March 17th, 2017 at 11:40:25 PM
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I wondering if anybody could help me with this.

If there were a simple game of three card poker where the player gets three cards and the dealer gets three cards, there's only the initial bet and the highest hand wins: What would be the house edge if the dealer only played with a J or better.

Hope this question makes sense.

Cheers

kingcreights

If there were a simple game of three card poker where the player gets three cards and the dealer gets three cards, there's only the initial bet and the highest hand wins: What would be the house edge if the dealer only played with a J or better.

Hope this question makes sense.

Cheers

kingcreights

March 18th, 2017 at 9:58:48 AM
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Quote:kingcreightsI wondering if anybody could help me with this.

If there were a simple game of three card poker where the player gets three cards and the dealer gets three cards, there's only the initial bet and the highest hand wins: What would be the house edge if the dealer only played with a J or better.

Hope this question makes sense.

Cheers

kingcreights

I would speculate that there wouldn't be a House Edge, there would be a Player Advantage. The reason why is that the dealer would be folding hands that the dealer could conceivably win. For instance:

Player: 10-6-5

Dealer: 10-7-5

The dealer, "Would not qualify," and the initial, 'Ante,' bet would be paid to the player.

Last edited by: Mission146 on Mar 18, 2017

Vultures can't be choosers.

March 18th, 2017 at 11:44:19 AM
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Quote:Mission146I would speculate that there wouldn't be a House Edge, there would be a Player Advantage. The reason why is that the dealer would be folding hands that the dealer could conceivably win. For instance:

Player: 10-7-4

Dealer: 10-6-5

The dealer, "Would not qualify," and the initial, 'Ante,' bet would be paid to the player.

Actually, I think the opposite is true; not sure but would like to know. We (CM and I ) did a similar evaluation for my game, and the J or better requirement had a higher house edge than Q or better. K or better became a player advantage, because the house would fold too many hands they should play. (On OftM, anyway; it could be different on 3CP, as you say)

I don't have the spreadsheets with me, so perhaps CM can verify if I'm remembering this correctly.

Regardless , I don't follow your example with a 10 high. The player would win that one if the dealer did qualify, so it costs the player 1/2 a win. But the dealer would not conceivably win that hand.

"If the house lost every hand, they wouldn't deal the game."

March 18th, 2017 at 1:34:17 PM
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Quote:beachbumbabsActually, I think the opposite is true; not sure but would like to know. We (CM and I ) did a similar evaluation for my game, and the J or better requirement had a higher house edge than Q or better. K or better became a player advantage, because the house would fold too many hands they should play. (On OftM, anyway; it could be different on 3CP, as you say)

I don't have the spreadsheets with me, so perhaps CM can verify if I'm remembering this correctly.

Regardless , I don't follow your example with a 10 high. The player would win that one if the dealer did qualify, so it costs the player 1/2 a win. But the dealer would not conceivably win that hand.

Sorry, I put the hands backwards, the dealer was supposed to have the better, Ten-High, I'll switch it.

What it sounds like to me the OP is saying is that both Player and Dealer get three cards, there is only one initial bet (No, 'Play,' bet) and that the best hand wins. Perhaps I am misunderstanding the OP.

If I am not misunderstanding the OP, then it is almost like Casino War just with 3CP Poker Rankings. That being the case, neither side would ever want to fold because they would be folding a hand that could conceivably win. For instance, if the dealer would have 6-4-2, then folding and automatically losing to the player would be bad because the player could only have 5-4-2.

Maybe the OP meant to say the HE if the PLAYER needed a Jack-High to qualify, that would have a House Edge. Alternatively, maybe he meant that it would be, 'No Action,' if the dealer didn't have at least a Jack. I took it as the player would win, hence only the initial bet.

Vultures can't be choosers.

March 18th, 2017 at 1:39:55 PM
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Pretty sure the ante bet would be refunded to the player if dealer doesn't qualify. So yeah there'd be a HE.

"What would Brian Boitano do?"

March 18th, 2017 at 1:53:41 PM
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Quote:RSPretty sure the ante bet would be refunded to the player if dealer doesn't qualify. So yeah there'd be a HE.

In that case, yes, I thought the OP meant if the player would be paid on the bet in such an event.

