4 of a kind probability n-hands

This probability question has me stumped. I've not found an answer on this blog. Can anyone help?

For video poker Jacks or Better, single hand, the probability of hitting 4 of a kind after the draw is known to me 0.002358. But, if I am playing that same game on a 3-hand machine, what is the probability of hitting a 4 of a kind on any of the 3 hands? If all 3 hands were independent (which they are not) that's easy. It's the conditional probability that has me stumped. Once the 3-hand question is answered, can you extend that to n-hands? My probability theory is a little rusty on this one. Thanks.

You mean, the probability that at least one of the hands will be 4 of a kind?

I don't think there's an easy way to calculate this. The way I see it, you have to determine the probability of each possible deal, then, for each of those deals, calculate the probability of NOT getting 4 of a kind on one hand, cube it (or raise it to the Nth power, for N hands), subtract this from 1 (to calculate the probability of getting at least one 4 of a kind on that deal), and multiply by the probability of getting that particular deal in the first place; add up the results for all of the deals.

For example, of the 2,598,960 possible deals, 64 of them have three 2s, a 6, and a 9, so you discard the 6 and the 9; the probability of getting the fourth 2 in a particular hand is 1/47, so the probability of getting at least 1 in 3 hands is 1 - (1 - 1/47)^3, and the overall probability value from this deal is 64 / 2,598,960 x (1 - (1 - 1/47)^3).

Remember to take into account the possibility that you may not play for the 4 of a kind. For example, if you are dealt three 6s and two 9s, you keep the full house rather than playing for the 4 of a kind, so the probability of getting a 4 of a kind from that deal is zero.

I think that overly complicated the problem. Considering just the 3 hand question for simplicity, as I see it this is a conditional probability problem. As such,

P(4 kind on at least one of the 3 hands) = P(4 kind on dealt hand 1) + P(4 kind on hand 2 given what is held on hand 1) + P(4 kind on hand 3 given what is held on hand 1).

Isn't the probability of 4 of a kind on the dealt hand just the (known) probability of 4 of a kind on a single hand game which is 1:423 or 0.002358? I'm also thinking the 2nd and 3rd hand probabilities are the same since the same cards are kept for both. If so, the above simplifies to

P(4 kind on at least one of the 3 hands) = 0.002358 + 2P(4 kind on hand 2 [or 3] given what is held on hand 1).

If my thinking that this is a conditional probability problem is wrong someone please let me know. Is my logic messed up? Anyone else care to jump in? Thanks.

Quote: JHH

I think that overly complicated the problem. Considering just the 3 hand question for simplicity, as I see it this is a conditional probability problem.

Yes, but there are over 134,000 possible conditions for the initial deal you have to consider.

Look at it this way:
Suppose you have three coins, and you toss each one - and each coin that is not heads, you toss a second time. What is the probability that at least one coin will end up heads?
For one coin, it's 3/4, so if you treat them as independent events, it is 1 - (1 - 3/4)^3 = 63/64.
However, if the first toss is the same for all three coins - similar to how all of the initial deals in multi-hand video poker are the same - then 1/2 of the time, all three will be heads, and the other 1/2 of the time, all three will be tails; if all three are tails, the probability that at least one of the second tosses will be heads is 7/8, so the overall probability of at least one head is 1/2 + (1/2 x 7/8) = 15/16.

If the probability of getting quads in one hand is 1/424, I bet the probability of getting quads in at least one of three hands (dealt from same deck, trying only for quads) is very close to 1 - (423/424)^3 =~ 1 in 141.7

Assuming the other hands don’t have quads should have only a tiny effect on the hand being analyzed. I think so anyway !

^ That sounds very logical. If you get dealt quads then obviously you've got them! In all the other cases you are going to play the three hands the same way - i.e. hold a number of cards and re-draw. Assuming the re-draws are equally likely this makes hand 1 have the same chance of getting Quads as only playing one hand. You can then add the chances that hand 2 now makes quads and similarly hand 3.

There is technically some small probability that multiple hands will get Quads so you have to subtract roughly 3xp^2 and p^3, and adjust for the effect of being dealt quads in the first place. However for practical purposes the answer is nearly 3 times more likely, as the 1 in 141.7 above suggests.

This wouldn't be so true if the event, say any Pair or better, was fairly likely.

