understanding split EV for infinite deck
August 2nd, 2012 at 6:15:00 AM
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I'm currently trying to understand how split EVs are calculated for Blackjack. I understand that the infinite deck approximation is the most simplest, since there are no card identity or removal effects present. The game rules are dealer stands on soft 17, double after split is allowed, and resplit up to 4 times allowed.
For further simplicity, I chose a hand that one would never split, i.e. 55 vs 2. Hence the resplit option takes no effect. Thankfully, the Wizard provides a EV table on his website.
For example when I split 55, this is what I would expect to get:
When I sum up the EV of all those possible results, weighted with their corresponding probability, I get -0.0953. This is the EV for a single new hand after the split. Since I have two new hands, I would need to double this for a split EV of -0.1907 disregarding rounding errors from copying the table. However the table reports splitting 55 vs 2 with an EV of -0.2902. What am I missing here ?
For further simplicity, I chose a hand that one would never split, i.e. 55 vs 2. Hence the resplit option takes no effect. Thankfully, the Wizard provides a EV table on his website.
For example when I split 55, this is what I would expect to get:
| probabilty | hand | action | EV |
|---|---|---|---|
| 4/52 | 5, 2 | hit | -0.1092 |
| 4/52 | 5, 3 | hit | -0.0218 |
| 4/52 | 5, 4 | hit | +0.0744 |
| 4/52 | 5, 5 | double | +0.3589 |
| 4/52 | 5, 6 | double | +0.4706 |
| 4/52 | 5, 7 | hit | -0.2534 |
| 4/52 | 5, 8 | stand | -0.2928 |
| 4/52 | 5, 9 | stand | -0.2928 |
| 16/52 | 5, 10 | stand | -0.2928 |
| 4/52 | 5, A | hit | -0.021 |
When I sum up the EV of all those possible results, weighted with their corresponding probability, I get -0.0953. This is the EV for a single new hand after the split. Since I have two new hands, I would need to double this for a split EV of -0.1907 disregarding rounding errors from copying the table. However the table reports splitting 55 vs 2 with an EV of -0.2902. What am I missing here ?
August 2nd, 2012 at 7:14:10 AM
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Quote: MangoJ...I chose a hand that one would never split, i.e. 55 vs 2. Hence the resplit option takes no effect...
I agree with your EV of -0.1907, diregarding rounding errors from copying the table. The Wizard's table is the EV's for always resplitting pairs to four hands regardless of the optimal strategy. His value for splitting 55 vs 2 only once would be -0.1935.
August 2nd, 2012 at 7:33:17 AM
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Ahh, thank you very much! I didn't consider that the resplit is "forced".
So when considering "infinite resplit" (4 times resplit seems reasonable close to that given the low probability of a resplit), calculation would be x = (−1.6172 + x) / 13 solving to x= -0.1348 and EV = 2x = -0.2696.
Thank you ! Now I got reasonable close to the number :)
So when considering "infinite resplit" (4 times resplit seems reasonable close to that given the low probability of a resplit), calculation would be x = (−1.6172 + x) / 13 solving to x= -0.1348 and EV = 2x = -0.2696.
Thank you ! Now I got reasonable close to the number :)
August 2nd, 2012 at 7:47:57 AM
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For my infinite-deck analysis I think I assumed infinite re-splitting. So, if you split 55 vs 2 the first time I assume you don't suddenly realize your error and then double with another 5, but would keep making that mistake as long as you kept getting more fives.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 2nd, 2012 at 7:54:18 AM
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Thanks! I also now got the infinite resplit EV correct (i think). it shoudl be the solution of EV/2 = (-1.6172 + EV) / 13 for EV = -0.294, where the magic number -1.6172 / 13 is the summed table without the "5, 5" row. Since infinite resplit is a little bit worther, 4 times resplit should be a bit higher.
August 2nd, 2012 at 8:12:55 AM
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Does Bicycle cards sell an infinite deck? I want to verify the Wiz's results. Never hurts to double check.
