low house edge or maybe player edge games, but need your confirmation

Hi. I see many low house edge games in 5dimes.com's bonus casino. Some of them may have a low player's edge but i am just not sure so I need some confirmation.

For example, they offer a single deck Blackjack game that has an EV of -0.0899% according to Wizard's calculator. That game also offers an increased insurance payout, 11 to 5 or odds of 3.2. I'm wondering how much house edge does this shave off, would it turn to a player's edge? because insurance can be advantagous when the player does not hold any 10 card. i.e.

3.2*(16/49)=1.044898 or a player advantage of 4.4898%

I'm suspecting the amount of house edge shaved off would be the advatange of the bet multiplied by the probability of the positive EV event happening.i.e.

0.044898*(4/52)(35/51)(34/50)=0.0016117, or 0.16117%

and this number should be multiplied by the size of the insurance bet compared to the original bet, which is

1/(11/5)=0.454545

and the final result should be

0.0007326 or 0.07326%?

Are my calculations correct?


They also run a promotion on Tuesday where they pay 1.05 for pai gow poker rather than 0.95. According to Wizardofodds.com, the probability of player winning is 28.61% while tie is 41.48%.

the return of the game is therefore 2.05*.2861+1*.4148=1.001305, or a player advantage of 0.13%. However, the probabilities of winning and tie are based on the Trump Plaza house way, while 5dimes uses the Flamingo house way. Does that change the odds? If i just use the house way button, would the % of winning and tieing be the different ?

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Quote: Ibeatyouraces

11 to 5 is actually 2.2 to 1 not 3.2 to 1.

odds of 3.2 means it includes the original stake, for example when you are paid even money it means odds of 2.0 or pays 1 to 1

Quote: Dipsy

3.2*(16/49)=1.044898 or a player advantage of 4.4898%

Quote: Dipsy

odds of 3.2 means it includes the original stake, for example when you are paid even money it means odds of 2.0 or pays 1 to 1

If the 3.2 includes your original 1 unit bet, you need to make the math reflect that. You're only making 2.2 units of profit...

It's how you word it.

This example is 3.2 for 1, or 2.2 to 1. You can't just say '3.2' because the assumption is '... to 1'.

In a casino, there are only two bets I can think of that are worded, and are paid, as "X for 1". Those would be slot machines and the Craps FireBet. In both of those cases, the casino takes your original wager, and you get paid using other coins/chips. Even old coin-op machines, if you won, you did not get your original quarter back. All other games, when you win, your original bet is either returned, or can remain in play as a new bet.

Note that some casinos will word some bets (particularly the junk in the middle of Craps) as X for 1. But they are usually paid as (x-1) to 1. I.E. Your original bet remains up. THAT is a practice that does nothing but confuse a novice player, and therefore slow the game down, so you gotta wonder, WHY?

Note that I've also seen some FireBets worded as 'X to 1' but are paid as (X+1) for 1.

in sports odds of 3.2 is equivalent to 2.2 to 1, so yeah. but i'm more interested in the answers to my thread than the wordings.

Quote: Dipsy

in sports odds of 3.2 is equivalent to 2.2 to 1, so yeah.

I forgot about sports betting. It uses "for" terminology because the casino / sportsbook takes your money immediately. If you win, the total paid includes your original bet. (On a side note, I wish they'd stop calculating horse races based on $2 bets...)

Quote: Dipsy

...but i'm more interested in the answers to my thread than the wordings.

I've been wrestling with this a LOT.

I think, for one thing, that the 11:5 insurance payoff, on a single deck, single player game IS a positive player expectation. But only if, as you're saying, you're not holding any tens:
3.2*(16/49)=1.0448979 as you suggest.

If you're holding one ten:
3.2*(15/49)=0.9795918 so no bet.

However, if you do the math using an infinite deck, (as so many lazy people do):
3.2*(4/13)=0.9846152.

Perhaps the person that implemented this at that website, just didn't do the right math?



So the big question is this: Does this overcome the edge on the basic game?

I think not, but it certainly makes it close!