If you have three 2's, the probability of getting a fourth 2 is 1 in 47.

Quote: ChumpChange

If you have three 2's, the probability of getting a fourth 2 is 1 in 47.

As you draw two cards your chances are 2/47 (i.e. the fourth two can either be the 6th or 7th card in the original deck).

Charliepatrick and Ace2, I think you are getting close but there is still something missing in my mind and that is the conditional probability aspect. If we can agree that the probability of at least one 4 of a kind of the 3 hands is

P(at least one 4 kind) = P(4 kind hand 1)+P(4 kind hand 2 given held on hand 1)+P(4 kind hand 3 given held on hand 1) = 0.002358 + 2P(4 kind hand 2 given held on hand 1)

then I think the conditional probability for the P(4 kind hand 2 given held on hand 1) becomes

P = P(4 kind | held 0) + P(4 kind | held 1) + P(4 kind | held 2) + P(4 kind | held 3) + P(4 kind on deal).

I can calculate (or look up on Wizard of Odds) all those individual probabilities and the result comes out to be 0.088378 for the 3 hand problem. That seems a bit on the high side intuitively, which makes me think my logic is incorrect. If I go further and extend that to the n-hand problem things really go haywire as for n=10 the probability is greater than 1 which of course is a red flag that I've done something wrong. At this point, I am not so worried about the numbers but the logic, as what I have outlined seems to make sense. Thanks for the help.

Quote: charliepatrick

^ That sounds very logical. If you get dealt quads then obviously you've got them! In all the other cases you are going to play the three hands the same way - i.e. hold a number of cards and re-draw. Assuming the re-draws are equally likely this makes hand 1 have the same chance of getting Quads as only playing one hand. You can then add the chances that hand 2 now makes quads and similarly hand 3.

There is technically some small probability that multiple hands will get Quads so you have to subtract roughly 3xp^2 and p^3, and adjust for the effect of being dealt quads in the first place. However for practical purposes the answer is nearly 3 times more likely, as the 1 in 141.7 above suggests.

This wouldn't be so true if the event, say any Pair or better, was fairly likely.

Let's take a look at this by assuming that we hold trips and draw for a quad.

One Hand Quad Probability:

1 - (46/47 * 45/46) = 0.04255319148

Probability of Missing Quads on All Three Hands:

(46/47 * 45/46)^3 = 0.87769569363

Probability of Hitting at Least 1 of 3 Quads:

1 - 0.87769569363 = 0.12230430637

Probability of hitting quad on single hand hold multiplied by three:

0.04255319148 * 3 = 0.12765957444

________________________

The reason for this disparity is because the probability of hitting multiple quads in a three-handed scenario doesn't translate exactly from a probability standpoint. We can resolve the discrepancy by looking at the average number of quads that will be hit.

In the single-handed scenario, this remains: 0.04255319148

Now, we go to the multi-handed scenario to look at the average number of quads that we will hit. First, the probability of catching quads on all three hands:

(0.04255319148)^3 = 0.00007705421

Which we then multiply by three because that is the number of quads we hit:

0.00007705421 * 3 = 0.00023116265

The next thing that we will look at is the probability of hitting quads on any two out of the three hands. This can happen in three ways: Hands 1 & 2, Hands 1 & 3 or Hands 2 & 3:

To determine this: We simply use the probability of missing quads once and the probability of hitting them (relative to a single hand) twice and multiply by the three ways it can happen:

(46/47 * 45/46) * (0.04255319148) * (0.04255319148) * 3 = 0.00520115966

The number of Quads hit in these scenarios would be two, so we multiply our result by two:

0.00520115966 * 2 = 0.01040231932

Finally, we look at the probability of hitting quads on just one hand, which can also happen three ways (the three hands) so we take the probability of missing quads on a single hand twice, multiply those together, multiply the single-hand probability of hitting quads, then multiply all of that by three:

(46/47 * 45/46) * (46/47 * 45/46) * (0.04255319148) * 3 = 0.11702609245

The number of quads hit in this scenario is one, so this remains the same. We now add together the expected number of quads for each of the three scenarios:

0.11702609245 + 0.01040231932 + 0.00023116265 = 0.12765957442

Again:

Probability of hitting quad on single hand hold multiplied by three:

0.04255319148 * 3 = 0.12765957444

0.12765957442 = 0.12765957444 (With small errors due to rounding)

CONCLUSION

Okay, so the first thing that we will identify is how many times more likely it is on this particular hold:

0.12230430637/0.04255319148 = 2.87415120033x more likely to hit at least one quad on this particular hold.