August 2nd, 2012 at 8:50:33 AM
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Quote: buzzpaffDoes Bicycle cards sell an infinite deck? I want to verify the Wiz's results. Never hurts to double check.
Sure. 20 decks of ordinary decks should do. Just pull each card out of a separate 52-card deck.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
August 2nd, 2012 at 8:59:41 AM
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Once again the man behind the curtain proves he is the WIZ.
January 5th, 2014 at 8:30:03 PM
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In the Wiz's Appendix 1 he lists an EV for splitting 10,10 vs. 9, at -.2067 (minus 20.67 %) for the player. In the same Appendix, the Wiz lists hitting 10 vs. 9 as .1165 for the player. If hitting 10 vs. 9 has a positive EV then I contend that splitting 10,10 vs. 9 should return a positive EV (not -.2067). Why, because splitting 10,10 vs. 9 is similar to hitting 10 vs. 9 for 2 hands. It's actually a little better than playing two hands of hitting 10 vs. 9 because of the ocassional resplit for a third advantageous hand of 10 vs. 9. So why does the Wiz list -.2067 while I believe the EV for splitting 10,10 vs. 9 should be a little more than 2 X .1165 (about .25).
January 5th, 2014 at 8:58:26 PM
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DavidY,
Welcome to the forum! I think your question is interesting, though very complex (and above my abilities to answer). Could you please edit it into about 6-8 paragraphs, though? It's very difficult to follow as posted, and I think you might get more responses if it were easier to read. It would just take a moment; there's a button on the lower right of the window your post is in. Thanks!
Welcome to the forum! I think your question is interesting, though very complex (and above my abilities to answer). Could you please edit it into about 6-8 paragraphs, though? It's very difficult to follow as posted, and I think you might get more responses if it were easier to read. It would just take a moment; there's a button on the lower right of the window your post is in. Thanks!
If the House lost every hand, they wouldn't deal the game.
January 5th, 2014 at 10:26:18 PM
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An even simpler way to phrase my statement is: "I believe many of the Wiz's EV percentages for splitting in Appendix 1 are wrong." For example splitting 10,10 vs. 9 lists -.2067 when I believe the correct EV is about .25. This is a huge discrepancy. Of course I know the correct BS is standing (.7584).
January 5th, 2014 at 10:35:34 PM
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Quote: DavidYormarkAn even simpler way to phrase my statement is: "I believe many of the Wiz's EV percentages for splitting in Appendix 1 are wrong." For example splitting 10,10 vs. 9 lists -.2067 when I believe the correct EV is about .25. This is a huge discrepancy. Of course I know the correct BS is standing (.7584).
This has been addressed, it assumed infinite splits. In other words, if you get another 10, you will keep splitting. A large part of the +EV from hitting a 10 against a 9 comes from all the times you catch a 10.
So, your chances of getting a 20 are greatly diminished. You are going to bust A LOT, and even bust multiple hands a lot.
January 6th, 2014 at 2:44:07 AM
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Quote: AxiomOfChoiceThis has been addressed, it assumed infinite splits. ....
At the beginning of Appendix 1, the Wizard writes, "The following tables display expected returns for any play in blackjack based on the following rules: dealer stands on a soft 17, an infinite deck, the player may double after a split, split up to three times except for aces..."
Here is a summary of my results, and my three splits allowed number agrees with the Wizard's:
Splits allowed EV 10,10 vs 9
one 0.23306
two -0.04047
three -0.20673
infinite -0.60742
January 6th, 2014 at 11:03:17 AM
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Thank you ChesterDog for pointing out to AxiomOfChoice that "infinite deck" does not mean "infinite splits". Before I start an exhaustive test to find out if the Wiz's (and ChesterDog's) EV (-.2067) for splitting 10,10 vs. 9 with the rules laid out in Appendix 1 (split up to three times, etc) is correct or wrong, I must first find out if we are talking about the same rules. Please correct me, if necessary, on each of the following rules:
1) Splitting 10,10 includes splitting any combination of 10 value cards (10,J,Q,K), so I would be allowed to split 10,K.
