One More Question About Jacks or Better

I just have one more question.
I was wondering why the versions with the different pay tables have different odds if the pay tables still have the same hierarchy. For example: How come 90-9-6 Jacks or Better has 500,790,444 Royal Flushes and 9-6 Jacks or Better has 493,512,264 Royal Flushes.


Thanks.

Not quite sure what you're asking, but it could be because of the different strategy with 90-9-6 JOB, you go for the royal flush more and thus have more chances to hit it. Those odds are for perfect strategy play.

So you're saying that this is because the EV of royal flush is higher?
I'm still confused, though, because if you look at the wizard's distribution tables:
they still account of all possible scenarios, even the less paying ones, at least thats what I understood.

The EV of the straight flush is higher. So, with perfect strategy, you would go for the SF more often, possibly disregarding a 10JAs if you also had the 9s, for example. So with perfect play, the combination of royal flush chances is lessened. I think that's what you are getting it, but maybe somebody else can explain it better.

Quote: teddys

The EV of the straight flush is higher. So, with perfect strategy, you would go for the SF more often, possibly disregarding a 10JAs if you also had the 9s, for example. So with perfect play, the combination of royal flush chances is lessened. I think that's what you are getting it, but maybe somebody else can explain it better.

Actually, you're getting more royals with the 90 SF (500 million @ 90 vs. 493 million @ 50). My guess is that certain "2/3 to a royal" hands (especially ones including a T) gain enough EV through straight flushes that they eclipse hands they would otherwise be marginally below (and couldn't possibly make a royal); e.g., JTs would likely leapfrog QJo, or KQs/KJs might leapfrog AKQJr (this is probably why you also make fewer straights, even though you go for straight flushes more).

That said, you would never hold JT9s over AJTs, though (from Wizard's optimal strategy, AJTs is ~1.2868 while JT9s is ~0.6429 at best, and I doubt a 90 SF will double the EV of JT9s).

/ Although you'd hold AJT9s if the fifth card is a T, K or Q

Quote: teddys

The EV of the straight flush is higher. So, with perfect strategy, you would go for the SF more often, possibly disregarding a 10JAs if you also had the 9s, for example. So with perfect play, the combination of royal flush chances is lessened. I think that's what you are getting it, but maybe somebody else can explain it better.

Yes, but then with perfect strategy, wouldnt we be counting only the appropriate outcomes?
But it seems that we are counting all outcomes, even the ones who do not yield an optimal payout.

Quote: Mic

Yes, but then with perfect strategy, wouldnt we be counting only the appropriate outcomes?
But it seems that we are counting all outcomes, even the ones who do not yield an optimal payout.

Wildqat said it better than I could. We are not counting all outcomes, only the ones that yield an optimal payout. The stated probabilities are for hands you could receive on the draw with optimal play. Since every game has a different strategy, the probability of those hands on the draw will be lessened or increased per the game. Once again, hope that explains ... I feel like I am not doing a good job of it, though.

Quote: teddys

Wildqat said it better than I could. We are not counting all outcomes, only the ones that yield an optimal payout. The stated probabilities are for hands you could receive on the draw with optimal play. Since every game has a different strategy, the probability of those hands on the draw will be lessened or increased per the game. Once again, hope that explains ... I feel like I am not doing a good job of it, though.

Yes, but I was saying that I feel we are not looking at only the optimal outcomes but at all possible outcomes, because otherwise we wouldn't have the same amount of total outcomes for different versions of the game.

No, we are only looking at the optimal outcomes, but you can't just add them up. It goes like this:

Say you're dealt a pat full house. The optimal play, of course, is to draw zero (hold five). This gives you one result after the draw: full house.

The next hand, you're dealt two pair. The optimal play is to hold the two pair and draw one. This has 47 possible outcomes: four full houses and 43 two pair.

Now you have to reconcile these two hands. You can't just say, "I got one full house from the pat full house, and four full houses and 43 two pairs from the two pair, so now I have five full houses and 43 two pairs." You have to equalize the amount of outcomes, as if you played the two pair 47 times and got the four full houses and 43 two pairs, and you played the full house 47 times and came up with 47 full houses, so now you have 51 full houses and 43 two pairs.

The same goes for drawing more cards: the more cards you draw, the more outcomes, and the more you have to compensate for the hands for which you draw fewer cards. The table below sums it up:

Cards held	Possible outcomes	Scaling factor
0	1,533,939	5
1	178,365	43
2	16,215	473
3	1,081	7,095
4	47	163,185
5	1	7,669,695

Going back to the example, the full house's outcome gets multiplied by 7,669,695, and the two pair's outcomes each get multiplied by 163,185, so that you have 7,669,695 + 4 × 163,185 = 8322435 full houses, and 7016955 two pairs.

This puts every dealt hand on equal footing. Each of the 2,598,960 starting hands contributes 7,669,695 outcomes to the total, making a total of 19,933,230,517,200 ending hands, which is the same total as the Wizard's tables.

As the paytable changes, the optimal play for some hands changes. This means that for some given hands, some outcomes are possible for one paytable and different outcomes are possible for another paytable because you're holding different cards. The total amount of outcomes remains the same, though.

This is why I love this site. Where else are you going to get that kind of analysis, for free, and quickly, no less?

Thanks, Wildqat.

Thanks! I finally understand.

So the way I need to approach it is:
1. Draw a hand
2. Compute all 32 combinations and choose the one with the highest EV
3. Count all the outcomes for the combination with the highest EV and multiply by the scaling factor

?

Quote: Mic

Thanks! I finally understand.

So the way I need to approach it is:
1. Draw a hand
2. Compute all 32 combinations and choose the one with the highest EV
3. Count all the outcomes for the combination with the highest EV and multiply by the scaling factor

So if one wanted to know just the total probability (P = n/N) of getting a royal flush in draw poker, without regards to EV,
one would sum (weighted)up:
1)prob on the draw.
2)prob hold 0 cards.
3)prob hold 1 card.
4)prob hold 2 cards.
5)prob hold 3 cards.
6)prob hold 4 cards.
I know I've seen this at the WoO site, but I cant locate it.
Updated: It is in ask the wizard below.
https://wizardofodds.com/askthewizard/videopoker-probability.html