June 19th, 2010 at 7:08:20 PM
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I'm hoping the fine members of this forum can help me out with a question that I have been working on for a while in my spare time.

Appendix 1 from the Wizard's terrific wizardofodds site shows the Expected Return (ER) for every possible play in Blackjack. It contains the following description:

"To use this table look up the returns for any given play, the one with the greatest return is the best play. For

example suppose you have two 8's and the dealer has a 10. The return by standing is -0.5404, by hitting is -0.5398,

doubling is -1.0797, and by splitting is -0.4807. So splitting 8's you stand to lose the least, 48.07 cents per original dollar bet, and is thus the best play."

I wanted to know how to calculate the percentage of time a person will win with any given play. It is my

understanding that the expected return is calculated with the following formula where W% = Win Percentage and L% = Loss Percentage:

ER = (payoff * W%) - (bet * L%)

For example, for playing a single number on Roulette, with a payoff of 35 to 1, the ER per dollar bet is:

ER = (35 * (1/38)) - (1 * (37/38)) = 35/38 - 37/38 = -2/38 = -1/19 = -0.0526

So, for Standing and Hitting plays in Blackjack, I assume the ER formula is:

ER = (1 * W%) - (1 * L%)

By replacing L% with (1 - W%):

ER = (1 * W%) - (1 * (1 - W%)) = 1W% - 1 + W% = 2W% - 1

ER = 2W% - 1

By rearranging, I can determine the formula for W%:

W% = (ER + 1)/2

Since the bet and payoff are doubled for Double Down and Splitting plays, I assume the ER formula for these plays is:

ER = (2 * W%) - (2 * L%)

ER = 2W% - 2(1 - W%)

ER = 4W% - 2

And again, by rearranging, I can determine the formula for W%:

W% = (ER + 2)/4

Using these determined formulas, the W% of the possible plays for two 8's versus a dealer's 10 can be calculated:

Standing W% = (-.5404 + 1)/2 = 22.98%

Hitting W% = (-.5398 + 1)/2 = 23.01%

Doubling W% = (-1.0797 + 2)/4 = 23.01%

Splitting W% = (-.4807 + 2)/4 = 37.98%

However, if these formulas for calculating the percentage of time we will win with a particular play are correct, it means that the play with the greatest ER may not always necessarily be the play with the greatest W%.

For example, let's take something like a player's 11 against a dealer's 7. From the Wizard's appendix, the ER for hitting is 0.2921 and the ER for Doubling Down is 0.4629, indicating that Doubling Down is the best play.

However, looking at W% (assuming everything above is correct):

Hitting W% = (.2921 + 1)/2 = 64.61%

Doubling W% = (.4629 + 2)/4 = 61.57%

This indicates to me that although you win more money in the long run by Doubling Down on this play (.4629 > .2921), you slightly decrease your chances of winning that one particular hand by doing so (61.57% < 64.61%). This makes perfect sense to me, because you can only take one additional card when you Double Down. If you pull a 2 for a total of 13, you are stuck standing with a 13 against a dealer's 7, which is a situation where you would definitely take another card if you could do so.

I am looking for confirmation that these situations can and do occur. I can take that question up with a different forum, but, my conclusion is only correct if my math is correct. I'm hoping the members of this forum (or the great Wizard himself) can let me know if I have made a mistake here.

Thanks!

Appendix 1 from the Wizard's terrific wizardofodds site shows the Expected Return (ER) for every possible play in Blackjack. It contains the following description:

"To use this table look up the returns for any given play, the one with the greatest return is the best play. For

example suppose you have two 8's and the dealer has a 10. The return by standing is -0.5404, by hitting is -0.5398,

doubling is -1.0797, and by splitting is -0.4807. So splitting 8's you stand to lose the least, 48.07 cents per original dollar bet, and is thus the best play."

