce the same results.

And yeah, I know people are willing to play any 2 suited cards, willing to chase any straight no matter what, we see more multi-way pots, and that these factors make it more likely to see more hands made, but do these factors account for the whole difference?

I have yet to see any data to prove there is an actual difference, so this last bit is pure speculation. It’s just food for thought.

13

OK, I have boiled this down to one simple question…

The original situation was 9 players at a table, 2 spades in your hand, 2 on the flop.

Is it better to ignore the fact that the 16 remaining hole cards will contain at least 1 spade 98.5% of the time because that information isn’t perfect??

Or is it better to make a probabilistic estimation, even though that information can’t be perfect?

14

Hey SPG

I think I have figured out why you are getting 2 different answers for the same problem.

While you are correct in assuming that the other 16 cards have 2.5 spades on average, that is not always true.
The probability that the other 16 cards contain 2 spades is about 23% and the probability that they have 3 spades is 30%.

If you take these figures into consideration, you are likely to turn a flush only ablout 9.57% of the time. The probability of rivering a flush is slightly higher (I believe, because there is one card less to consider.) So in actual fact, you are likely to hit your flush only about 20% of the time when you are at a 9-handed table.

That figure will go up if you are 6-handed (I believe, as lesser outs are taken away)

Here’s how I got to 9.57% for the turn -
Consider 10 cases.
Case 1: The 16 other player cards contain no spade.
Then probability of hitting a flush is 9/34

Case 2: The 16 other player cards contain 1 spade.
Then probability of hitting the flush is 8/34

And so on.

Once this is done, then calculate the probability that case 1 occurs.
Multiply this by the probability that you hit a flush in case one. Then do the same for the rest of the cases. Finally, divide this number by total possibilities to get probability of turning a flush.

Mathematically speaking,
I’m using theorem of total probability
(P(Case 1) X P(hitting a flush in case 1) +
P(Case 2) X P(hitting a flush in case 2) …)
/ 47C16

Here
P(Case 1) = (38C16) X (9C0)
P(Case 2) = (38C15) X (9C1)
P(Case 3) = (38C14) X (9C2)
And so on

Hope this helps
JJ

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15

I think the main flaw in my thinking was in trying to eliminate all 16 hole cards just because there is a high probability that one of them is a spade.

Here’s my new new way to look at it…

What can we actually know about these 16 unseen cards?

Well, since there are only 9 spades available, they can’t all be spades. In fact, 7 of them can’t possibly be spades, the other 9 may or may not be spades. I’m gonna move these 7 to the known pile, meaning we now know 12 cards and don’t know 40 cards. Since 9 of these 40 are spades, we get…

((9/40)X100)= 22.5%

This isn’t much different than what we get with the conventional way of looking at it, which gives ((9/47)X100)=19.1%, but is probably more accurate.

NOTE: This too will be wrong 98.5% of the time, and is thus completly unsatisfactory.

16

The standard way poker odds are calculated is based on the assumption that we are the only player in the game. Having a 4,6 or 9 person table does not affect them in any way. Therefore, the assumption is that there are 47 cards left in the deck for purposes of calculating our odds (52 - our 2 hole cards and the 3 cards on the flop). We do not count any cards in other players’ hands, nor do we count the burn cards.

This is the accepted conventional approach and uses the most basic assumption, that all unknown cards are available. With increased information, you can begin to adjust the odds for the specific hand. For example, if you know player 1 only enters a pot holding an A, you can then remove 1 A from all calculations of draws you have requiring an A.

The method assumes every card that is unknown is available because you cannot know the distribution of any cards with any reasonable degree of accuracy without additional information. All assumptions you make without additional information reduce the accuracy of the model and therefore the model specifically chooses to eliminate them. The small number of cards in the deck make the inclusion of assumptions especially problematic because even small errors in your assumptions change the percentages significantly.

Hope this helps just a little. I feel like I went the long way around of saying “it is because it is”. The calculations simply count only what is perfectly known and discount everything else.

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17

Well yeah, that’s why they are incorrect 98.5% of the time.

In the example given: 9 player table, 2 spades in your hand, 2 on the flop, the conventional method assumes there is nothing we can know about the 16 hole cards in the other 8 hands, but this is simply wrong.

We can know with 100% certainty that at least 7 of them are not spades.

The actual distribution of spades in the remaining 9 cards will follow some sort of bell curve. The range will be 0-9, with both extreme results being fairly rare. What I’m looking for is the mode… the number which appears most frequently.

I calculated that there will be at least 1 spade in these other 16 hole cards 98.5% of the time, but I have no idea how to calculate the probabilities for the rest of the range.

18

It is true that standard poker odds assume that you are the only player at the table. However, that assumption is not always true.

Consider this example:

Here, conventional odds would say that Schanbacher was about 19% to win, whereas in actual fact he was about 3% to win, if you take into account the fact that Lamphere folded the other 2 tens.

While yes, the chances of such a hand are slim, in the off chance that it does happen, it implies that the standard odds are way off.

It is a similar story with the spades example mentioned earlier. In the off chance that the 16 cards in front of you contain 7 spades, then you’re left with 2 outs instead of the standard 9.

These scenarios are not impossible, so I did take them into consideration when I carried out my calculations to conclude that the chances of turning a flush is 9.57% instead of 19%, on a nine handed table.

Hope this helps
JJ

19

The probability that the 16 cards in front of you:

Have no spades is 1.47%
Have 1 spade is 9.26%
Have 2 spades is 23.15%
Have 3 spades is 30.25%
Have 4 spades is 22.69%
Have 5 spades is 10.08%
Have 6 spades is 2.64%
Have 7 spades is 0.39%
Have 8 spades is 0.02%
Have 9 spades is 0.0008%

Hope this helps
JJ

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20

Actually, it helps a lot! That’s exactly what I was looking for, thanks.

So the ACTUAL probabilities would be…

When the remaining 16 hole cards contain 2 spades: (7/31)X100=22.5%
When the remaining 16 hole cards contain 3 spades: (6/31)X100=19.4%
When the remaining 16 hole cards contain 4 spades: (5/31)X100=16.1%

And this answers my question… why is the conventional way used if it’s not accurate?

And the answer is… It will be accurate 30% of the time (in the 3 spades out in hole cards case) and within +/- 3% for the 2 spades and 4 spades cases, which means it’s correct or within 3% 75% of the time.

I now realize Joe did point this out earlier, I was just too thick to grasp it. Basically, the distribution of spades in the unseen 16 hole cards is the same as the distribution in the other 31 cards, so eliminating them shouldn’t change much. 9 outs in 47 cards isn’t that different than 6 outs in 31 cards.

Virtually everyone but me will have considered this thread a total waste of time, but hey, I learned something! Thanks!

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