OT: Math/poker question

[nitanee123]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

 

[WDLion]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

How are there 47 cards remaining?

 

[mfb5053]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

Burn card?

 

[nitanee123]

How are there 47 cards remaining?

52 cards minus the 5 (3 community plus the 2 I'm holding).

 

[mfb5053]

52 cards minus the 5 (3 community plus the 2 I'm holding).

Are you playing against yourself? I think you'll win

 

[WDLion]

52 cards minus the 5 (3 community plus the 2 I'm holding).

Are you the only player?

 

[michnittlion]

38 out of 47 chance the next card is not a heart.

If that card is not a heart, 37 out of 46 chance the next card is not a heart.

Thus, the probability of 2 non-hearts = (38*37)/(47*46) = 65%

1 - 65% = 35% chance you hit your flush.

 

[mrmk5110]

Are you the only player?

Doesn't make a difference in terms of odds as those are cards that he doesn't know if the other players have or don't have. The rest of the hearts could have already been dealt, or they may not have been dealt at all.

 

[4rosco]

38 out of 47 chance the next card is not a heart.

If that card is not a heart, 37 out of 46 chance the next card is not a heart.

Thus, the probability of 2 non-hearts = (38*37)/(47*46) = 65%

1 - 65% = 35% chance you hit your flush.

Link that shows that your method is the correct method to calculate the percentages

https://www.wikihow.com/Learn-Poker-Percentages

 

[nitanee123]

38 out of 47 chance the next card is not a heart.

If that card is not a heart, 37 out of 46 chance the next card is not a heart.

Thus, the probability of 2 non-hearts = (38*37)/(47*46) = 65%

1 - 65% = 35% chance you hit your flush.

This helps. I'm having a problem wrapping my head around calculating the odds of NOT getting it vs. GETTING it.

On the turn, I'm coming up with 19.14% chance of getting it, which corresponds to the 80.86% chance of NOT getting it (as you said, 38/47). I just can't grasp the odds on the river. If I didn''t hit it on the turn, then I should I not get it 19.56% times on the river (9/46)? And if that is the case then why cannot I just add the %'s together? (I'm just trying to type this out to help me figure this out.)

 

[michnittlion]

This helps. I'm having a problem wrapping my head around calculating the odds of NOT getting it vs. GETTING it.

On the turn, I'm coming up with 19.14% chance of getting it, which corresponds to the 80.86% chance of NOT getting it (as you said, 38/47). I just can't grasp the odds on the river. If I didn''t hit it on the turn, then I should I not get it 19.56% times on the river (9/46)? And if that is the case then why cannot I just add the %'s together? (I'm just trying to type this out to help me figure this out.)

Yeah, thinking about things from "the not getting the Flush direction" is a little goofy. My mind, when calculating probabilities, does work more in that direction. From your direction, you can also think about it like this:

There are 2 possible ways to get the flush: (1) Heart on the Turn (if this happens, the River Card is irrelevant, at least as regards your flush), and (2) Non-heart on the Turn, Heart on the River.

The odds of event (1) happening are 9 of 47

The odds of event (2) happening are (38 out of 47) multiplied by (9 out of 46).

Those sum up to 35%.

 

[Connorpozlee]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

There are 46 other cards after the flop (your two, three in the flop, and the burn card. 52-6 is 46). After the turn, there are 44 cards left (the previous six cards, the turn card, and the burned card. 46-2 is 44).
Does that help at all?

 

[nitanee123]

Yeah, thinking about things from "the not getting the Flush direction" is a little goofy. My mind, when calculating probabilities, does work more in that direction. From your direction, you can also think about it like this:

There are 2 possible ways to get the flush: (1) Heart is on the Turn (if this happens, the River Card is irrelevant, at least as regards your flush), and (2) Non-heart on the Turn, Heart on the River.

The odds of event (1) happening are 9 of 47

The odds of event (2) happening are (38 out of 47) multiplied by (9 out of 46).

Those sum up to 35%.

I definitely see how your math works. Its event 2 that is tripping me up. thank for the responses.

 

[WDLion]

Doesn't make a difference in terms of odds as those are cards that he doesn't know if the other players have or don't have. The rest of the hearts could have already been dealt, or they may not have been dealt at all.

I get your point but my definition of remaining cards would be those still held by the dealer.

 

[Vic Vaselino]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

If your luck runs like mine ,you'll hit the flush and still lose.

 

[nitanee123]

There are 46 other cards after the flop (your two, three in the flop, and the burn card. 52-6 is 46). After the turn, there are 44 cards left (the previous six cards, the turn card, and the burned card. 46-2 is 44).
Does that help at all?