(How many times does the dealer get an ace, where you don't have any tens? Multiply that by the basic edge to get your answer.)


FYI: Double deck, you hold no tens:
3.2*(32/101)=1.0138613 - still a positive EV!

Quote: DJTeddyBear

(How many times does the dealer get an ace, where you don't have any tens? Multiply that by the basic edge to get your answer.)

The short answer to that is:
(36/52) * (35/51) * (4/50) = 0.038009

Multiply that by the payoff odds of insurance:
0.038009 * 1.0448979 = 0.0397155

To be honest, even though I said to multiply by the basic edge, I'm really not sure that's what you do with this number. I kinda think you add it to the EV of -0.0899% to get an overall positive EV of 3.88165%

Quote: DJTeddyBear

Perhaps the person that implemented this at that website, just didn't do the right math?

No, obviously they have 11 to 5 insurance on all their blackjack games. All their games have low house edge and maybe even a player's edge. Which should encourage more action.

Quote: DJTeddyBear

The short answer to that is:
(36/52) * (35/51) * (4/50) = 0.038009

Multiply that by the payoff odds of insurance:
0.038009 * 1.0448979 = 0.0397155

To be honest, even though I said to multiply by the basic edge, I'm really not sure that's what you do with this number. I kinda think you add it to the EV of -0.0899% to get an overall positive EV of 3.88165%

If this is correct it will be a crazy game! but i'm not sure

Quote: DJTeddyBear

Quote:

How many times does the dealer get an ace, where you don't have any tens? Multiply that by the basic edge to get your answer.

The short answer to that is:
(36/52) * (35/51) * (4/50) = 0.038009

Multiply that by the payoff odds of insurance:
0.038009 * 1.0448979 = 0.0397155

To be honest, even though I said to multiply by the basic edge, I'm really not sure that's what you do with this number. I kinda think you add it to the EV of -0.0899% to get an overall positive EV of 3.88165%

Actually, it's slightly lower. I gave the dealer four chances to pull an ace, even if you hold an ace or two.

The number of times the dealer has an ace when you don't have a ten is:
(32/52)*(31/51)*(4/50) + (32/52)*(4/51)*(3/50) + (4/52)*(3/51)*(2/50) = 0.0330015

0.0330015 * 1.0448979 = 0.0344832

Add the EV of -0.0899% to get 0.0335842

Mind you, I'm still not sure if this is the correct way to calcuate it, but I think so...

The bet in question will yield a positive overall EV, it is only on the order of 0.1%. First you have to get the probability of drawing two non-tens to a dealer's Ace. There are three mutually exclusive cases.

I. Both of the player's cards are not Aces. The probability is 35/52 * 34/51 * 4/50 = 0.0358974.

II. One of the player's cards is an Ace. There are two distinguishably different ways of drawing this so the probabililty is 2 * 4/52 * 34/51 * 3/50. = 0.00615385.

III. Both of the player's cards are Aces. There is only one distinguishable way of drawing this so the probability is 4/52 * 3/51 * 2/50 = 0.000180995.

These cases are mutually exclusive so the probabilities add to yield a probability of 0.0424133.

Given that the first three cards are non-tens, the payout from the insurance bet is 11/5 * 16/49 - 33/49 = 0.0448980. or 4.48980%.

The total EV is 0.0448980 * 0.0424133 = 0.00190427.

Subtracting this from the overall EV of the game of -0.08999% yields 0.00190427 - 0.0008999 = 0.001 or +0.1%.

So the game does have a positive EV, but it is not large. If the player's bank is $10,000, betting Kelly (bank * advantage) would require a $10 bet. I believe this 5Dimes game has a $1 minimum. You could reduce your chances of going broke before doubling your bank and reduce the number of bets to double your bank by always betting Kelly, down to the $1 minimum. You will probably see a lot of large swings before you make your goal. The source of the volatility come from the fact that the favorable insurance bet occurs only once in about 24 hands. When it does occur you will still lose about twice as often as you win.

If you go with a $1000 bank, which certainly would be in line with my pocketbook, then the probability is almost 50% that you would go broke before doubling your bank. Good Luck.