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Old 04-01-2017, 11:50 AM   #1
doylebrunson1337
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Destroying Villains gun by destroying my gun (Prisoners dilemma)

The game is 5 card draw with discarding 1 time and three players are playing.

Dealer has 4567T
Small blind hand 22KTA
and big blind has 4567T

Small blind decides to pat with 14% equity (cause why chase trips when the payroll is after a better hand?)

bb discards 1 card to a straight
dealer does the same.

They both have 16 outs into making a better pair or a str8.

37 cards in the deck and 16 outs = 43%!

So when small blind pats he goes 57%X57% right? That is 32,2%!

So how the f does my opponent hit 43% each when I am holding 32,2? and if it is reducing them both to 27% each then what happened to the 16%?

This reminds me of the prisoners dilemma and if both of them take the wrong turn and discard 1 I am actually then playing against 1 hand rather than two but I get 1/2 more money since patting 14 into 32,2% increases my value to 2,35:1.

The prisoners need to know what the other one is going to do before discarding his card.

This is not fact! I am asking for advice on how this hand goes. I admit that I am stupid, I actually got banned in another forum for this post and they deleted every post I made there even in other topics that did not have with this to do at all...

Last edited by doylebrunson1337; 04-01-2017 at 11:58 AM.
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Old 04-02-2017, 02:27 PM   #2
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

I only briefly read your post, but if I'm understanding right you're saying:
you have 32.2% to win
each of your opponents has 43% to hit their draw
and you're confused because 32.2 + 43 + 43 > 100 right?

Well, they each have 43% to beat *you*. When both of them hit, they both beat you but only one of them wins the hand. So each of their actual equities < 43%.
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Old 04-03-2017, 09:59 AM   #3
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

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Originally Posted by RustyBrooks View Post
I only briefly read your post, but if I'm understanding right you're saying:
you have 32.2% to win
each of your opponents has 43% to hit their draw
and you're confused because 32.2 + 43 + 43 > 100 right?

Well, they each have 43% to beat *you*. When both of them hit, they both beat you but only one of them wins the hand. So each of their actual equities < 43%.
If I understood correctly I am still 14% but by patting I only increase my hands value by something like 230%. They still have 43% of winning but instead of playing one opponent (that would have 43% against my 57%) I add one more and by patting pair of deuces everybody dies? And by that I mean by patting my hand I can actually cast a spell that makes outs vanish from the deck?

Excuse my bad English.

This is so confusing, I have mailed and got an answer from my old high school teacher to go and talk about this hand for an hour booked next week.

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Old 04-03-2017, 10:07 AM   #4
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by RustyBrooks View Post
I only briefly read your post, but if I'm understanding right you're saying:
you have 32.2% to win
each of your opponents has 43% to hit their draw
and you're confused because 32.2 + 43 + 43 > 100 right?

Well, they each have 43% to beat *you*. When both of them hit, they both beat you but only one of them wins the hand. So each of their actual equities < 43%.


edit: I read your post again and understood it all. Except for " instead of playing one opponent (that would have 43% against my 57%) I add one more and by patting pair of deuces everybody dies? Who is the winner here?"

edit2: And 14% of 32,2% is 43% too.
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Old 04-03-2017, 10:59 AM   #5
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

There are several questionable statements in the OP. Isnít the number of villain outs 20 if hero stands pat? One or both V can pair to beat your 22 but why would you not discard at least the T to possibly get two pair or a set unless standing pat has other good features (e.g., representing a straight or better - I donít play this game)?

With 20 outs each villainís equity is 54%. Your equity is then Pr(both miss) which is approximately 0.46^2 = 21% (actually 0.46*(1-20/36)=0.204).Total equity is 2*0.54 -.54^2 + 0.21 = 1.0. Using 16 outs, the probabilities will also sum to 1.0 when you correctly sum. As Rusty noted, only one villain wins when both hit unless they draw the same rank out and then they split the pot.
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Old 04-03-2017, 02:44 PM   #6
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by statmanhal View Post
There are several questionable statements in the OP. Isn’t the number of villain outs 20 if hero stands pat? One or both V can pair to beat your 22 but why would you not discard at least the T to possibly get two pair or a set unless standing pat has other good features (e.g., representing a straight or better - I don’t play this game)?

