Hawksed
• Posts: 7
Joined: Dec 19, 2009
December 28th, 2009 at 8:37:49 PM permalink
I would like to compute my average hourly wagers at a \$10 Blackjack table assuming I consistently bet the minimum but take advantage of every split and double down opportunity. Assume the basic strategy is followed with these casino rules: 6 decks, doubling after spits allowed, surrender not allowed, and dealer hits on soft 17.
boymimbo
• Posts: 5994
Joined: Nov 12, 2009
December 29th, 2009 at 9:02:03 AM permalink
Using the Wizard's basic strategy chart and allowing for an infinite deck, on the initial two cards, you will play an average of 1.1256 units. This is before you take into account additional bets due to splitting. That will take a bit of work to figure out, but since you can only split 56 of the 2197 possible dealer-player combinations (or 2.6%, less if you figure in dealer blackjacks) and we figure you double and split 1/3rd of the split hands as a rough guestimate, you get an average bet of 1.1342 units as an approximation.

Anyone else want to contribute?
----- You want the truth! You can't handle the truth!
Hawksed
• Posts: 7
Joined: Dec 19, 2009
December 29th, 2009 at 7:09:02 PM permalink
Thanks for your helpful response. So if I assume that 70 hands per hour are played, then my average expected cash outlay is 1.1342 x 70 x \$10 which equals \$793.94. If the casino edge is .58% than I can expect to lose \$4.60 per hour over the long haul. Is my math correct?
boymimbo
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Joined: Nov 12, 2009
December 29th, 2009 at 8:49:27 PM permalink
I don't think so. I think the House Advantage is based on the initial bet and not the sum of all bets. So, for this calculation, I don't think you need to think about the average amount bet based on doubling and hitting. The Expected Value is based on the initial bet and the house advantage is based on the sum of all outcomes based on the optimal strategy.

But I'm only 90% sure.
----- You want the truth! You can't handle the truth!
pocketaces
• Posts: 158
Joined: Nov 11, 2009
December 29th, 2009 at 9:56:55 PM permalink
Quote: boymimbo

I don't think so. I think the House Advantage is based on the initial bet and not the sum of all bets. So, for this calculation, I don't think you need to think about the average amount bet based on doubling and hitting. The Expected Value is based on the initial bet and the house advantage is based on the sum of all outcomes based on the optimal strategy.

But I'm only 90% sure.

This is correct. The other figure which the OP is alluding to is called the 'element of risk'. In blackjack (and in all other games involving raises) the element of risk is less than the house edge. Multipliing the element of risk by the average bet including raises will yield the same expected loss as the traditional method, which uses the house edge.

According to the Wizard's house edge comparison page, in blackjack the element of risk is about 0.05 percentage points less than the house edge. This is assuming Atlantic City rules (S17 8-deck game). With the non-surrender H17 game the average bet will be a bit larger and the element of risk a bit lower (relative to the house edge) than the S17 game.

Of course both the house edge and element of risk will be overall higher under the inferior H17 rules, but basic strategy does call for a bit more doubling in the H17 game. Hence the discrepancy.
Last edited by: pocketaces on Dec 30, 2009
Hawksed
• Posts: 7
Joined: Dec 19, 2009
December 30th, 2009 at 10:25:43 AM permalink
Forgive me for being mathematically challenged. But if I understand you correctly, all I need to do is use element of risk (.53) instead of the house edge (.58) to calculate my expected average loss per hour of play. So then the average bet of \$11.342 (\$10 x 1.1342) times 70 games played per hour times the element of risk yields \$4.21 in average hourly losses.
pocketaces
• Posts: 158
Joined: Nov 11, 2009
December 30th, 2009 at 11:32:52 AM permalink
Quote: Hawksed

Forgive me for being mathematically challenged. But if I understand you correctly, all I need to do is use element of risk (.53) instead of the house edge (.58) to calculate my expected average loss per hour of play. So then the average bet of \$11.342 (\$10 x 1.1342) times 70 games played per hour times the element of risk yields \$4.21 in average hourly losses.

Assuming the 0.58% house edge, I think the element of risk would be more like 0.51%. As I said we cannot simply apply the 0.05 spread to the game with the higher house edge. Its close but it needs to be adjusted. Using your estimate of an 11.342 average bet I get an expected loss of \$4.11

But what we have been saying is figuring this out is unnecessary. The house edge figure tells us all we need to know. We multiply your starting bet by the number of hands and the house edge. This gives us an expected loss of \$4.06. Close to our other figure, which suggests the element of risk and/or average bet is only slightly off.

Note that the house edge accounts fully for taking every double and split possibility dictated by basic strategy. There is no need to worry about the average bet when doing any expected value calculation. Because they are not widely available for all rule sets, we needed to estimate both figures. There is no need to do so using the simpler method involving the house edge. The house edge for just about any rules has already been calculated and available to use.
Hawksed