High Pair vs 3 to Royal

[parnellllll]

Have been playing 3yrs or so and about 80k hands of 9/6 Jacks or Better a year. I religiously hold high pair over 3 to Royal and have never once had 4 of a kind. Does anyone know how many statistically I should have had? I want to get caught up!

[onemoretry]

Without knowing how many times you've actually made that draw to a high pair, it's pretty difficult to say how many quads, statistically, you "should" have had. The likelihood of hitting a quad drawing to a pair is about 1 in 360.

[Jstark]

So with approximately 240,000 hands, you'd have to figure out the probability to get dealt a high pair along with 3 to a royal at the same time. Example: K K Q 10 6. Then it would be much easier to figure out.

[advantage playe]

if u r on a progressive and the royal pay is high enough , it is correct to throw away the high pair and hold 3 to a royal ! 1081 to 1 to hit royal I think .

[parnellllll]

No progressive, but i can see what is being said about how many times i incur this situation per session. Off top of my head, maybe 5 or 10 times. So looks like i might need another few years to draw 2 to four of a kind!! Yuk

[Carcounter]

I'm confused. You said you play 80k hands a year and have been playing for 3 years. That's 240k hands and you have never filled in a quad from holding a high pair? That can't be possible. Am I missing something.

[Jstark]

I'm confused. You said you play 80k hands a year and have been playing for 3 years. That's 240k hands and you have never filled in a quad from holding a high pair? That can't be possible. Am I missing something.

He specified that he hasn't while also having 3 to a royal in the same dealt. That's a rarer occurrence.

[BobDancer]

So with approximately 240,000 hands, you'd have to figure out the probability to get dealt a high pair along with 3 to a royal at the same time. Example: K K Q 10 6. Then it would be much easier to figure out.

True. If you're going to do this analysis, it's better to start with the RF3. The ones with three high cards (AKQ, AKJ, AQJ) have a higher probability of having a high pair in the hand than the ones with only two high cards (AKT, AQT, AJT, KQT, KJT, QJT).

For the ones with 3 high cards, (there are 12 of them, including suits), they each have 9 chances to pair up one of the existing high cards --- multiplied by the 38 possibilities for the fifth card. The fifth card cannot make two pair, trips, or a 4-card royal. So starting from (AKQ) of hearts and the A of spades, the fifth card may be any 2, 3, 4, 5, 6, 7, 8, or 9 (32 cards), plus any non heart ten or jack (6 more cards), leaving us with 38.

Multiplying that out, you get 12 * 9 * 38 = 4104 possibilities

You have to add in the possibility of getting the pair not in the royal cards --- such as (AKQ) of hearts and two black jacks. So there are the 12 starting hands multiplied by 3 ways to get a pair of jacks that doesn't include a heart. So we add 36 to the 4104 possibilities and get 4140 different ways to get a RF3 with three high cards and a high pair in the same hand.

Figuring out how many ways you can get an RF3 with two high cards and a high pair in the hand is an exercise that should be simple enough to figure out once you see the methodology above.

Once you find out the number of different ways to get a high pair and an RF3 in the same hand, remember there are 2,598,960 starting hands, it's easy to figure out how many of these combinations you should have had in 240,000 starting hands.

Without finishing the exercise, I STRONGLY suspect that the original poster hit one or more quads and either didn't notice at the time or has forgotten. Quads are "only" 125 coins in that game and unless you're paying $48 or more per hand, it won't come with a W2G.

If I'm made an arithmetic or analytic mistake in the above, I invite correction.

[Vman96]

So with approximately 240,000 hands, you'd have to figure out the probability to get dealt a high pair along with 3 to a royal at the same time. Example: K K Q 10 6. Then it would be much easier to figure out.

True. If you're going to do this analysis, it's better to start with the RF3. The ones with three high cards (AKQ, AKJ, AQJ) have a higher probability of having a high pair in the hand than the ones with only two high cards (AKT, AQT, AJT, KQT, KJT, QJT).

For the ones with 3 high cards, (there are 12 of them, including suits), they each have 9 chances to pair up one of the existing high cards --- multiplied by the 38 possibilities for the fifth card. The fifth card cannot make two pair, trips, or a 4-card royal. So starting from (AKQ) of hearts and the A of spades, the fifth card may be any 2, 3, 4, 5, 6, 7, 8, or 9 (32 cards), plus any non heart ten or jack (6 more cards), leaving us with 38.

Multiplying that out, you get 12 * 9 * 38 = 4104 possibilities

You have to add in the possibility of getting the pair not in the royal cards --- such as (AKQ) of hearts and two black jacks. So there are the 12 starting hands multiplied by 3 ways to get a pair of jacks that doesn't include a heart. So we add 36 to the 4104 possibilities and get 4140 different ways to get a RF3 with three high cards and a high pair in the same hand.

Figuring out how many ways you can get an RF3 with two high cards and a high pair in the hand is an exercise that should be simple enough to figure out once you see the methodology above.

Once you find out the number of different ways to get a high pair and an RF3 in the same hand, remember there are 2,598,960 starting hands, it's easy to figure out how many of these combinations you should have had in 240,000 starting hands.

Without finishing the exercise, I STRONGLY suspect that the original poster hit one or more quads and either didn't notice at the time or has forgotten. Quads are "only" 125 coins in that game and unless you're paying $48 or more per hand, it won't come with a W2G.

If I'm made an arithmetic or analytic mistake in the above, I invite correction.

Your methodology looks good to me. And when you follow the math all the way through, i got 3 to a Royal plus a high pair happening every 1 in 243.075 deals on average.

So for 240,000 deals, the player should see this about 987.35 times on average.

The probability of converting a pair into a quad is 45/16215 = 1 in 360.333.

So he would be in a 2.74 "cycle" drought and and we will see this happen roughly 6.43% of the time by the binomial distribution.

So it's unlikely, but it's far, far away from appropriately suggesting that he's more likely to be mistaken. I've definitely have played over 100,000 deals myself, and I'm pretty sure I'm waiting on this one to happen too.

And OP, you're in a 2.74 cycle drought with those numbers, so you "should" have had 2.74 of these on average.

[Jstark]

Had this happen to me last night, but got the opposite result on one hand played. Dealt K Q 10 10 8 , hold the royal draw only to watch the other two 10's pop up. Stuff happens.