2) Splitting One time means up to a total of 2 hands, splitting two times means up to a total of 3 hands, and so on.
3) "infinite deck" means that I always have a 16/52 (.30769) chance of receiving a 10 value card (no removal effect). I know in an eight deck game, if I receive one 10 value card and the dealer receives a nine, then I have a 127/414 (.30676) chance of receiving a 2nd 10 value card on the next card dealt; but in an infinite deck game, in that same situation, I have a 16/52 (.30769) chance.
1) Splitting 10,10 includes splitting any combination of 10 value cards (10,J,Q,K), so I would be allowed to split 10,K.
2) Splitting One time means up to a total of 2 hands, splitting two times means up to a total of 3 hands, and so on.
3) "infinite deck" means that I always have a 16/52 (.30769) chance of receiving a 10 value card (no removal effect). I know in an eight deck game, if I receive one 10 value card and the dealer receives a nine, then I have a 127/414 (.30676) chance of receiving a 2nd 10 value card on the next card dealt; but in an infinite deck game, in that same situation, I have a 16/52 (.30769) chance.
January 6th, 2014 at 11:25:38 AM
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Quote: DavidYormark...Please correct me, if necessary, on each of the following rules:
1) Splitting 10,10 includes splitting any combination of 10 value cards (10,J,Q,K), so I would be allowed to split 10,K.
2) Splitting One time means up to a total of 2 hands, splitting two times means up to a total of 3 hands, and so on.
3) "infinite deck" means that I always have a 16/52 (.30769) chance of receiving a 10 value card (no removal effect). I know in an eight deck game, if I receive one 10 value card and the dealer receives a nine, then I have a 127/414 (.30676) chance of receiving a 2nd 10 value card on the next card dealt; but in an infinite deck game, in that same situation, I have a 16/52 (.30769) chance.
Yes. Splitting 10, 10 includes splitting any combination of 10 value cards. Splitting one time means splitting to two hands, splitting two times means splitting to three hands, etc. And "infinite deck" would mean that you always have a 16/52 chance of receiving a 10 value card.
January 6th, 2014 at 12:16:04 PM
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Quote: DavidYormarkThank you ChesterDog for pointing out to AxiomOfChoice that "infinite deck" does not mean "infinite splits".
I was just quoting the Wizard's earlier post in this thread:
Quote: WizardFor my infinite-deck analysis I think I assumed infinite re-splitting. So, if you split 55 vs 2 the first time I assume you don't suddenly realize your error and then double with another 5, but would keep making that mistake as long as you kept getting more fives.
January 6th, 2014 at 2:14:38 PM
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Now that we got the rules of this situation (splitting 10,10 vs. 9) straight, I will test out the EV both using a computer simulation (using an infinite deck with appendix 1 rules) and actually physically dealing out playing cards for at least 2000 hands (1000 hands with 1 deck and then 1000 hands with 8 decks). I will allow a maximum of 3 splits (total of 4 hands). My prediction is the only situation that could return a negative EV for the player is when I use 1 deck. With 8 decks or an infinite deck it doesn't matter how many times I resplit 10's because the effect of card removal is negligible in 8 deck and nil in infinite deck. In other words, AxiomOfChoice's previous comment to me that resplitting is exhausting the valuable 10's that I need to produce powerful 20's is significant only when one plays single deck blackjack where if you end up with the maximum of 4 hands you have reduced the total number of 10 value cards from 16 to 12. In 8 deck blackjack that same situation reduces the total number of 10 value cards from 128 to 124 which is too negligible a reduction to turn a winning situation (hard 10 vs 9) into a losing situation (-.2067). ChesterDog just told me that I always have a 16/52 chance of getting a 10 value card so my question to all of you is what difference does it make if I resplit 10's a million times if I always have the same chance (16/52) of getting another 10 for a powerful 20. There is no exhausting of the 10's because we have already agreed that we are using an infinite deck. I will report back here within a few days with all my results. In the meantime feel free to comment.