The burned card isn't counted, but I re-read michnitt's post and then did the math on a piece of paper and it now I understand it.

thanks

 

[michnittlion]

I get your point but my definition of remaining cards would be those still held by the dealer.

It's a bit like quantum mechanics. Until you have precisely measured where the 6 of Hearts (or any other card) is --- it could be any of the outstanding cards.

Its location is a "wave function" (of sorts).

 

[not so fast my friend]

The burned card isn't counted, but I re-read michnitt's post and then did the math on a piece of paper and it now I understand it.

thanks

The 38.7 vs 35 percent is because in the 38.7, you are "double counting" the cases where you make a heart on both the turn and river.

 

[nitanee123]

The 38.7 vs 35 percent is because in the 38.7, you are "double counting" the cases where you make a heart on both the turn and river.

thanks...that makes sense!

 

[wbcincy]

thanks...that makes sense!

Yep, you'll notice the odds of hitting it both times (19.14% X 19.56%) is the extra 3.7%.

 

[PSU2UNC]

There are 46 other cards after the flop (your two, three in the flop, and the burn card. 52-6 is 46). After the turn, there are 44 cards left (the previous six cards, the turn card, and the burned card. 46-2 is 44).
Does that help at all?

The burn cards don't matter. Just as the other players cards don't matter.

 

[Connorpozlee]

The burn cards don't matter. Just as the other players cards don't matter.

Got it. Makes sense.

My theory: Bet ‘em like you got ‘em.

 

[Grant Green]

It's a bit like quantum mechanics. Until you have precisely measured where the 6 of Hearts (or any other card) is --- it could be any of the outstanding cards.

Its location is a "wave function" (of sorts).

Agreed. The odds calc is theoretical. Your odds could be worse than 35 if all other hearts are held by other players, but could be better if the deck holds all the other hearts.

Next topic, reverse implied odds?

 

[JR4PSU]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

Well, there is certainly no room for 4% in your sig pics dress. So we know it isn't there.

 

[JR4PSU]

Are you the only player?

You only count the cards you know, not the ones you don't know. You only know those in your hand and those on the board. So only 5 cards known with 47 out there. The next card out of the dealers hand can be any of those 47 cards.

 

[JR4PSU]

I get your point but my definition of remaining cards would be those still held by the dealer.

You don't know what cards the dealer is holding. They can be any of the remaining 47 cards that you do not know about.

 

[PSUSignore]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

The back of the envelope math I use in my head is # of outs * 4 before the turn, and # of outs * 2 on the river. That would mean if you're on a flush draw you have 9 outs * 4 = 36% chance to hit your flush prior to seeing the turn, and 9* 2 = 18% chance to hit your flush prior to seeing the river. The pot odds compared to the odds of making your hand then dictate whether you should call or fold, if you are looking at it strictly from a mathematical standpoint. These odds are not precise though, but this method is easy to calculate in your head as long as you have a relatively good idea of the number of outs you have. Keep in mind that you might accidentally count too many outs if your opponent is on the same, but higher, flush draw.

 

[Grant Green]

The back of the envelope math I use in my head is # of outs * 4 before the turn, and # of outs * 2 on the river. That would mean if you're on a flush draw you have 9 outs * 4 = 36% chance to hit your flush prior to seeing the turn, and 9* 2 = 18% chance to hit your flush prior to seeing the river. The pot odds compared to the odds of making your hand then dictate whether you should call or fold, if you are looking at it strictly from a mathematical standpoint. These odds are not precise though, but this method is easy to calculate in your head as long as you have a relatively good idea of the number of outs you have. Keep in mind that you might accidentally count too many outs if your opponent is on the same, but higher, flush draw.

I do the same 4X calc (then maybe subtract a percent or two).

"Keep in mind that you might accidentally count too many outs if your opponent is on the same, but higher, flush draw"

^^reverse implied odds!!^^

 

[Pinkhippo PeanutButter]

I do the same 4X calc (then maybe subtract a percent or two).

"Keep in mind that you might accidentally count too many outs if your opponent is on the same, but higher, flush draw"

^^reverse implied odds!!^^

Opponents could also score a 4 of a kind & beat a flush, or a full house even. I tend to forget/neglect those possibilities when looking a a flush.

 

[PSUSignore]

I do the same 4X calc (then maybe subtract a percent or two).

"Keep in mind that you might accidentally count too many outs if your opponent is on the same, but higher, flush draw"

^^reverse implied odds!!^^

If the OP really want to read into this stuff, books by David Sklansky do a good job of getting into the mathematics. It's been a while since I've played regularly, and maybe there are better books today than the ones I bought 10+ years ago, but I really liked his materials and have a couple of his books.