With 20 outs each villain’s equity is 54%. Your equity is then Pr(both miss) which is approximately 0.46^2 = 21% (actually 0.46*(1-20/36)=0.204).Total equity is 2*0.54 -.54^2 + 0.21 = 1.0. Using 16 outs, the probabilities will also sum to 1.0 when you correctly sum. As Rusty noted, only one villain wins when both hit unless they draw the same rank out and then they split the pot.
If both VILLAINS are chasing str8 or better by discarding 1 card they both have 8 outs for str8 and 8 outs for a better pair then 22. I don't play 5 card draw either but I have played a lot of chigaco with two discard and one of my tactics is to pat pair of 22 and holding TKA and play for one point after first discard then I throw away 22 and aim for the Chicago play offs. I have won a lot of time 1 point for having a pair when both my opponents has worse than pair of 2.

Also I watched a Jason Statham movie called "Revolver" and got inspired I quote "Best way to win in chess is to sacrifice your own pieces so that your opponent thinks you are dumb and he is smart", the reason why this hand popped into my head a couple of days later.

Both of VILLAINS hitting a better hand (first VILLAIN draws a better hand and the last VILLAIN hits a better hand) is 43%X43% = something around 18% so it is not very much likely to happen and for the chance of that happening the first drawing villain must hit to complete 50% of that outcome.

But my question is how/why do I get 2,3:1 patting a pair of 2? I should have less. By playing ABC-tags I could also bet 1/5 of the pot after the discard and win vs half their range that is better than mine making pair of 2 a 64% before the action began but that is not the point or discussion I want in this thread.

edit: Math has always been the worst class of mine during school so maybe I am blabbering but that 43% becomes less since there is another VILLAIN behind him drawing to same is a fact I understood but who is eating equity when first one misses? If HERO wins the equity when first VILLAIN misses he gets more than doubled equity even when the other VILLAIN has same amount of outs. (It think)

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Old 04-03-2017, 03:30 PM   #7
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Actually by patting HERO has something around 3% equity and that if both VILLAINS discard 5 card each.
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Old 04-03-2017, 05:08 PM   #8
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Right, it is 16 outs - I forgot the villains have same hands so there are only 8 ways for them getting a better pair. But as I said, the math is the same and there are no "lost outs".
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Old 04-03-2017, 08:01 PM   #9
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by statmanhal View Post
Right, it is 16 outs - I forgot the villains have same hands so there are only 8 ways for them getting a better pair. But as I said, the math is the same and there are no "lost outs".
Ok so both villain gets less equity than their drawing hit rate? I understand that now.
I just can't figure out how my equity goes down from 57% vs one opponent to 32% against two opponents, its is only 25% less by adding another 43% gun.
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Old 04-03-2017, 08:32 PM   #10
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

I don’t quite know what’s bothering you. As you correctly state in the OP, if each villain has a 43% chance of hitting his outs (not necessarily win),the chance you win is Pr(V1 and V2 do not improve) = 0.57*0.57 = 0.32 (there’s a slight dependency which I ignored).

The total equity for all players is then

0.43 + 0.43 – 0.43^2 + 0.32 = 1.0

0.43 +0.43 -0.43^2 = 0.68 is based on Pr(V1 hits his outs)+Pr(V2 hits his outs) – Pr(both V1 and V2 hit their outs). Again, I assume independence which is not true but makes for easier explanation.