January 6th, 2014 at 2:19:22 PM
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Quote: DavidYormarkIn other words AxiomOfChoice's previous comment to me that resplitting is exhausting the valuable 10's that I need to produce powerful 20's
I never said that. My point was that if you split every time you draw another 10, you end up with no 20s (except for the rare time that you draw a stiff and then hit to a 20)
In other words, in the infinite-split case, you are throwing out your 20's. It has nothing to do with using up cards, it has to do with the fact that every time you get one you split it and re-draw.
In the split-to-4-hands case, you are throwing out your first 3 20's, should you be fortunate enough to get that many.
January 6th, 2014 at 8:57:53 PM
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Yes, with infinite splitting you can never end up with a 2 card 20, therefore the player is probably at a negative EV even with an infinite deck. In single deck with more than 1 split allowed, I would expect a negative EV. In single deck with infinite splitting, the dealer will destroy you. Remember, the Wiz's appendix 1 states infinite deck and up to 3 splits (max of 4 hands). Under those conditions I will be surprised if his -.2067 is correct. Believe me, I will find the correct answer within a week and will fully admit I am wrong if that's the case. Axiom, I see where you quote the Wiz, "For my infinite-deck analysis I think I assumed infinite re-splitting". If the Wiz did use infinite splitting then it contradicts the stated rules at the top of Appendix 1.
January 6th, 2014 at 10:23:48 PM
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Quote: DavidYormarkYes, with infinite splitting you can never end up with a 2 card 20, therefore the player is probably at a negative EV even with an infinite deck. In single deck with more than 1 split allowed, I would expect a negative EV. In single deck with infinite splitting, the dealer will destroy you. Remember, the Wiz's appendix 1 states infinite deck and up to 3 splits (max of 4 hands). Under those conditions I will be surprised if his -.2067 is correct. Believe me, I will find the correct answer within a week and will fully admit I am wrong if that's the case. Axiom, I see where you quote the Wiz, "For my infinite-deck analysis I think I assumed infinite re-splitting". If the Wiz did use infinite splitting then it contradicts the stated rules at the top of Appendix 1.
Because it's infinite-deck, this is all pretty simple to figure out combinatorially. A simulation is probably a waste of time (although, of course, you should get the same answer). I would do it, but, honestly, I trust the Wizard's numbers here.
-.2067 seems reasonable to me. Why do you think it's wrong? In the infinite-split case, you end up with a stiff 5/9 of the time, which is an underdog against a 9. Even with the split-to-4-hands case, you end up with a stiff a disproportionately large part of the time. It's very different from hitting a 10, because when you hit a 10 you keep your 20 when you get it... and that happens a lot (4/13 of the time) and obviously a 20 is a huge favorite over a 9.
Even in the split-to-4 case, it is nothing like hitting a 10. When you hit a 10 you get 20 4/13 of the time. That is a huge part of your EV (since you are a massive favorite with a 20 over a 10)
January 6th, 2014 at 10:57:54 PM
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My appendix 1 is based on a maximum of three splits. As mentioned before, I assume that once a a player splits the first time, he will keep doing it if he gets the opportunity to re-split, even if it is not the best play. However, the spreadsheet I link to at the bottom is based on infinite re-splits, to keep the spreadsheet as simple as possible.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 30th, 2014 at 1:26:10 PM
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I was wrong; you were right. The Wiz's and ChesterDog's EVs are correct. AxiomOfChoice made a good attempt to explain why, but it did not penetrate through my thick skull. My mind was stuck on the idea that whenever I split, I was creating another advantageous situation (hitting 10 vs. 9 for a .11653 EV). The fact is, hitting 10 vs. 9 is a .11653 EV, only if you stand on a 20 when you get one of the many 10 value cards in the deck. Obviously the Wiz's Appendix 1 (hit 10 vs. 9 EV .11653) applies to 8-2, 7-3, and 6-4 where one is not allowed to split (you can only split pairs). But that same EV applies to any total of 10 vs. 9 (5,5 or a single 10 value card after the first split) if you stand on 20 when you get another 10 value card. The more you split, the more you lower your EV (all the way down to a whopping -.60742 with infinite resplitting). It didn't matter if I dealt real cards or ran computer simulations; one deck or infinite deck. The only parameter that significantly effected EV was how many splits the player was allowed to take.