The average number of quads hit in the three-handed scenario is exactly the probability of hitting quads in the single-handed scenario multiplied by three, however, the probability of hitting more than zero quads goes down as a result compared to one single-hand draw * 3. Whether or not this difference is small I leave up to anyone else to decide.

The average number of quads (per hand played) will always be the expected number of quads on single-hand multiplied by three. In terms of probability multiplied by three, the result will be closer to 3x as likely for draws of zero, one or two cards being held because it is much less likely that multiple hands will get quads in these scenarios.

In fact, let's go ahead and look at quads holding two:

Single-Hand:

nCr(2,2)*nCr(45,1)/nCr(47,3) = 0.0027752081406105

That gives us the base probability that we need for both hitting and missing quads as well as the probability of getting quads on one hand. The number of quads one expects to receive doing this three times is simply:

0.0027752081406105*3 = 0.0083256244218315

The first thing that we will do is look at the probability of all three hands hitting quads:

(0.0027752081406105)^3 = 0.0000000213740432

We will then multiply this probability by three as that is the number of quads we will get:

0.0000000213740432 * 3 = 0.0000000641221296

The next thing that we will do is look at the probability of hitting quads on two of three hands. Once again, there are three ways that can happen:

(1-0.0027752081406105) * (0.0027752081406105) * (0.0027752081406105) * 3 = 0.00002304121

The number of quads hit in this scenario will be two, so:

0.00002304121 * 2 = 0.00004608243

Finally, we will look at the probability of only hitting quads once, which can also happen in three different ways:

(1-0.0027752081406105) * (1-0.0027752081406105) * (0.0027752081406105) * 3 = 0.00827947786

Add them together:

0.00827947786 + 0.00004608243 + 0.0000000641221296 = 0.00832562441

Remembering:

That gives us the base probability that we need for both hitting and missing quads as well as the probability of getting quads on one hand. The number of quads one expects to receive doing this three times is simply:

0.0027752081406105*3 = 0.0083256244218315

0.0083256244218315 = 0.00832562441 (Errors due to rounding)

The 0.0083256244218315 is the probability of getting quads in the single-handed scenario multiplied by three compared to:

1 - (1-0.0027752081406105)^3 = 0.00830254045

If we take the probability of getting more than zero quads in the three-handed scenario and divide it by the probability of drawing to quads in the single-handed scenario, the result is:

0.00830254045/0.0027752081406105 = 2.99168207548

Once again, the average number of quads per hand remains perfect to (single-hand probability * 3), but the overall probability of getting quads in any of the three hands is a little less than that because some of it is taken up by catching quads on multiple hands.

We see that our probability disparity for hitting any number of quads, relative to the probability of a single hand hitting quads multiplied by three, is less of a disparity due to how unlikely it is that multiple quads will be caught when drawing holding only a pair.

Holding one card or zero cards and attempting to draw quads will be almost exactly three times the probability because catching quads on multiple hands holding zero or one (based on a three-handed game) will be an extremely rare event.

Solution

For those inclined to solve the overall problem, this is how it would be done:

1.) The probability of being dealt quads remains the same. Because of this, there is no discrepancy in probability between the three-handed and single-handed game and there is no discrepancy in the average number of quads hit (because there never is anyway) due to the fact that three is three times one.

2.) The probability of hitting quads based on holding three cards that are the same is completed above.

3.) The probability of hitting quads based on holding a pair is completed above.

4.) Determine the three-handed probability of hitting any number of quads based on holding one card. Remember, of course, that this can happen by way of hitting the other three of the card that you are holding or by hitting all four of another rank that is not the rank being held.

5.) Same as #4, but for holding zero cards.

6.) Determine how frequent the holds applicable to steps #2-5 are, multiply those by the corresponding probability of drawing to quads with each step, add them together and then add them to the probability of being dealt quads (step 1).

Of course, Step 6 is the tough one. Probably would have to be done by simulation unless someone has already done it somewhere.