Vultures can't be choosers.

March 18th, 2017 at 3:54:23 PM
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I wasn't referring to the existing 3 card poker game with an ante bet. I'm just trying to give an example of an imaginary simple game so I can get an indication of the maths behind a qualifying hand. maybe imagine casino war with three cards

I just want to know if there was simple game where you put down an intial single bet and you get three cards and dealer gets three cards what is the house edge if the dealer will only play if he has a jack or better.other wise the hand is a stand off.

All the best

Kingcreights

I just want to know if there was simple game where you put down an intial single bet and you get three cards and dealer gets three cards what is the house edge if the dealer will only play if he has a jack or better.other wise the hand is a stand off.

All the best

Kingcreights

March 18th, 2017 at 4:03:31 PM
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I'm going to try this and waste a lot of time if I don't do this right, but I should get close.

The first thing that I am going to do is eliminate, of the 22,100 possible hands, all hands that have a card higher in rank than a ten in them. Fortunately, that part is easy:

(36/52)*(35/51)*(34/50) = 0.32307692307

That's going to be very important because now I have to subtract certain other hand probabilities from that in order to arrive at how many combinations there are. In terms of how many possible hands that would be, we see:

22100 * .32307692307 = 7139.99999985, so, 7,140.

For the remainder of these calculations, we are going to be basing it on 7,140 and subtracting certain types of hands from that. We will use the same logic here:

https://wizardofodds.com/games/three-card-poker/appendix/1/

With the remaining cards, which are essentially the same as a deck that only runs 2-10, which means that there are nine ranks and four suits. That tells us the number of combinations is:

(36*35*34)/(3*2*1) = 7,140

Which agrees with the above number, so that's fantastic.

The first thing that we will do is eliminate the straight-flush, the span of which can run 2-3-4 all the way to 8-9-10, which comprises seven total spans. (A-2-3 is gone because it would mean an Ace). 7 * 4 (suits) = 28

7,140 - 28 = 7,112

This is simple, the number of possible Trips equals the number of cards in the deck. There are 36 cards that remain in the deck, therefore there are 36 different Trips. Alternatively, nCr(4,3)= 4 which chooses three suits out of four and then you multiply by the number of ranks, which is nine, so 9 * 4 = 36

7,112 - 36 = 7,076

For the straights, from the straight-flush section we know that there are seven possible spans for a straight. A straight has three cards, each may be one of four suits. However if all three suits are the same then the player has a straight flush. So the number of suit combinations is (4)^3-4 = 64-4 = 60. So there are 7*60=420 possible straights.

7,076 - 420 = 6,656

There are four possible suits for the Flush. For each suit there are nCr (9,3) = 84 ways to draw three ranks out of nine. However, from the straight flush section we know that there are seven combinations which result in three connected ranks, giving the player a straight flush. So the combinations giving a straight, but not a straight flush, are:

nCr(9,3)-7 = 84 -7 = 77.

Therefore, the number of flushes is 77 * 4 = 308

6656-308 = 6,348

There are nine possible ranks for the pair and eight for the singleton, ergo, there are 72 ways to pick the ranks. Within the pair, there are nCr(4,2)=6 ways to pick two suits out of four. For the rank of singleton there are four possible suits. So the total suit combination is 6 *4 = 24. The total number of pair combinations is 72 * 24 = 1,728

6348-1728 = 4,620

4620/22100 = 0.2090497737556561

Thus, 0.2090497737556561 is the probability of the dealer getting a non-qualifying hand.

The probability of the dealer getting a qualifying hand is 1-0.2090497737556561 = 0.7909502262443439

In general, there are four game states:

Player Qualifies, Dealer Qualifies:

(17480/22100) * (17479/22099) = 0.6255947782490107

When that game state is met, the dealer and player will win, lose and tie equally. Therefore, no House Edge is derived from that game state.

Player Does Not Qualify, Dealer Qualifies:

(4620/22100) * (17480/22099) = 0.1653554479953332

The dealer would win in such a situation, anyway, so no additional House Edge is derived from that situation.