Since 0.68 is hero losing and since villains have same hands each villain has the equivalent of 0.34 equity assuming flush not possible.
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Old 04-03-2017, 08:53 PM   #11
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by statmanhal View Post

The total equity for all players is then

0.43 + 0.43 – 0.43^2 + 0.32 = 1.0

0.43 +0.43 -0.43^2 = 0.68 is based on Pr(V1 hits his outs)+Pr(V2 hits his outs) – Pr(both V1 and V2 hit their outs). Again, I

Since 0.68 is hero losing and since villains have same hands each villain has the equivalent of 0.34 equity assuming flush not possible.
This is the thing that I was thinking about! Are we also counting (Villain1 misses his outs +villain2 hits his outs)? Also, is the outcome different if HERO's hand has the button instead of being the small blind?

Since they have 34% each with a 43% hit rate then when villain1 misses HERO gains 34% +32% = 66% (value of the coin I mean investing with 14% in 1$ and then 5x its value meaning that we can call and win a 66cent bet from villain2 after draw) meaning that the probable outcome in this hand is that HERO go's from 0/3 to 2/3 and then lose most of the times and the other scenario, 1/3 of the time HERO goes 0/3 to 0/3 when first villain1 hits??? (It would even out the 1% missing in HERO's hand) Again, is it different to pat deuces from small blind and patting them on the button?

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Old 04-04-2017, 07:57 AM   #12
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

I would like to change positions on the players in this hand. SB and Dealer both has 4567T and HERO is in BB with 22TKA also meaning that Dealer raises to 30 cent and gets called by SB and HERO raises the pot to 1,80$ so the pot is 5,40$ giving the dealer 36ish%% pot odds when calling and SB gets something around the sum of its hitting equity and HERO practically shots himself in the head getting 1/6 of that raise but with doubled equity. Then when VILLAIN1 misses HERO must decide what cards to throw or what to do so HERO can get as much as equity possible from the small blind when SB misses his draw, 33,9% was his equity and now I want to eat it so by patting after a guy that lost.

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Old 04-04-2017, 10:29 AM   #13
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by doylebrunson1337 View Post
This is the thing that I was thinking about! Are we also counting (Villain1 misses his outs +villain2 hits his outs)? Also, is the outcome different if HERO's hand has the button instead of being the small blind?
To get each villainís win probability directly, consider the following, again ignoring card removal (just complicates the math but same approach applies):

Villain 1 wins if he hits his outs and villain 2 doesnít: Pr= 0.43*0.57 =0.245

Villain 1 wins if both he and villain 2 hit their outs but V1 has better or equal hand: Pr = 0.43*0.43*0.50=0.092

V1 equity = 0.245+0.092 = 0.34, as I showed earlier.

I don't see how the position of the players affects the math.

I'll leave the remaining questions to someone else, since they are not very clear to me.
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Old 04-04-2017, 11:58 AM   #14
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

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Originally Posted by statmanhal View Post
To get each villain’s win probability directly, consider the following, again ignoring card removal (just complicates the math but same approach applies):

Villain 1 wins if he hits his outs and villain 2 doesn’t: Pr= 0.43*0.57 =0.245

Villain 1 wins if both he and villain 2 hit their outs but V1 has better or equal hand: Pr = 0.43*0.43*0.50=0.092

V1 equity = 0.245+0.092 = 0.34, as I showed earlier.

I don't see how the position of the players affects the math.

I'll leave the remaining questions to someone else, since they are not very clear to me.
edit: Miscalculated

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Old 04-04-2017, 12:19 PM   #15
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by statmanhal View Post
To get each villain’s win probability directly, consider the following, again ignoring card removal (just complicates the math but same approach applies):

Villain 1 wins if he hits his outs and villain 2 doesn’t: Pr= 0.43*0.57 =0.245

Villain 1 wins if both he and villain 2 hit their outs but V1 has better or equal hand: Pr = 0.43*0.43*0.50=0.092

V1 equity = 0.245+0.092 = 0.34, as I showed earlier.

I don't see how the position of the players affects the math.