Roughly (see https://wizardofodds.com/games/video-poker/analyzer/ https://wizardofodds.com/games/poker/) playing one hand
(i) Dealt Quads = 0.000240
(ii) Playing regular Jacks+ strategy, Quads made = 0.002363
This means approx 10% of Quads are dealt, and 90% are drawn, so the first guess would be 2.8x.002363 rather than 3x.
NB I am assuming each draw in made from the remaining deck ignoring the original five cards drawn. So the chances of getting, say the fourth "A", remains the same for all subsequent hands.
(a) 0.000240: Quads dealt
(b) 0.002123: Quads drawn by first hand (subtract 240 from 2363)
(c) 0.002118: Quads drawn by second hand but not the first (2123*(1-.000240-.002123))
(d) 0.002113: Quads drawn by third hand but not the first nor second (2123*(1-.000240-.002123-.002118))
Total
0.006594.

If you just go for Quads regardless then the chance of making it on the first hand is 1 in 377 (0.002653). Note you throw away straights/flushes/Pair from FH or 2P. The end figure is probably nearer 0.0075.

PS: In response to previous reply 6594 is 2.79x2363 which is close to the numbers coming out (as I imagine most drawn quads will come from Trips.). Using the 2.79 figure on the second bit would imply 0.007402.

Quote: Mission146

Solution

For those inclined to solve the overall problem, this is how it would be done:

1.) The probability of being dealt quads remains the same. Because of this, there is no discrepancy in probability between the three-handed and single-handed game and there is no discrepancy in the average number of quads hit (because there never is anyway) due to the fact that three is three times one.

2.) The probability of hitting quads based on holding three cards that are the same is completed above.

3.) The probability of hitting quads based on holding a pair is completed above.

4.) Determine the three-handed probability of hitting any number of quads based on holding one card. Remember, of course, that this can happen by way of hitting the other three of the card that you are holding or by hitting all four of another rank that is not the rank being held.

5.) Same as #4, but for holding zero cards.

6.) Determine how frequent the holds applicable to steps #2-5 are, multiply those by the corresponding probability of drawing to quads with each step, add them together and then add them to the probability of being dealt quads (step 1).

Of course, Step 6 is the tough one. Probably would have to be done by simulation unless someone has already done it somewhere.

Well, I guess I might as well do steps 4 & 5 because they are easy enough:

Probability of getting quads holding one card

In this scenario, we can assume that the card being held is a high card of some kind AND that we obviously would not be throwing away a pair, trips or quads (otherwise that would be the preferable hold).

As a result, there are two ways that we can get quads:

1.) By drawing the other three of the card that we are holding.

2.) By drawing all four of some other rank.

Okay, so now we look at a hand such as Ad-5c-8c-4h-10s

In this hand, we would almost universally (probably universally) be holding the Ace, so there are four ranks for which we cannot possibly get quads because we are discarding one of that rank for each.

That means that there are nine ranks for which we can get quads, which are:

A-2-3-6-7-9-J-Q-K

The probability of catching Quad Aces on a single hand hold is:

nCr(3,3)*nCr(44,1)/nCr(47,4) = 0.0002466851680543

The probability of catching any Quads of the other eight ranks is:

nCr(4,4)*nCr(43,0)/nCr(47,4) = 0.0000056064810921 * 8 = 0.0000448518487368

We add these probabilities together for a total probability of catching quads of any kind:

0.0000448518487368 + 0.0002466851680543 = 0.0002915370167911

That is our probability of catching quads on one hand, which means the expected number of quads doing this three times is:

0.0002915370167911 * 3 = 0.0008746110503733

The first thing that we will do is determine our probability of missing quads on all three hands, which will also give us our probability of catching quads on more than zero hands:

(1-0.0002915370167911)^3 = 0.999125643906344330088959311188543321402506520969

Therefore, the probability of catching quads on more than zero hands is:

1- 0.999125643906344330088959311188543321402506520969 = 0.000874356093655669911040688811456678597493479031

Relative to the probability of hitting quads on a single hand:

0.000874356093655669911040688811456678597493479031/0.0002915370167911 = 2.99912547394345885945412334103921

As expected, it is slightly less than three times likely, due to the possibility of catching quads on multiple hands on the same hold in our three-handed scenario. Let us now figure out our expected number of quads, on average:

The first thing that we will do is look at the probability of catching quads on all three of the hands, which we will then multiply by three as that is the number of quads we would get:

(0.0002915370167911)^3 * 3 = 0.0000000000743365448

This result reflects the fact that this will almost never happen.