Player Does Not Qualify, Dealer Does Not Qualify:

(4620/22100)*(4619/22099) = 0.0436943257603229

Half of these situations actually benefit the player, because the dealer would then beat the player with a non-qualifying hand. Half of these situations actually benefit the dealer, because the player would beat the dealer when the dealer has a non-qualifying hand. Therefore, there is no effect on the House Edge.

Here is where the House Edge comes in:

Player Qualifies, Dealer Does Not Qualify:

(17480/22100)*(4620/22099) = 0.1653554479953332

The player would literally always win and never lose in this situation, but now the player always pushes.

NO HOUSE EDGE, NO QUALIFY RULE:

(0.6255947782490107 * 5 * .5) + (0.6255947782490107 * -5 * .5) = 0

(0.1653554479953332 * -5) = -0.82677723997

(0.0436943257603229 * 5 * .5) + (0.0436943257603229 * -5 * .5) = 0

(0.1653554479953332 * 5) = 0.82677723997

That would be a fair game, the House Edge would be zero.

However, with the, 'Dealer Qualify,' rule, the last part does not happen and the result of such an event is:

(0.1653554479953332 * 0) = 0

Which leaves:

0 + -0.82677723997 + 0 + 0 = -0.82677723997

Which means that the player is expected to lose 82.677723997 cents for every five dollars bet, which makes the House Edge:

.82677723997/5 = 0.16535544799

Which is, of course, roughly the probability that the subject matter hand (Player Qualifies, Dealer Doesn't) comes up.

The first thing that I am going to do is eliminate, of the 22,100 possible hands, all hands that have a card higher in rank than a ten in them. Fortunately, that part is easy:

(36/52)*(35/51)*(34/50) = 0.32307692307

That's going to be very important because now I have to subtract certain other hand probabilities from that in order to arrive at how many combinations there are. In terms of how many possible hands that would be, we see:

22100 * .32307692307 = 7139.99999985, so, 7,140.

For the remainder of these calculations, we are going to be basing it on 7,140 and subtracting certain types of hands from that. We will use the same logic here:

https://wizardofodds.com/games/three-card-poker/appendix/1/

With the remaining cards, which are essentially the same as a deck that only runs 2-10, which means that there are nine ranks and four suits. That tells us the number of combinations is:

(36*35*34)/(3*2*1) = 7,140

Which agrees with the above number, so that's fantastic.

The first thing that we will do is eliminate the straight-flush, the span of which can run 2-3-4 all the way to 8-9-10, which comprises seven total spans. (A-2-3 is gone because it would mean an Ace). 7 * 4 (suits) = 28

7,140 - 28 = 7,112

This is simple, the number of possible Trips equals the number of cards in the deck. There are 36 cards that remain in the deck, therefore there are 36 different Trips. Alternatively, nCr(4,3)= 4 which chooses three suits out of four and then you multiply by the number of ranks, which is nine, so 9 * 4 = 36

7,112 - 36 = 7,076

For the straights, from the straight-flush section we know that there are seven possible spans for a straight. A straight has three cards, each may be one of four suits. However if all three suits are the same then the player has a straight flush. So the number of suit combinations is (4)^3-4 = 64-4 = 60. So there are 7*60=420 possible straights.

7,076 - 420 = 6,656

There are four possible suits for the Flush. For each suit there are nCr (9,3) = 84 ways to draw three ranks out of nine. However, from the straight flush section we know that there are seven combinations which result in three connected ranks, giving the player a straight flush. So the combinations giving a straight, but not a straight flush, are:

nCr(9,3)-7 = 84 -7 = 77.

Therefore, the number of flushes is 77 * 4 = 308

6656-308 = 6,348

There are nine possible ranks for the pair and eight for the singleton, ergo, there are 72 ways to pick the ranks. Within the pair, there are nCr(4,2)=6 ways to pick two suits out of four. For the rank of singleton there are four possible suits. So the total suit combination is 6 *4 = 24. The total number of pair combinations is 72 * 24 = 1,728

6348-1728 = 4,620

4620/22100 = 0.2090497737556561

Thus, 0.2090497737556561 is the probability of the dealer getting a non-qualifying hand.

The probability of the dealer getting a qualifying hand is 1-0.2090497737556561 = 0.7909502262443439

In general, there are four game states:

Player Qualifies, Dealer Qualifies:

(17480/22100) * (17479/22099) = 0.6255947782490107

When that game state is met, the dealer and player will win, lose and tie equally. Therefore, no House Edge is derived from that game state.