I'll leave the remaining questions to someone else, since they are not very clear to me.
By adding all outcomes (according to you) for VILLAIN winning then HERO goes
76,5%^2 X 82% = 48% basically saying that patting deuces doubled their outs and then I took them back. If I only had pair of threes instead, I could offer them one more draw. ( I think)
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Old 04-04-2017, 01:42 PM   #16
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

i find what statmanhal has posted has been helpful and correct with regard to what OP requested help with. I find much of what the OP posts confusing.

although I'm reiterating what has already been posted, perhaps it will be clearer to OP.



the probability (generalized)

when 2 people 'draw' with a 43% chance of success (and for simplicity we use the assumption that the draws are independent which is to say the first draw doesn't have an impact on the next draw).

18.5% is the probability that both draws will be successful .43*.43 =.185

24.5% is the probability that one draw will be successful and the other draw won't be. (there are 2 distinct instances of this: draw1 successful and draw2 unsuccessful and draw1 unsuccessful and draw2 successful)

the 24.5% and the 18.5% together equal the 43% chance of a successful draw. .185 +.245 =.43

32.5% is the probability of 2 unsuccessful draws 1-.43=.57, .57 *.57= .325

the check:
.185 +.245 +.245 +.325 =1

equities (generalized)

when 2 players, each with a single 43% draw wager on success and a 3rd player wagers on neither draw being successful then their respective equities are:

33.75%
33.75%
32.5%

check: .3375 +.3375 +.325 =1

33.75% because 24.5% of the time he will win 100% of the pot because he will have the only successful draw and 18.5% of the time he will win just 50% of the pot because both draws will be successful resulting in the pot being split 50%/50%. .245 *1 + .185 *.5 = .3375

32.5% because the 32.5% of the time both draws are unsuccessful he wins 100% of the pot


the order that they draw or don't draw is not significant (because of independence)

the probabilities don't change unless new information is presented. the equities don't change unless the probabilities change.
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Old 04-04-2017, 03:05 PM   #17
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

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Originally Posted by ngFTW View Post
i find what statmanhal has posted has been helpful and correct with regard to what OP requested help with. I find much of what the OP posts confusing.

although I'm reiterating what has already been posted, perhaps it will be clearer to OP.



the probability (generalized)

when 2 people 'draw' with a 43% chance of success (and for simplicity we use the assumption that the draws are independent which is to say the first draw doesn't have an impact on the next draw).

18.5% is the probability that both draws will be successful .43*.43 =.185

24.5% is the probability that one draw will be successful and the other draw won't be. (there are 2 distinct instances of this: draw1 successful and draw2 unsuccessful and draw1 unsuccessful and draw2 successful)

the 24.5% and the 18.5% together equal the 43% chance of a successful draw. .185 +.245 =.43

32.5% is the probability of 2 unsuccessful draws 1-.43=.57, .57 *.57= .325

the check:
.185 +.245 +.245 +.325 =1

equities (generalized)

when 2 players, each with a single 43% draw wager on success and a 3rd player wagers on neither draw being successful then their respective equities are:

33.75%
33.75%
32.5%

check: .3375 +.3375 +.325 =1

33.75% because 24.5% of the time he will win 100% of the pot because he will have the only successful draw and 18.5% of the time he will win just 50% of the pot because both draws will be successful resulting in the pot being split 50%/50%. .245 *1 + .185 *.5 = .3375

32.5% because the 32.5% of the time both draws are unsuccessful he wins 100% of the pot


the order that they draw or don't draw is not significant (because of independence)

the probabilities don't change unless new information is presented. the equities don't change unless the probabilities change.
Ok thx for clearing it all out. I need both of my villains draw to 14 outs each here to make ze profit. I was just curious about the equity trade between mine and villains hands during the pat/discards on how much value the coin could get. But I think you made one miscalculation
Quote:
Originally Posted by ngFTW View Post

the check:
.185 +.245 +.245 +.325 =1
Correct me if I am wrong (I am sure that I could be wrong but I am seeking knowledge as u see, math was my biggest fail in high school and I later buy that book just to see that there aint no advanced gamble theory) the 32,5% don't need to hit, it already has hit being pat thus should not counted in the probability when villains are discarding and I see the 32,2% as a base that adds equity on it without going "X" with the other probability. Cuz arent we going 2/3 of the time 2/3 after first villain1 draws and miss??