For catching quads on any two of the three hands, we recall that there are three ways that can happen and that we take the probability of missing quads once, multiplied by the probability of hitting quads twice, multiplied by the three ways it can happen:

(0.0002915370167911) * (0.0002915370167911) * (1-0.0002915370167911) * 3 = 0.000000254907159933542137887330369964207519562907

Which we will again multiply by two as that is the number of quads we receive in this scenario:

0.000000254907159933542137887330369964207519562907 * 2 = 0.000000509814319867084275774660739928415039125814

Finally, we determine the probability of hitting quads on only one of the hands and multiply by the three ways that can happen:

(0.0002915370167911) * (1-0.0002915370167911) * (1-0.0002915370167911) * 3 = 0.000874101161716888095492089552000035792480437093

We then add these results together to get the expected number of quads:

0.000874101161716888095492089552000035792480437093 + 0.000000509814319867084275774660739928415039125814 + 0.0000000000743365448 = 0.000874611050373299979767864212739964207519562907

0.0008746110503733 = 0.000874611050373299979767864212739964207519562907 (Negligible difference due to rounding)

As we can see, the expected NUMBER of quads with this three-handed hold is the same as single-handed expected number of quads multiplied by three, but the overall probability of catching more than zero quads is not quite three times as great. As we expected, it is closer to three times as great than holding a pair or holding trips because it is significantly less likely that multiple hands will result in quads.

The post after this will go over the case holding zero cards and the post after that will summarize everything.

The next thing that we will do is look at the probability of catching quads on a total redraw. We imagine a hand such as:

2d-4h-7c-9s-10d

Probability of Quads Holding Zero

The first thing that must occur to us is that Quads are now impossible for five of the thirteen ranks because we have thrown away one of each of those cards. Therefore, there are eight ranks for which quads are still possible:

nCr(4,4)*nCr(43,1)/nCr(47,5) = 0.0000280324054607 * 8 = 0.0002242592436856

That reflects the probability and expected number of quads for a single hand draw. Doing it three times:

0.0002242592436856*3 = 0.0006727777310568

Reflects the number of quads we expect to receive on three attempts.

The next thing that we will do is look at the probability of hitting more than zero quads doing this three-handed:

1 - (1-0.0002242592436856)^3 = 0.000672626865710157302256534806874397888709910016

We will see how this relates to the probability of catching quads on a single-hand:

0.000672626865710157302256534806874397888709910016/0.0002242592436856 = 2.99932727256115157843732347164736

We see that this is slightly closer to the single-hand probability multiplied by three. One might have expected it to be even closer than this, but in truth, holding one card and drawing to quads isn't that much more likely than holding zero cards and drawing to quads.

Okay, so now we look at our probability of hitting quads on all three hands and multiply by the number of quads, which is three.

(0.0002242592436856)^3 * 3 = 0.0000000000338354778

The next thing we do is look at our probability of hitting two out of the three quads, which can happen three ways:

(0.0002242592436856) * (0.0002242592436856) * (1-0.0002242592436856) * 3 = 0.000000150842789657469289565695216806333870269952

We multiply this by two because that is the number of quads we will hit:

0.000000150842789657469289565695216806333870269952 * 2 = 0.000000301685579314938579131390433612667740539904

Finally, we look at the probability of hitting exactly one quad out of three, which can happen three ways:

(0.0002242592436856) * (1-0.0002242592436856) * (1-0.0002242592436856) * 3 = 0.000672476011642007218740019362703193666129730048

We will now add these together for the expected number of quads:

0.000672476011642007218740019362703193666129730048+0.000000301685579314938579131390433612667740539904+0.0000000000338354778 = 0.000672777731056799957319150753136806333870269952

And thus:

0.000672777731056799957319150753136806333870269952 = 0.0006727777310568 (Difference due to rounding)

SUMMARY AND CONCLUSION

The original question had to do with the probability of hitting quads in a three-handed video poker game as compared to a single-handed video poker game.

We have determined that hitting quads in a three-handed video poker game is not 3x as likely as it would be in a single-handed video poker game, however, the expected number of quads remains consistent with the expected number in a single handed video poker game multiplied by three.