no splits .758357
one split .23306
two splits -.04047
three splits -.20673
infinite splits -.60742
I guess old age is setting in. Probability and Statistics can be very confusing. I apologize for my remarks about questioning the Wiz's reliability.
no splits .758357
one split .23306
two splits -.04047
three splits -.20673
infinite splits -.60742
I guess old age is setting in. Probability and Statistics can be very confusing. I apologize for my remarks about questioning the Wiz's reliability.
January 30th, 2014 at 1:29:08 PM
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If you play Blackjack with an infinite deck, who deals ? Easter Bunny or Santa Claus ?
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 1:31:30 PM
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Quote: BuzzardIf you play Blackjack with an infinite deck, who deals ? Easter Bunny or Santa Claus ?
You really need to give this one up.
January 30th, 2014 at 1:33:04 PM
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I will when I get an answer. Or do you play against an infinite deck ?
This is a serious question. I am patent pending on a poker game where the players use an infinite deck and the probabilities are exactly the same as those of a 52
deck card.
This is a serious question. I am patent pending on a poker game where the players use an infinite deck and the probabilities are exactly the same as those of a 52
deck card.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 1:36:35 PM
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Quote: BuzzardI will when I get an answer. Or do you play against an infinite deck ?
You have gotten an answer. Infinite deck analysis is often good for making quick and simple estimates, particularly when the effect of removal of any card is very small.
January 30th, 2014 at 1:37:48 PM
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Well, let me re-phase that. WOULD you play a poker game that used an infinite deck ?
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 1:41:03 PM
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Sure, why not? I've always wanted to get 7 of a kind....
January 30th, 2014 at 1:45:30 PM
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No, I said the probalities are the same. Same prob of starting hand, ranking of hands, etc. And I believe the probabilities of a 2 card hand winning will be approximately the same.
But need a nuclear calculator to figure it out.
220 X 1326 to the 9th power X 210 is a hell of a big number. That's just for starts.
But need a nuclear calculator to figure it out.
220 X 1326 to the 9th power X 210 is a hell of a big number. That's just for starts.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 2:32:05 PM
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Quote: BuzzardNo, I said the probalities are the same. Same prob of starting hand, ranking of hands, etc. And I believe the probabilities of a 2 card hand winning will be approximately the same.
But need a nuclear calculator to figure it out.
220 X 1326 to the 9th power X 210 is a hell of a big number. That's just for starts.
With an infinite deck, there is not the same probabilities of a starting hand.
A pair of aces: 4/52 * 3/51 with a normal deck = 1 in 221
With a infinite deck: 4/52 * 4/52 = 1 in 169
As all the probabilities of starting hands add up to 1, and if pairs are more common in an infinite deck, then unpaired hands must be less common.
Therefore not all starting hands have the same probability as they do with a normal deck. Assuming Texas Hold'em
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 30th, 2014 at 2:44:38 PM
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I beg to differ. With a proper filter, say that the player's second card could not be the same as his first card, the odds would remain 1 in 221. Now the probability for more than one of ten players having a pair of aces increases, but in my scenario, each player's probabilities of having a pair of Aces would remain 1 in 221.
Now for that nuclear calculator. All players in a 10 handed game is 221 to the 10th power I would think ? That's a pretty big number, isn't it.
Remember I am mathematically challenged.
Now for that nuclear calculator. All players in a 10 handed game is 221 to the 10th power I would think ? That's a pretty big number, isn't it.
Remember I am mathematically challenged.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 2:56:30 PM
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Okay, for the flop, can I deal any ace? Effectively you are saying each player has their own deck for their hole cards.