Player Does Not Qualify, Dealer Qualifies:

(4620/22100) * (17480/22099) = 0.1653554479953332

The dealer would win in such a situation, anyway, so no additional House Edge is derived from that situation.

Player Does Not Qualify, Dealer Does Not Qualify:

(4620/22100)*(4619/22099) = 0.0436943257603229

Half of these situations actually benefit the player, because the dealer would then beat the player with a non-qualifying hand. Half of these situations actually benefit the dealer, because the player would beat the dealer when the dealer has a non-qualifying hand. Therefore, there is no effect on the House Edge.

Here is where the House Edge comes in:

Player Qualifies, Dealer Does Not Qualify:

(17480/22100)*(4620/22099) = 0.1653554479953332

The player would literally always win and never lose in this situation, but now the player always pushes.

NO HOUSE EDGE, NO QUALIFY RULE:

(0.6255947782490107 * 5 * .5) + (0.6255947782490107 * -5 * .5) = 0

(0.1653554479953332 * -5) = -0.82677723997

(0.0436943257603229 * 5 * .5) + (0.0436943257603229 * -5 * .5) = 0

(0.1653554479953332 * 5) = 0.82677723997

That would be a fair game, the House Edge would be zero.

However, with the, 'Dealer Qualify,' rule, the last part does not happen and the result of such an event is:

(0.1653554479953332 * 0) = 0

Which leaves:

0 + -0.82677723997 + 0 + 0 = -0.82677723997

Which means that the player is expected to lose 82.677723997 cents for every five dollars bet, which makes the House Edge:

.82677723997/5 = 0.16535544799

Which is, of course, roughly the probability that the subject matter hand (Player Qualifies, Dealer Doesn't) comes up.

Vultures can't be choosers.

March 18th, 2017 at 4:19:24 PM
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If I understand the simple game you make a bet, no decision required.

Good/Good : If you both get J32 or better then it's 50-50 who gets the best hand - no House Edge

Bad/Bad : If you both get T97 or worse then it's a no bet - no House Edge.

Good / Bad : If Player gets J32 or better while Dealer gets T97 or worse then the Player doesn't win anything.

Bad / Good : If Dealer gets J32 or better while the Player gets T97 or worse then the Player loses the bet, whereas if it was a fair game the Dealer wouldn't win anything.

Thus the game's House Edge comes from Bad / Good losing 1.

Roughly P(J32+) = 17480/22100

Roughly P(T97-) = 4620/22100

So House Edge = 16.53% (ouch!)

Good/Good : If you both get J32 or better then it's 50-50 who gets the best hand - no House Edge

Bad/Bad : If you both get T97 or worse then it's a no bet - no House Edge.

Good / Bad : If Player gets J32 or better while Dealer gets T97 or worse then the Player doesn't win anything.

Bad / Good : If Dealer gets J32 or better while the Player gets T97 or worse then the Player loses the bet, whereas if it was a fair game the Dealer wouldn't win anything.

Thus the game's House Edge comes from Bad / Good losing 1.

Roughly P(J32+) = 17480/22100

Roughly P(T97-) = 4620/22100

So House Edge = 16.53% (ouch!)

March 18th, 2017 at 5:20:45 PM
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Quote:charliepatrickIf I understand the simple game you make a bet, no decision required.

Good/Good : If you both get J32 or better then it's 50-50 who gets the best hand - no House Edge

Bad/Bad : If you both get T97 or worse then it's a no bet - no House Edge.

Good / Bad : If Player gets J32 or better while Dealer gets T97 or worse then the Player doesn't win anything.

Bad / Good : If Dealer gets J32 or better while the Player gets T97 or worse then the Player loses the bet, whereas if it was a fair game the Dealer wouldn't win anything.

Thus the game's House Edge comes from Bad / Good losing 1.

Roughly P(J32+) = 17480/22100

Roughly P(T97-) = 4620/22100

So House Edge = 16.53% (ouch!)

It rounds up to 16.54%, I believe, but thanks for the confirmation!

Vultures can't be choosers.