Would you rather play at True +5...

Assuming you could play only one spot, with no bet limit, and got no heat, would you rather play at True +5 for 10 rounds, or at True +10 for 5 rounds?

RomaneeConti

RomaneeConti,

Since you didn't specify the conditions, I assumed a heads-up DD, S17, DAS game with 75% pen for a HiLo counter.

According to CVCX, at +5 he enjoys an edge of 3.00% with a standard deviation of 1.177, while at +10 he enjoys an edge of 6.26% with a standard deviation of 1.222. Note that these numbers will all change significantly if we change the conditions.

Case (a): Thus, for 10 bets (say of M5 units each) at +5 he will earn 10*M5*3.00%/100% = 0.300*M5, with a std. dev. of 1.177*M5*sqrt(10) = 3.722*M5. Thus, about 68% of the time he'll end up between -3.422*M5 and +4.022*M5.

Case (b): Conversely, for 5 bets (say of M10 units each) at +10 he will earn 5*M10*6.26%/100% = 0.313*M10, with a std. dev. of 1.222*M10*sqrt(5) = 2.732*M10. Thus, about 68% of the time he'll end up between -2.419*M10 and +3.045*M5.

Now we have to assume something about his bet sizing. I see three logical alternatives:

1. Same max bet, so M5=M10. This is common for HiLo, since in most cases the HiLo counter maxes out at +4. Here, Case (b) wins, due to the higher EV (0.313 max bets vs. 0.3 max bets) and the lower volatility.

2. M10 = 2*M5. Here, the counter simply keeps raising his bet with the count. This is the strategy suggested by Katrina Walker in her Spanish 21 book. Now, if we rewrite the Case (a) results in terms of M10, the EV becomes 0.15*M10 and the 1-std.-dev. range becomes -1.711*M10 to +2.011*M10. Now we see that Case (b) has a higher EV (0.313 vs. 0.15) but also a higher volatility.

3. Optimal Betting. According to CVCX, under these very favorable conditions, the player should place his max bet at +7, and his +5 bet should be 80.8% of max: thus, if the max bet is $1000, the +5 bet should be $808. Therefore, here M5 = 0.808*M10. This case will be similar to #2 above: Case (b) will have both higher EV and higher volatility.

If you're asking for opinions, I'd go with Case (b): the 5 +10 bets.

Hope this helps!

Dog Hand

Assuming you could play only one spot, with no bet limit, and got no heat, would you rather play at True +5 for 10 rounds, or at True +10 for 5 rounds?

According to bjstats.com, for 6 deck, s17, das, ls, 75%, the advantages are 7.093% and 3.366% for TC = +10 and +5, respectively. The standard deviations are 1.178 and 1.160. Therefore, using optimal betting the EVs are (optimal bet = bankroll * advantage / variance):

+10: BR * 5 * 7.093% / (1.178^2) = BR * 25.56%
+5: BR * 10 * 3.366% / (1.160^2) = BR * 25.01%

I forgot to actually make the bets!

My numbers should be:

+10: BR * 5 * 7.093% * 7.093% / (1.178^2) = BR * 1.81%
+5: BR * 10 * 3.366% * 3.366% / (1.160^2) = BR * 0.84%

So five hands at +10 is more than twice as good as 10 hands at +5.

The reason the advantage is included twice is because it included in the optimal bet calculation as well as the EV calculation.

optimal bet = bankroll * advantage / variance
EV = bet * advantage

"Case (b): Conversely, for 5 bets (say of M10 units each) at +10 he will earn 5*M10*6.26%/100% = 0.313*M10, with a std. dev. of 1.222*M10*sqrt(5) = 2.732*M10. Thus, about 68% of the time he'll end up between -2.419*M10 and +3.045*M5."

The very last entry should be M10, not M5. (I'll let "Katarina" slide! :-))

Nice job!

Don

Theory indicates that in positive expectation situations that the swings will be less. Therefore at high plus counts it would be expected that the probabality of winning sucessive bets in a row (a streak) would be positive. To gain the most positive expectation my suggestion would be to divide the investment into 8 (allowing for splits & doubles) and then parley each winning bet until the end of the 10 trials. I know I haven't figuered all the
finite possibilities with this scenerio but the probalities are small to expect an outcome of 4way splits with doubledowns and insurance in succesion. Ultra conserative BR management would call for this decision. Place your bets....

Since part of the first +1 true count usually negates the house edge (depending on the game), +5 - +9 would be a closer set of circumstances for this question. Or even +6 - +11. It's not linear. Based on your question as written, I would take +10, especially in a 4 or 6 deck game.

Theory indicates that in positive expectation situations that the swings will be less.

What theory is that? The variance is actually higher at higher counts.

Therefore at high plus counts it would be expected that the probabality of winning sucessive bets in a row (a streak) would be positive.

The probability of any event is between 0 and 1 inclusive. Therefore, it is always either 0 or positive. If you mean to say that it is probable, you are simply wrong. Even if you have a 70% chance of winning a bet (you don't in blackjack), the probability of winning two consecutive bets is only 49%.

To gain the most positive expectation my suggestion would be to divide the investment into 8 (allowing for splits & doubles) and then parley each winning bet until the end of the 10 trials.

This flies in the face of optimal betting. At +5, your optimal bet would be around 2.5% of your bankroll. Win or lose, you adjust your next bet to be 2.5% of your new bankroll.