But the flop? Infinite deck? Then it means it's more likely to get trips or pair up your whole cards, as there are more cards of your suit left.
Thus, for Texas style games, at least, the game has changed.
Not sure why you need 1/222^10. That just gives you the chance of every player getting dealt a pair of aces in the hole.
But the flop? Infinite deck? Then it means it's more likely to get trips or pair up your whole cards, as there are more cards of your suit left.
Thus, for Texas style games, at least, the game has changed.
Not sure why you need 1/222^10. That just gives you the chance of every player getting dealt a pair of aces in the hole.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 30th, 2014 at 2:56:55 PM
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Quote: BuzzardIf you play Blackjack with an infinite deck, who deals ?
In infinite deck analysis no one deals any cards. That's the actual beauty, it's works like a 13-sided dice telling you which card you got. The second dealt card (and any other dealt card) has then the *same* probability of showing any value as the first or any other card in the game.
Yes - it is a different game, and should not be named "Blackjack" but "infinite deck Blackjack". The "infinite deck Blackjack" is *much* more simpler to calculate (for the very reasons above), and gives good *approximate* results compared to the real game of "Blackjack" whenever the card removal effect is not important.
Whether or not you can neglect card removal effect is a different story. For overall EV estimates card removal is important (because the Blackjack push significantly affects the overall EV, and the games EV is somewhat close to .5% - so any change on the scale of say .05% is significant).
But for decision making alone, i.e. finding the strategy when you should hit or stand, double down, split or surrender, the card removal effect is quite small (Note that there is no decision to make at a Blackjack push). Hence the card removal effect is small for decision making as the majority of hands all options give well varying EVs (say 10%), and picking the right option is still easy even if the estimated EV is off by a relatively small amount (say .05%).
January 30th, 2014 at 3:00:27 PM
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" Not sure why you need 1/222^10. That just gives you the chance of every player getting dealt a pair of aces in the hole."
That's just for starters.
That's just for starters.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 3:05:04 PM
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Quote: Buzzard" Not sure why you need 1/222^10. That just gives you the chance of every player getting dealt a pair of aces in the hole."
That's just for starters.
What are you trying to accomplish here exactly? I am so confused as to why you are quoting these numbers. Are you trying to calculate something? What's your point?
I'm not trying to give you a hard time; I just have no clue what we are talking about.
January 30th, 2014 at 3:05:20 PM
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' But the flop? Infinite deck? Then it means it's more likely to get trips or pair up your whole cards, as there are more cards of your suit left. "
Come on now, I said no 7 of a kind. Same filer removes all 10 cards that are in players hands, and removes from consideration each board card as it is dealt.
Probabilities of making a hand with your 2 cards remain the same.
But now the AQ in your hand that makes a flush might lose to another player's AK of the same suit.
But your starting hand probabilities, flop, turn , river, all are the same. Now what is 221 to the 10th, anybody ?
Come on now, I said no 7 of a kind. Same filer removes all 10 cards that are in players hands, and removes from consideration each board card as it is dealt.
Probabilities of making a hand with your 2 cards remain the same.
But now the AQ in your hand that makes a flush might lose to another player's AK of the same suit.
But your starting hand probabilities, flop, turn , river, all are the same. Now what is 221 to the 10th, anybody ?
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 3:06:19 PM
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Quote: Buzzard" Not sure why you need 1/222^10. That just gives you the chance of every player getting dealt a pair of aces in the hole."
That's just for starters.
Actually, it's pretty small : 1 in 277,927,878,692,682,183,940,201
But I'm not sure what you are trying to prove with it anyways. I am sure you do, though :)
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 30th, 2014 at 3:09:25 PM
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Quote: Buzzard' But the flop? Infinite deck? Then it means it's more likely to get trips or pair up your whole cards, as there are more cards of your suit left. "
Come on now, I said no 7 of a kind. Same filer removes all 10 cards that are in players hands, and removes from consideration each board card as it is dealt.
It's no longer an infinite deck, then. It's a finite deck.
