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nick777 wrote:

Just remember what you have on your site, you don’t have a winning system, you have an odds probability chart for 1 handed blackjack play.

Multi-hand blackjack and 1-hand blackjack use the same strategy chart, assuming same rules. This should be obvious, so I doubt wizard of odds explicity states this. I certainly do not on my site. Instead I assume the reader can has some basic knowledge.

nick777 wrote:

And what you are trying to say, is that your return will be the same regardless of whether you are playing 1 hand or 2 hands, can we agree on that, that is what you’re saying ?

The house edge does not change. Assuming a constant bet size and wagering, the average return will be approximately the same. The variance will differ, but we aren’t discussing variance.

nick777 wrote:

1 – What are the odds of you getting 2 10’s(or 2 splitting cards with 10 value)

If we assume a dealer 6 has been played, then 64/207 * 63/206 = ~9.5%
A more simple approximation is (4/13)^2 = ~9.5%

nick777 wrote:

2 – In 2 handed play, what are the odds of getting 2 hands each with 2 10’s(or 2 splitting cards with 10 value)

Roughly the above squared… 9.5% * 9.5%

nick777 wrote:

What you have been saying this whole time is that your odds are the same, well i don’t agree

THE OPTIMAL STRATEGY AND HOUSE EDGE ARE THE SAME. Obviously your odds of drawing all matching hands decrease as you increase the number of hands. If you played a million hands at once. You’d never draw the same cards in all one million hands, yet optimal strategy and house edge would remain the same.

nick777 wrote:

What are the odds of a player with a 10 beating the dealer with a 6 up card.

It depends on strategy. If we assume the player follows optimal strategy and does not split/resplit if he draws a 10, then… there is roughly a 61% chance of a win and 33% chance of loss. The expected return is ~0.28 x Bet Size.

nick777 wrote:

What are the odds of a player with a 20 total beating the dealer with a 6 up card.

There is roughly an 80% chance of a win and 10% chance of a loss. The expected return is ~0.70 x Bet Size.

nick777 wrote:

Your whole argument from the start has been that the odds will be the same, regardless of whether you have 1 splitting hand or 2 splitting hands, each with a value of 10.

If by “the odds” you mean optimal strategy, house edge, and overall expected return with overall bet size and wagering constant, then yes.

nick777 wrote:

What i am saying is that the probability of you losing with 4 hands hitting on a 10 vs a dealer with a 6 make it much more profitable than 2 hands hitting on a 10 vs a dealer with a 6, 20+% more profitable, and because of that fact(4 favoured draws vs dealer 6, instead of 2 favoured draws vs dealer 6)it is more profitable than standing with the 20 over the long run. If you don’t do the correct math it will never make sense to you, in total it will come out to be around 14% more profitable of a play, depending on the number of decks.

You mean losing all 4 hands simultaneously? Why would you worry about winning or losing all hands simulatenously unless computing variance.

The expected return on the 10 vs 6 hands is ~0.28 x Bet Size per Hand x Num Hands.

The expected return on the 20 vs 6 hands is ~0.70 x Bet Size per Hand x Num hand.

There are twice as many hands when splitting, making it 0.70 vs 0.56.

nick777 wrote:

5 – What are the odds of you losing both splitting hands with a 10 draw vs dealer 6

What are the odds of you losing all 4 splitting hands with a 10 draw vs dealer 6

The odds of simultaneously winning/losing all hands are (odds of winning or losing per hand)^ (num hands). So yes, the odds of a simulatenous win/loss in all hands decreases as the number of hands increases.

Was this your big revelation?! You do realize that losses count, even if you don’t lose all hands simulataneously don’t you? What would happen, if you played a million hands at once and had a ~zero chance of a simultaneous loss in all hands. Do you think the expected return would be 100% because you would never lose? Don’t you see how silly this is?

You are looking like a forum troll again in my mind. Each time, you seem to take a more ridiculous position than the last time. First it was splitting 10s has a higher expected return than standing in all situations. That’s an understandable mistake that some gamblers make. You’ve gone on to support this by computing averages without considering losses and posting several silly formulas. Now you are saying multihand blackjack has different optimal strategy and house edge from standard blackjack and justfying it by only looking at simulatenous losses in all hands??!! Why don’t you also only consider wins that occur simultaneously in all hands. Don’t you think the odds of simultaneously winning 4 hands are lower than simultaneously winning 2 hands?

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If you want to look at winning/losing all hands when computing overall return, then the correct way to do it is to sum the following:

-4 * Odds of losing 4 hands
-3 * Odds of losing 3 hands and pushing 1
-2 * (Odds of losing 3 hands and winning 1 + odds of losing 2 hands and pushing 2).
…

Do a Google search on permutations and combinations to get valid formulas . Of course, the far easier way to compute overall return is simply looking at average return per hand * num hands.