We have now determined the probability of hitting more than zero quads based on various holds for both a single-handed game and for a three-handed game. Those probabilities are as follows:

Holding Three Cards

Single-Handed: 0.04255319148

Three-Handed (Overall) = 0.12230430637

Three-Handed All Three Hands: 0.00007705421

Three-Handed Two of Three Hands: 0.00520115966

Three Handed One of Three Hands: 0.11702609245

Holding Three Cards Average Number of Quads

Single-Handed Three Times: 0.12765957444

Three-Handed (Overall): 0.12765957444

Three-Handed All Three Hands: 0.00023116265

Three Handed Two of Three Hands: 0.01040231932

Three Handed One of Three Hands: 0.11702609245

Holding Two Cards

Single Handed: 0.0027752081406105

Three Handed (Overall): 0.0083025404440432

All Three Hands: 0.0000000213740432

Two of Three Hands: 0.00002304121

One of Three Hands: 0.00827947786

Holding Two Cards Average Number of Quads

Single Handed Three Times: 0.0083256244218315

Three-Handed Overall: 0.0083256244121296 (Difference due to rounding)

All Three Hands: 0.0000000641221296

Two of Three Hands: 0.00004608243

One of Three Hands: 0.00827947786

Holding One Card

Single-Handed: 0.0002915370167911

Three-Handed (Overall): 0.000874356093655669911040688811456678597493479031

Three Handed All Three: 0.0000000000247788483

Three Handed Two of Three: 0.000000254907159933542137887330369964207519562907

Three Handed One of Three: 0.000874101161716888095492089552000035792480437093

Holding One Card Average Number of Quads

Single-Handed Three Times: 0.0008746110503733

Three Hands Overall: 0.000874611050373299979767864212739964207519562907

Three Hands All Three: 0.0000000000743365448

Three Hands Two of Three: 0.000000509814319867084275774660739928415039125814

Three Hands One of Three: 0.000874101161716888095492089552000035792480437093

Holding Zero Cards

Single-Handed: 0.0002242592436856

Three-Hands Overall:0.00067262686571015728802958505792

Three Hands All Three: 0.0000000000112784926

Three Hands Two of Three: 0.000000150842789657469289565695216806333870269952

Three Hands One of Three: 0.000672476011642007218740019362703193666129730048

Expected Number of Quads

Single-Handed Three Times: 0.0006727777310568

Three Hands Overall: 0.000672777731056799957319150753136806333870269952

Three Hands All Three: 0.0000000000338354778

Three Hands Two of Three: 0.000000301685579314938579131390433612667740539904

Three Hands One of Three: 0.000672476011642007218740019362703193666129730048

SOLVING FOR A GAME

In order to solve this for a given game or paytable, the first thing that you will have to do is determine the following:

1.) What is the Frequency of Being Dealt Quads?

2.) What is the frequency that you will hold 3OaK and draw?

3.) What is the frequency that you will hold a pair and draw?

4.) What is the frequency that you will hold a single card and draw?

5.) What is the frequency that you will hold nothing and draw?

When these frequencies become known, then you can multiply those frequencies by the probabilities above and add them together to determine the overall probability of finishing more than zero hands with a 4OaK playing a three-handed game.

As we have determined, this probability will be less than 3x the probability of finishing with a 4OaK for a single-handed game, but the expected number of 4OaKs will be exactly the same as that for a single-handed game multiplied by three.

Once again, the reason for this discrepancy is because you can draw to multiple 4OaKs in one play three-handed.

The game and paytable is going to be relevant because that will impact the optimal strategy for the game, which impacts your hold frequencies.

For one example, in Double Double Bonus, you will break up a Full House of Aces Full in favor of holding only the Aces, whereas you would keep the Full House in Jacks or Better. Therefore, the frequency with which you will hold Three Aces goes up in DDB which impacts the overall frequency of holding 3OaK and drawing.

Similarly, in DDB, you break up 2P if one of the pair is a pair of Aces, but in JoB, you would simply keep the two pair. Consequently, you will draw to a Pair of Aces more frequently on DDB which also means that you are drawing to a pair more frequently, in general.

With that, the game and paytable will impact your hold frequencies and each game and paytable must be determined for these frequencies individually, although, some games might be exactly the same for one or more paytables.