Being pat makes them chase me giving me according to your 33,75% each to villain1 and 2,,,, 0,6625^2 = 43%. Does it mean in theory that instead of being 57% over one hand that is 43% (cuz if we remove one villain1 we are 57% in lead) I add a buddy and offer myself as the 43% dog and gently give them my 57% (I am going to ask as many question as possible before getting it all clear ) (If it won't work, then how about a str8draw to villain2 that is not like villain1s str8draw but they have the same outcome of 16 possible outs, then there is the 30%ish chance of first villain taking the second villains out and that adds that % to my pat hand)

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Old 04-04-2017, 03:39 PM   #18
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Just realised. My pat is only 30,8% and not 32,2. So that post I wrote above I mean I get 42% and not 43%. Also that by not going drawing against 2x 33,75%(+1,4%) the actual chase is between the VILLAINs while I passively enjoy the scene. So by according to law of cosmos (not math) they must both hit to win because they both have the winning equity in their hand (not realising that villain1 should discard 5 and villain2 discards 1 card then they both get 44% and smoke me).

edit: imo

edit2: "Karma can only be divided by cosmos" - Homer J Simpson.

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Old 05-11-2017, 11:26 AM   #19
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Wow, this thread is over my head, I always assumed that using probability theory for draw poker is never completely valid and never completely accurate and always some kind of compromise between the two.

For example the only valid solution for draw poker ends up with all opponents playing random cards which is absurd.

But applying hands in hind site is results oriented thinking which is better than the alternative but could be very tricky in creating invalid conclusions.

I admit that I never actually tried to solve a draw poker game before using probability. Is their some inductive component I'm not picking up on that validates the hindsite results?

Im afraid that I didn't get very far in the posts before getting lost. where I'm sure I missed dome assumptions somewhere.

What am I missing?

Edit to add:
Well, I'm missing the rules for one, so maybe i'm basing my confusion on a false asumption about the rules.
  • What's the betting structure?
    FL, SL, PL all seem like legit variants.
  • I asssume there are only two betting rounds is that correct?
  • does the SB draw first and act first?
  • If so, how does the SB know the other two players are on a draw?
  • BTW this is draw where no cards are exposed and not stud where 4 cards are exposed correct?
  • whats the max you can draw?
    sorry I only know the home game variety of draw poker.

Last edited by TakenItEasy; 05-11-2017 at 11:54 AM.
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Old 05-13-2017, 03:17 AM   #20
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

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Wow, this thread is over my head, I always assumed that using probability theory for draw poker is never completely valid and never completely accurate and always some kind of compromise between the two.

For example the only valid solution for draw poker ends up with all opponents playing random cards which is absurd.

But applying hands in hind site is results oriented thinking which is better than the alternative but could be very tricky in creating invalid conclusions.

I admit that I never actually tried to solve a draw poker game before using probability. Is their some inductive component I'm not picking up on that validates the hindsite results?

Im afraid that I didn't get very far in the posts before getting lost. where I'm sure I missed dome assumptions somewhere.

What am I missing?

Edit to add:
Well, I'm missing the rules for one, so maybe i'm basing my confusion on a false asumption about the rules.
  • What's the betting structure?
    FL, SL, PL all seem like legit variants.
  • I asssume there are only two betting rounds is that correct?
  • does the SB draw first and act first?
  • If so, how does the SB know the other two players are on a draw?
  • BTW this is draw where no cards are exposed and not stud where 4 cards are exposed correct?
  • whats the max you can draw?
    sorry I only know the home game variety of draw poker.
FL would be easiest in this case so go for that 5/10. The game is 5 card draw and lets just say that small blind is Stu Ungar and he knows what they are holding. Max draw is 5 cards. With 42 cards left in the deck and in this case with 10 blockers, then drawing 5 cards results in a pair 36-44% of the time.

edit: My Question. When HERO pats 2,35:1 and first VILLAIN misses then that VILLAIN should have none value in his hand, so who gets it?