Quote:
Probabilities of making a hand with your 2 cards remain the same.
Yes, I think that is true, as there is the same number of unknown cards.
Quote:
But now the AQ in your hand that makes a flush might lose to another player's AK of the same suit.
Check out Pacific Poker from the BC lottery. That game is much as you describe.
Quote:But your starting hand probabilities, flop, turn , river, all are the same. Now what is 221 to the 10th, anybody ?
See above.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 30th, 2014 at 3:09:37 PM
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Thanks for the number LOL Now how big a calculator do I need to see if Aces will win the same percentage of hands as with a real deck of cards.
I think it 30% in 10 handed game, with the assumption everybody stays in all the way.
I think it 30% in 10 handed game, with the assumption everybody stays in all the way.
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 30th, 2014 at 3:10:49 PM
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Does anyone else get the idea that Buzz is just trying to trick us into doing free math for his new game?
I want a cut of the royalties. Or, at least, I want to be invited to your private island.
I want a cut of the royalties. Or, at least, I want to be invited to your private island.
January 30th, 2014 at 3:13:15 PM
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It's still an infinite deck, just with a filter. That's why all 10 players can have Aces. Now for all 10 players to have same Aces, only way all 10 can then have quad Aces,
it's like 221 X 1326 to the 9th, I THINK ?
it's like 221 X 1326 to the 9th, I THINK ?
Shed not for her
the bitter tear
Nor give the heart
to vain regret
Tis but the casket
that lies here,
The gem that filled it
Sparkles yet
January 31st, 2014 at 12:04:32 AM
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As I said, this game you describe is very similar to Pacific Poker. I suggest you and your patent attorney check it out to see if it is the same. Here the house cards, flop, turn and river are not copied. The other 39 cards can be dealt out to the province wide players in any combination.
I would hesitate to call it 'an infinite deck', as the next card out of the deck does depend on previous dealt cards, so there are effects of card removal. But whatever you want to call it.
As two players may have quad aces if they are both dealt a pair of aces, the value of your aces must necessarily be smaller... then is a 1 in X chance of turning pocket bullets into Quads, and the value of my pocket aces against all other hands depends on that quads not being matched.
If Played Y also has the same 1 in X chance of turning pocket Aces into Quads, I will tie with him. Thus, split the pot.
Therefore the value of the hand is lower. Basically some of my outright wins for any two starting cards will become splits. Thus, worth less overall. There's no comparative gain on the other side where the counterfeited cards can help me win hands that I would not have otherwise. As you also stated, my AQs could hit a flush. In a non-counterfeited game, it will be good. In this game, AKs will beat it. So again, with everyone staying in, the AQs is worth less overall, and loses more often that in the straight game. More often may not be a very big difference, but it is different.
Therefore, my conclusion is that with this game, as there are NOT removed cards, the results when playing against each other are not the same as in the normal game of Texas Hold Em.
I would hesitate to call it 'an infinite deck', as the next card out of the deck does depend on previous dealt cards, so there are effects of card removal. But whatever you want to call it.
As two players may have quad aces if they are both dealt a pair of aces, the value of your aces must necessarily be smaller... then is a 1 in X chance of turning pocket bullets into Quads, and the value of my pocket aces against all other hands depends on that quads not being matched.
If Played Y also has the same 1 in X chance of turning pocket Aces into Quads, I will tie with him. Thus, split the pot.
Therefore the value of the hand is lower. Basically some of my outright wins for any two starting cards will become splits. Thus, worth less overall. There's no comparative gain on the other side where the counterfeited cards can help me win hands that I would not have otherwise. As you also stated, my AQs could hit a flush. In a non-counterfeited game, it will be good. In this game, AKs will beat it. So again, with everyone staying in, the AQs is worth less overall, and loses more often that in the straight game. More often may not be a very big difference, but it is different.
Therefore, my conclusion is that with this game, as there are NOT removed cards, the results when playing against each other are not the same as in the normal game of Texas Hold Em.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
