Video Poker Forum

I am curious, what is the most likely scenario of achieving 4-of-a-kind?

Is it more likely to get 3 on the flop and draw the 4th one? Or is it more likely to get 2 on the flop and draw the other 2?

(There are other scenarios, such as drawing all 4, but I am pretty sure those are less likely than the two I mentioned.)

The better hand on the deal is obviously easier to complete, however, the better hand is harder to be dealt with.

P(Dealt 3 of a kind) = 4.83%, Chance to complete it is 2/47 = 4.26%, overall 0.21%
P(Dealt a pair) = 43.8%, chance to complete it is (47 + 46 + 46) / 47^3 = 0.13%, overall 0.057%

But you don't have to worry too much about it. All kinds of 4 of a kind happen about 1 in every 400 to 450 hands averagely in different varieties of JoB games.

Edit...

Thank you hophoofer, that was exactly what I was looking for! This is just out of simple curiosity.

So the "dealt 3" scenario is 3.68 times more likely than being dealt 2 and drawing the final 2(?) That ratio is higher than I expected, but it makes sense.

Where did you get your "P()" numbers used to make your calculations? Wizard of Odds? (I want to lookup P(dealt 4) to compare that scenario too. (It should simply be that number for that scenario.))

PS- You have a typ-o in your calculation: it should be (47+46+*45*) rather than (47+46+46). No biggy, it still rounds to .13% so your calculation is still correct.

Hey, after figuring out P(dealt 4) there are only 2 more scenarios!: Being dealt 0 cards (holding nothing) then being drawing all 4, and finally holding 1 card then picking up the final 3 after drawing more cards.

For the first scenario (holding 0 cards) wouldn't it just be *double* P(dealt 4)? This is because you are drawing 10 cards total rather than 5 cards when you get it immediately. (Oh - plus should you multiply by the chance that you will be dumping all 5 initial cards?)

However, I have no clue how to calculate the last scenario! (Holding just 1 card then drawing the final 3.) There is no baseline "P()" value for a single card. Any ideas?

MikeY42 wrote: ↑

Thu Jun 25, 2020 10:11 pm

PS- You have a typ-o in your calculation: it should be (47+46+*45*) rather than (47+46+46). No biggy, it still rounds to .13% so your calculation is still correct.

Good catch, thank you.

Actually the number is (1/47)*(1/46) + (1/47)*(45/46)*(1/45) + (46/47)*(1/46)*(1/45)
= (45 + 45 + 46)/(45 * 46* 47) = 0.14%

For other scenarios you mentioned, you can use the same way to calculate it.

Another scenario in JOB is holding 2 to the royal can result in 2 different quads. Example: hold J Q and get either quad Jacks or quad Queens.

Somehow the "Jacks or Better" requirement got added after the initial post. The numbers for other games are different. Why? Because on a hand like AAA44, you have a 1/47 chance of ending up with a quad in DDB, but a 0/47 chance in JoB

On hands that start out K'KQJ'x, in JoB you always hold KK, in other games you always hold a suited KQJ, and in still others it depends on what x is. (I'm assuming that x doesn't create a 4-to-the-royal or trips or two pair).

Be careful, MikeY42, of a conclusion that says one result occurs 3.68 times as often as another. That's WAY too much precision for something that differs by game, and even differs by pay schedules in a game. If you want to say, quads happes from dealt trips much more often than from dealt pairs, great. That's true. But exactly how much is so dependent on a number of other variables.

BobDancer wrote: ↑

Fri Jun 26, 2020 7:00 pm

Somehow the "Jacks or Better" requirement got added after the initial post. The numbers for other games are different. Why? Because on a hand like AAA44, you have a 1/47 chance of ending up with a quad in DDB, but a 0/47 chance in JoB

On hands that start out K'KQJ'x, in JoB you always hold KK, in other games you always hold a suited KQJ, and in still others it depends on what x is. (I'm assuming that x doesn't create a 4-to-the-royal or trips or two pair).

Be careful, MikeY42, of a conclusion that says one result occurs 3.68 times as often as another. That's WAY too much precision for something that differs by game, and even differs by pay schedules in a game. If you want to say, quads happes from dealt trips much more often than from dealt pairs, great. That's true. But exactly how much is so dependent on a number of other variables.

Bob, I think you meant TDB here.

Jstark wrote: ↑

Sat Jun 27, 2020 8:17 am

Bob, I think you meant TDB here.

Good catch. Actually I meant DDB but made a typo.

In DDB, you have 2/47 chances for the quad because you hold AAA. In TDB you have 1/47 because you should hold AAA4.

Had I actually meant TDB, I would have included a hand like an unsuited 66789 where you have a 3/1081 chance of getting a quad in JoB and DDB, but a 0% chance in TDB because in that game the correct play is 6789.