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Old 05-13-2017, 06:00 AM   #21
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by doylebrunson1337 View Post
FL would be easiest in this case so go for that 5/10. The game is 5 card draw and lets just say that small blind is Stu Ungar and he knows what they are holding. Max draw is 5 cards. With 42 cards left in the deck and in this case with 10 blockers, then drawing 5 cards results in a pair 36-44% of the time.

edit: My Question. When HERO pats 2,35:1 and first VILLAIN misses then that VILLAIN should have none value in his hand, so who gets it?
For identical straight draws thats a Hell of card placement read.
sounds like dealer is grouping known cards from last had at the top using false shuffles/strips

so anyway, lets just treat it as a showdown
if three way just make sure all 3 equities = 100%
V1= Y%
V2= Y%
H = pot - 2Y%

if one folds, then the equities are recalculated for 2 equities =100%
V2 = Y%
H = pot - Y%
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Old 05-13-2017, 09:00 AM   #22
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

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Originally Posted by TakenItEasy View Post
For identical straight draws thats a Hell of card placement read.
sounds like dealer is grouping known cards from last had at the top using false shuffles/strips

so anyway, lets just treat it as a showdown
if three way just make sure all 3 equities = 100%
V1= Y%
V2= Y%
H = pot - 2Y%

if one folds, then the equities are recalculated for 2 equities =100%
V2 = Y%
H = pot - Y%
So when HERO pats 32,2% and VILLAIN1 draws 1 not using his full equity then HERO should be standing on 1,75:1 more resulting in that it does not matter if VILLAIN2 draws 1 or 5 right? When VILLAIN1 draws 1 there is 24% chance that his hand will hit and the others won't but the part I don't get if HERO's total equity ends up as 43% or 57% after V1 1-card draw and when VILLAIN2 draws 1 or 5 cards.

edit: I am not super sure on if V1 gives hero 1,75:1 but that is his total equity when drawing 1 in hopes that V2 won't hit his hand. V1 draws 5 = 1,79:1 to V1. <---- 4% is a really big number here.

Last edited by doylebrunson1337; 05-13-2017 at 09:09 AM.
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Old 05-14-2017, 09:52 PM   #23
TakenItEasy
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by doylebrunson1337 View Post
So when HERO pats 32,2% and VILLAIN1 draws 1 not using his full equity then HERO should be standing on 1,75:1 more resulting in that it does not matter if VILLAIN2 draws 1 or 5 right? When VILLAIN1 draws 1 there is 24% chance that his hand will hit and the others won't but the part I don't get if HERO's total equity ends up as 43% or 57% after V1 1-card draw and when VILLAIN2 draws 1 or 5 cards.

edit: I am not super sure on if V1 gives hero 1,75:1 but that is his total equity when drawing 1 in hopes that V2 won't hit his hand. V1 draws 5 = 1,79:1 to V1. <---- 4% is a really big number here.
no, you need to recalculate v2 equity then hero gets the rest. Don't forget V1 blank draw for 16 cards removed from the stub.
v2 eq = 16/36*pot
H eq = (1-16/36)*pot

Last edited by TakenItEasy; 05-14-2017 at 10:02 PM.
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Old 05-16-2017, 09:14 AM   #24
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Re: Destroying Villains gun by destroying my gun (Prisoners dilemma)

Quote:
Originally Posted by TakenItEasy View Post
no, you need to recalculate v2 equity then hero gets the rest. Don't forget V1 blank draw for 16 cards removed from the stub.
v2 eq = 16/36*pot
H eq = (1-16/36)*pot
Ok! So it is basically HERO AKof flip vs 2 lower pocket pairs?
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