Calculating experimental return from split/double-down hands...
June 4th, 2015 at 4:22:39 PM
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Hi all, new forum member here...
I'm working on my own Monte Carlo simulation, and referring to well-known tables of expected value such as found in Appendix 5 at Wizard of Odds in order to compare my results with the well-established ones.
One thing that I'm not too sure of is if I'm recording the percent return correctly for a completed hand in which you split and/or doubled-down. I'm seeing significant deviance from the published tables, to the player's detriment, and I'm not sure if that's due to a bug in the simulations, or just calculating the return wrong for such hands.
My gut tells me that the way to calculate the percent return for such a hand is (units won)/(units wagered)-1, where both units won (which includes the total wager in the case of a win or push) and units wagered is the total ending wager after splitting/doubling. That is, I include the additional bet made in order to split or double down in the "units wagered" place.
Some sources are ambiguous about this, and could be interpreted to suggest that the correct formula would be (units won)/(original bet)-1, where original bet doesn't take into account the extra wager(s) you made to split or double. This doesn't seem correct to me, as this would seem to magnify your apparent return and minimize your variance; you would often appear to win multiple times your original bet, but never appear to lose more than 100% of it. Surely by increasing your amount at risk that value has to be included in the percent return formula.
Can someone confirm that the first formula is the correct one, or if there is a different way of calculating this that's not obvious to me?
I'm working on my own Monte Carlo simulation, and referring to well-known tables of expected value such as found in Appendix 5 at Wizard of Odds in order to compare my results with the well-established ones.
One thing that I'm not too sure of is if I'm recording the percent return correctly for a completed hand in which you split and/or doubled-down. I'm seeing significant deviance from the published tables, to the player's detriment, and I'm not sure if that's due to a bug in the simulations, or just calculating the return wrong for such hands.
My gut tells me that the way to calculate the percent return for such a hand is (units won)/(units wagered)-1, where both units won (which includes the total wager in the case of a win or push) and units wagered is the total ending wager after splitting/doubling. That is, I include the additional bet made in order to split or double down in the "units wagered" place.
Some sources are ambiguous about this, and could be interpreted to suggest that the correct formula would be (units won)/(original bet)-1, where original bet doesn't take into account the extra wager(s) you made to split or double. This doesn't seem correct to me, as this would seem to magnify your apparent return and minimize your variance; you would often appear to win multiple times your original bet, but never appear to lose more than 100% of it. Surely by increasing your amount at risk that value has to be included in the percent return formula.
Can someone confirm that the first formula is the correct one, or if there is a different way of calculating this that's not obvious to me?
June 4th, 2015 at 5:46:46 PM
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(total units won or lost)/(total initial wagers)
Even a game like Ultimate Texas Hold'em that has a requirement for two initial wagers (Ante & Blind) has an edge based on a one-unit wager. That is, if you wager $25 on Ante and $25 on Blind then the 2.158% edge applies to $25, not $50.
Even a game like Ultimate Texas Hold'em that has a requirement for two initial wagers (Ante & Blind) has an edge based on a one-unit wager. That is, if you wager $25 on Ante and $25 on Blind then the 2.158% edge applies to $25, not $50.
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June 4th, 2015 at 6:57:28 PM
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Quote: matunosHi all, new forum member here...
I'm working on my own Monte Carlo simulation, and referring to well-known tables of expected value such as found in Appendix 5 at Wizard of Odds in order to compare my results with the well-established ones.
One thing that I'm not too sure of is if I'm recording the percent return correctly for a completed hand in which you split and/or doubled-down. I'm seeing significant deviance from the published tables, to the player's detriment, and I'm not sure if that's due to a bug in the simulations, or just calculating the return wrong for such hands.
First of all, welcome to the forums!
A double-down hand is almost trivial to analyze by combination mathematics (or 'analytically') so getting exact agreement with Appendix 5 results for double-down hands is a very appropriate way to check your simulations.
Splitting pairs, however, is the last frontier for rigorously calculating composition-dependent blackjack probabilities. Calculation of outcomes for scenarios where pairs -and especially pairs of low rank (22-66) - are split and resplit is a methodological nightmare. The Wizard uses a recurrence methodology which is approximate, not rigorous, to provide outcomes for Splitting pairs - but he is the best person to explain his own work, so I won't try to do that. However, If your MC simulation results get to the point at which they reproduce the Appendix 5 Double, Hit and Stand calculations for all other types of hands, but still disagree on the split pairs hands then please realize that many of us would be interested in hearing more!
Good luck with your work and we look forward to chatting more with you.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
June 4th, 2015 at 7:06:22 PM
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If my initial bet is 1 unit, I am dealt an A,8 versus a dealer's 8. I double down, per basic strategy.
If I win, I'm paid 2 units, plus I get back my wager, which was 2 units (original bet plus double down wager). For the purposes of the formula you've given, have I won 2 units or 4 units?