I wanted to know how to calculate the percentage of time a person will win with any given play. It is my

understanding that the expected return is calculated with the following formula where W% = Win Percentage and L% = Loss Percentage:

ER = (payoff * W%) - (bet * L%)

For example, for playing a single number on Roulette, with a payoff of 35 to 1, the ER per dollar bet is:

ER = (35 * (1/38)) - (1 * (37/38)) = 35/38 - 37/38 = -2/38 = -1/19 = -0.0526

So, for Standing and Hitting plays in Blackjack, I assume the ER formula is:

ER = (1 * W%) - (1 * L%)

By replacing L% with (1 - W%):

ER = (1 * W%) - (1 * (1 - W%)) = 1W% - 1 + W% = 2W% - 1

ER = 2W% - 1

By rearranging, I can determine the formula for W%:

W% = (ER + 1)/2

Since the bet and payoff are doubled for Double Down and Splitting plays, I assume the ER formula for these plays is:

ER = (2 * W%) - (2 * L%)

ER = 2W% - 2(1 - W%)

ER = 4W% - 2

And again, by rearranging, I can determine the formula for W%:

W% = (ER + 2)/4

Using these determined formulas, the W% of the possible plays for two 8's versus a dealer's 10 can be calculated:

Standing W% = (-.5404 + 1)/2 = 22.98%

Hitting W% = (-.5398 + 1)/2 = 23.01%

Doubling W% = (-1.0797 + 2)/4 = 23.01%

Splitting W% = (-.4807 + 2)/4 = 37.98%

However, if these formulas for calculating the percentage of time we will win with a particular play are correct, it means that the play with the greatest ER may not always necessarily be the play with the greatest W%.

For example, let's take something like a player's 11 against a dealer's 7. From the Wizard's appendix, the ER for hitting is 0.2921 and the ER for Doubling Down is 0.4629, indicating that Doubling Down is the best play.

However, looking at W% (assuming everything above is correct):

Hitting W% = (.2921 + 1)/2 = 64.61%

Doubling W% = (.4629 + 2)/4 = 61.57%

This indicates to me that although you win more money in the long run by Doubling Down on this play (.4629 > .2921), you slightly decrease your chances of winning that one particular hand by doing so (61.57% < 64.61%). This makes perfect sense to me, because you can only take one additional card when you Double Down. If you pull a 2 for a total of 13, you are stuck standing with a 13 against a dealer's 7, which is a situation where you would definitely take another card if you could do so.

I am looking for confirmation that these situations can and do occur. I can take that question up with a different forum, but, my conclusion is only correct if my math is correct. I'm hoping the members of this forum (or the great Wizard himself) can let me know if I have made a mistake here.

Thanks!

June 19th, 2010 at 7:12:01 PM
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deleted

June 19th, 2010 at 7:29:03 PM
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I didn't make it all the way thru your post, but I think

ER = (2 * W%) - (1 * L%)

should be

ER = (1 * W%) - (1 * L%)

You stand to gain one bet from a win, and lose one from a loss. The win should not include the return of your original bet. This leads to...

W% = (ER + 1)/2

So for an expected return of 0, W% = L% = 50%.

I'm guessing you'll need to make similar adjustments through the rest of your analysis.

ER = (2 * W%) - (1 * L%)

should be

ER = (1 * W%) - (1 * L%)

You stand to gain one bet from a win, and lose one from a loss. The win should not include the return of your original bet. This leads to...

W% = (ER + 1)/2

So for an expected return of 0, W% = L% = 50%.

I'm guessing you'll need to make similar adjustments through the rest of your analysis.

June 20th, 2010 at 5:52:18 AM
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PapaChubby is correct. A positive ER should equate to a Win Percentage greater than 50%.

I have corrected the original post.

I have corrected the original post.

June 20th, 2010 at 10:39:59 AM
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I think your assessment is correct. It makes sense to me that sometimes doubling or splitting will decrease your likelihood of winning the hand(s). But if your likelihood is still above 50%, it is desirable to get as much money as you can on the table. Doubling or splitting on these occasions is not something you would choose to do, except for the fact that you get to double your bet in the process.

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