Thank you for providing the wizard of odds website - it is excellent. My questions relate to the advantage an Ace gives. I understand that if I am dealt an Ace my expected return for that hand increases significantly. My first question would be if I am dealt an Ace for one of my two cards and my initial stake is $10, what would my expected return be for that hand?
My second question assumes that the answer to the first question is a significantly positive return. Since having an Ace in my hand significantly improves my hands return, would I increase my overall return of play by playing multiple hands? My reasoning goes along these lines, if I am playing one hand both myself and the dealer have an equal chance of receiving an Ace. If I were to play four hands, I have a fourfold chance of receiving an Ace in one of my hands compared to the dealer, so there is a very good chance one of my hands has a significantly positive return.
I've had a read through most of the questions and answers, hopefully I haven't missed this being addressed there. I appreciate any time you spend answering this question.
Jimmyfrez
No, it would not help to bet multiple hands. While you'll get more aces, you'll also get more lousy hands. It averages out in the end.
I understand - playing multiple hands is similar to making multiple bets in roulette. I've also got a couple of questions relating to counting cards. Assuming you've been counting and you've got a significant positive count, I understand the normal strategy is to increase your bet, the idea being that your betting more money when the odds are in your favour, and less when the odds are against you. Would it have the same effect to go from playing one hand to playing two or more hands when the odds are in your favour? Assuming this is correct, I would have thought there would be some advantages to doing it this way. I'd assume this would give you less volatility in your bankroll, and possibly attract less attention than increasing your bet. Perhaps a combination of increasing your bet and increasing the number of hands played would be the optimal strategy?
I am very interested in your Ace-Five method, as I believe it shows the power of the Ace (not only increases your chance of winning but also increases your chance of an increased payout). Without modifying basic strategy it still gives you approx a 0.3% edge. On one of your pages I remember reading an approximate breakdown of the advantages of counting (I think it was based on using the High Low method). In the breakdown it appeared that most of the advantage came from modifying basic strategy i.e. standing as opposed to hitting, and taking insurance. Given that the Ace-Five method gives almost as much of an edge (using a more normal spread of say 1-8), my conclusion is that the Ace-Five method is actually a better predictor of when to increase your bet, as it doesn't have any of the advantage of modifying basic strategy. I also conclude that if you could use the Ace-Five method for betting spread, as well as a separate ten count for modifying basic strategy you could obtain quite a distinct advantage. I also would have thought that keeping a ten count (against 2,3,4 & 6) would be more powerful for modifying basic strategy than using the high-low for most of the major strategy modifications. Would that be correct?
One last question. I've noticed that for the high-low method apparently the single most powerful strategy modification is to take insurance when the true count is +3 or higher. I just wanted to run through the math. By my (probably flawed) reasoning when the true count is +3 it means that on average the remaining decks have 16 tens but you would only be dividing by 49, rather than 52 (ignoring the Ace), which only gives 32.65% chance of drawing a ten. If you take the Ace into account in the count it is worse. Where am I going wrong with this calc?
Once again, brilliant website, one of the few I've seen that I put a lot of faith in.
Jimmyfrez
Quote: jimmyfrez
One last question. I've noticed that for the high-low method apparently the single most powerful strategy modification is to take insurance when the true count is +3 or higher. I just wanted to run through the math. By my (probably flawed) reasoning when the true count is +3 it means that on average the remaining decks have 16 tens but you would only be dividing by 49, rather than 52 (ignoring the Ace), which only gives 32.65% chance of drawing a ten. If you take the Ace into account in the count it is worse. Where am I going wrong with this calc?
For the insurance payout in Hi-Low, if the true count is +3, that means that in a deck of the NEXT 52 cards played in a shoe, on average, there are 17.5 10s and 14.5 3s to 6s.
17.5/52 = .336538
Expected insurance payout: .336528 x 2 - (1-.336528) = .009615 units (Player Advantage).
While your math is exactly right, you are pulling an exception to the rule in making the assumption that you are receiving no other cards. For example, if the count is +3, you know you have more 10s than 3s to 6s. But did you take into account the 2s, 7-9s and Aces played as well? The counting method is an approximation and you did an exact calculation.
That makes sense now. I did realise it was a little bit approximate, but knew I had to be going wrong with my calculation as I was well under 33.33%.
Also in response to one of my own questions where I ask whether playing two hands when the count is good as opposed to doubling your bet gives the same expected result. After thinking it through, I think the answer to this is no. I think by doing this it would increase the percentage of hands played when the count is in +ve return and therefore decreases the percentage of hands played when the count is in -ve return, so I think this would improve your results as compared to just playing straight basic strategy, but not as much as you would by increasing your bet. However feel free to correct me if I'm wrong, and it would be good to hear from someone a bit more advanced whether they agree with this assessment. Also if my reasoning is correct you could gain more of an advantage by doing both, that is changing to playing two hands as well as increasing your bet than you would by just increasing your bet?
I never have to worry about playing multiple hands at blackjack.
Its all I can do to muster the skills to play one hand. Making that poor dealer add up additional hands would be cruel, so I just play one hand.
Quote: jimmyfrezThanks Boymimbo,
That makes sense now. I did realise it was a little bit approximate, but knew I had to be going wrong with my calculation as I was well under 33.33%.
Also in response to one of my own questions where I ask whether playing two hands when the count is good as opposed to doubling your bet gives the same expected result. After thinking it through, I think the answer to this is no. I think by doing this it would increase the percentage of hands played when the count is in +ve return and therefore decreases the percentage of hands played when the count is in -ve return, so I think this would improve your results as compared to just playing straight basic strategy, but not as much as you would by increasing your bet. However feel free to correct me if I'm wrong, and it would be good to hear from someone a bit more advanced whether they agree with this assessment. Also if my reasoning is correct you could gain more of an advantage by doing both, that is changing to playing two hands as well as increasing your bet than you would by just increasing your bet?
I believe playing multiple hands does have a similar effect as bet increase when the count is high - and it is essetially wonging in when the count is high, and wonging out when it is low. When the count is high, there is an equal chance that you or the dealer will receive a BJ (higher than a low or neutral count), so increasing the hands you are playing increases your chances when the chances are good.