Similarly, if the hand results in a push, should the numerator in the formula you've given be 0 or 2? And finally if I lose, is it -2 or 0?
I'm normally pretty good with probabilities and expected values, but my brain is getting hung up on how the additional wager(s) are being accounted for.
If I win, I'm paid 2 units, plus I get back my wager, which was 2 units (original bet plus double down wager). For the purposes of the formula you've given, have I won 2 units or 4 units?
Similarly, if the hand results in a push, should the numerator in the formula you've given be 0 or 2? And finally if I lose, is it -2 or 0?
I'm normally pretty good with probabilities and expected values, but my brain is getting hung up on how the additional wager(s) are being accounted for.
June 4th, 2015 at 7:43:03 PM
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Quote: matunosI double down, per basic strategy.
If I win, I'm paid 2 units, plus I get back my wager, which was 2 units (original bet plus double down wager). For the purposes of the formula you've given, have I won 2 units or 4 units?
+2 The outcome is that your bankroll has gained 2 units.
Quote: matunosSimilarly, if the hand results in a push, should the numerator in the formula you've given be 0 or 2?
0 You neither win nor lose money.
Quote: matunosAnd finally if I lose, is it -2 or 0?
-2. You lose two bets
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
June 5th, 2015 at 12:31:53 PM
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Ev is calculated as (units won)/(original bet)
Units won means Units won not your 'payout'.
Ie you bet 1 , you get back 2, your win is 1. Ev is 1/1=100%
Ie, you bet 1, double and get back 4. Win is 2. Ev is 2/1=200%
Your formula of
(units won)/(units wagered)-1 where units won means payout is wrong.
It will give you the wrong results (both absolute and comparative between Double and Hit)
In all Double BS plays where if you did not double and hit you might hit also a 2nd time, it will Always give you the wrong comparative result.
For example Hit and Double 9 v 6.
With your Formula Hit will give a higher Ev than Double.
Because when you Double you actualy will less times than if you hit. Reason being that if you hit and got a 2 you would hit again whereas double you are stuck on 11.
Double has higher Ev though because even though you win less times, you win Double the amount which compesates over and above that.
Units won means Units won not your 'payout'.
Ie you bet 1 , you get back 2, your win is 1. Ev is 1/1=100%
Ie, you bet 1, double and get back 4. Win is 2. Ev is 2/1=200%
Your formula of
(units won)/(units wagered)-1 where units won means payout is wrong.
It will give you the wrong results (both absolute and comparative between Double and Hit)
In all Double BS plays where if you did not double and hit you might hit also a 2nd time, it will Always give you the wrong comparative result.
For example Hit and Double 9 v 6.
With your Formula Hit will give a higher Ev than Double.
Because when you Double you actualy will less times than if you hit. Reason being that if you hit and got a 2 you would hit again whereas double you are stuck on 11.
Double has higher Ev though because even though you win less times, you win Double the amount which compesates over and above that.
June 8th, 2015 at 5:37:49 PM
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Hi,
Yes, I think my issue is in trying to calculate expected *return* up front and aggregating that, as opposed to aggregating expected *value*, after which it's more clear why I would divide by the cost to play a hand (since I have now calculated the [approximate] expected value of all possible outcomes of playing a hand).
For analyzing which play is better in a given scenario, I suspect it would be more correct to analyze at each decision point. E.g. if I am doubling, my initial bet is already made; I am now trying to decide if the additional expected value obtained from doubling is worth the cost to double. Thus, I would calculate the expected value of standing and hitting (both of which are "free") versus the expected value of doubling, factoring in the additional cost. What I want to make sure of is that I'm not making doubling (or splitting) appear more valuable than they are, since I am taking on the risk of greater loss.
As an aside, I've also realized that I was only counting the end results against the initial hand, which means I'm not recording results against intermediate hands (e.g. hard totals of more than 2 cards, individual results for split hands, etc.), which seems like a bigger gap I need to cover.
At any rate, thanks for your help!
Yes, I think my issue is in trying to calculate expected *return* up front and aggregating that, as opposed to aggregating expected *value*, after which it's more clear why I would divide by the cost to play a hand (since I have now calculated the [approximate] expected value of all possible outcomes of playing a hand).
For analyzing which play is better in a given scenario, I suspect it would be more correct to analyze at each decision point. E.g. if I am doubling, my initial bet is already made; I am now trying to decide if the additional expected value obtained from doubling is worth the cost to double. Thus, I would calculate the expected value of standing and hitting (both of which are "free") versus the expected value of doubling, factoring in the additional cost. What I want to make sure of is that I'm not making doubling (or splitting) appear more valuable than they are, since I am taking on the risk of greater loss.
As an aside, I've also realized that I was only counting the end results against the initial hand, which means I'm not recording results against intermediate hands (e.g. hard totals of more than 2 cards, individual results for split hands, etc.), which seems like a bigger gap I need to cover.
At any rate, thanks for your help!