 

[BigWesley]

I've got a question for the math folks on this board. This is probable a simple answer but I'm stumped by it.

The game is No Limit Texas Hold 'Em and you are holding two of same suit (let's say you have King and 5 of Hearts). Flop comes out with two more hearts, say 9 and 10, therefore you are on a flush draw. So you have 9 outs to hit your flush with two more cards left to be shown.

So in my mind, I'm thinking there is a 19.14% (9 hearts [aka outs] out of 47 remaining cards) chance of hitting it on the turn PLUS 19.56% (9 hearts out of 46 remaining on river) for a total % of 38.7%. Everything I read, however, says there is only a 35% chance. Where is the almost 4% difference coming from?

Thanks

So did you lose the hand?

 

[Grant Green]

Opponents could also score a 4 of a kind & beat a flush, or a full house even. I tend to forget/neglect those possibilities when looking a a flush.

If you have low cards, you better consider the possibility of getting out-flushed at least. You need to take it into account when calculating your pot odds.

 

[Erial_Lion]

Played a resort poker game this afternoon here in Punta Cana. 9 players, $20, winner take all. End up heads up with about 45% of the chips against a woman that’s never played hold ‘em before. A4 offsuit in small, min raise and she calls...235 rainbow flop. Bet, raise, re-raise, all-in, call. She’s got Q3. Turn 5, river 5, dealer and I have to tell her she won.

I love poker!

 

[Grant Green]

Played a resort poker game this afternoon here in Punta Cana. 9 players, $20, winner take all. End up heads up with about 45% of the chips against a woman that’s never played hold ‘em before. A4 offsuit in small, min raise and she calls...235 rainbow flop. Bet, raise, re-raise, all-in, call. She’s got Q3. Turn 5, river 5, dealer and I have to tell her she won.

I love poker!

Ouch. Insane hand. So, she bet and re-raised the flop with middle pair? and then called an all-in? Yikes. She needed runner-runner to win, didn't she.

 

[2PBC22]

As a couple people have said, it is sometimes easier in probability to calculate the probability that something will not happen and then subtract it from 1 to get the probability of it happening. Since an event either must or must not happen, the combined probabilities always add up to 1.

Based on your situation you can:
1. Draw a heart and a heart: (9/47) * (8/46)=.0333
2. Draw a heart, then a non heart: (9/47) * (38/46)=.1582
3. Draw a non heart, then a heart: (38/47) * (9/46)=.1582
4. Draw 2 non hearts: (38/47)*(37/46)=.65

So your options are add 1,2, and 3 from above because those are desirable outcomes, or you can take 1 - .65 as has been said.

 

[Erial_Lion]

Ouch. Insane hand. So, she bet and re-raised the flop with middle pair? and then called an all-in? Yikes. She needed runner-runner to win, didn't she.

Yes, yes, and yes. This was literally her first time ever playing. She didn’t get the concept that a pair was good pre-flop, but not as much post. Fun time regardless.

Thought about having a Helmuth-like meltdown, but instead congratulated her. LOL

 

[PSU Mike]

Hypothetically, what items of clothing is the blonde sitting across from you still wearing in this scenario?

 

[PSUSignore]

Played a resort poker game this afternoon here in Punta Cana. 9 players, $20, winner take all. End up heads up with about 45% of the chips against a woman that’s never played hold ‘em before. A4 offsuit in small, min raise and she calls...235 rainbow flop. Bet, raise, re-raise, all-in, call. She’s got Q3. Turn 5, river 5, dealer and I have to tell her she won.

I love poker!

Ouch. There was pretty much no way you were going to get away from that hand post flop, especially against someone not experienced enough to know she should be folding based on how aggressive you were betting it. Tough break. I both love and hate playing noobies. Generally if you play smart you'll collect their money. But in the short term you have to deal with nonsensical calls like this, and the subsequent suck outs.

 

[caseofthemondays]

I once had K,J
Flop comes K,K,J. I bet and they go all in. Stacks close to even. Insta Call. They flip A,4.
Turn A, River A
I Lose Aces over kings to my Kings over Aces.
0.3% chance.
Poker is fun.

https://www.cardplayer.com/poker-tools/odds-calculator/texas-holdem

 

[Op2]

Played a resort poker game this afternoon here in Punta Cana. 9 players, $20, winner take all. End up heads up with about 45% of the chips against a woman that’s never played hold ‘em before. A4 offsuit in small, min raise and she calls...235 rainbow flop. Bet, raise, re-raise, all-in, call. She’s got Q3. Turn 5, river 5, dealer and I have to tell her she won.

I love poker!

In addition to telling her she won you should have added something like "It's customary for the winner to pay a mortgage payment for the loser as a consolation prize."