New to these mixed game streets but just wondering if someone can crunch the numbers behind this hand and what are the odds of this all happening.

https://twitter.com/nickschulman/sta...429745665?s=21

.16%

So using permutations is best since the order of the cards matter to his drawing decisions.

One simplifying assumption the drawing hand will always be drawing at the wheel so the draw will always be at least 2-1. This means there were always be at least 2 draws and up to 3 even though that isn't necessarily true.


So given that assumption there are 2 possibilities:

Villain draws 2-1 or villain draws 2-1-1. Notice villain cannot draw 2 to a wheel which is why I am completely ignoring that case.

So we're looking for 3 and 4 card sequences that look like this:

3 card:

742
724
472
427
274
247

4 card:

74x2
72x4
47x2
42x7
27x4
24x7

Since we know what 10 cards are at the start of the hand there are 52-10 = 42 cards remaining.

The total number of 3 card permutations are 42*41*40 = 68,880

The total number of 4 card permutations is 42*41*40*39 = 2,686,320

Now for every 3 card case above we only have 3 outs per card so 3*3*3 = 9 hands that contain 742 but we care what order they arrive in. We have 6 different ways they can arrive for a total of 9*6 = 54 permutations.

For every 4 card hand we have 3*3*x*3 where x can be any other card. Since there are 42 cards left in the stub but we have seen 2 more since we have drawn there are 40 cards left in the deck. Since we wouldn't need to draw again if one of those 40 cards were the 3 remaining outs we would need to make our hand x = 40-3 = 37.

So we have 3*3*37*3 ways to make our final hand but again we have 6 different ways those hands can arrive. Notice the x can't appear like this 7x42 since we stipulated the player would never draw 2 on the second draw. So 6*(3*3*37*3) = 5,944.

So we have all the information we need to make the calculation.

Probability we make our hand in the first 2 draws + probability we make our hand in all 3 draws = 54/68880 + 5,994/2,686,320 = .000784 + .0022 = .002984 * 100 = .2984% chance.

I have a problem with one of your mathematical statements being 3*3*3=9.

Haha omg yea whoops. No idea what I was thinking there. Would obviously be 27.

Also I think I have to take into consideration the probability we didn't make our hand on the 2nd draw so that will slightly scale down the 2nd probability. I'll come back later and see if I can get it right this time. Unless someone else beats me to it.

Poker Central calcd it as .16% and I got something very close but it was assuming he would only continue on if he caught at least a 2 on the 1st D2. That would probably be shortchanging it a little because perhaps he peels with 345 or the UI 356, I don’t know what he does

I don’t think anyone will try and beat you to it so you probably have all of the time in the world lol.

That said it’s probably way easier to calc using probabilities.

I was watching on the rail. One of the craziest things that's ever happened, and sad for those of us who are friends with Bryce :-(

If Arieh was trying to make a wheel only, the chances aren't that bad. There are also chances he will make #2 to chop starting with a 6. The problem is he needs to make 65432 to pitch the 6 and have a chance at the wheel. That is when the chances really get small. Also, Arieh will often make an 8 or 7 low and pat, when he could draw if he was only trying to make a wheel.

There is also the issue of dead cards, and the dead cards of the player who folded and were pitched, which were likely to be high cards. I don't know if it would be easy to get a correct answer analytically. The best way would probably be to run simulations.

If you want to practice taking % beats like that play Archie. Hands like that are like a regular Tuesday afternoon. I lost with a flush vs a D3, think that’s a .2% and have seen other crazy stuff

Only but one way to get the true %, call up Josh and tell him sorry that he came so close only to ultimately fail and be like btw would you have patted 96532?

Once you get the drawing assumptions down the calcs are relatively straightforward imo

Yea even though my calc was wrong the method I was using was going to be an over estimate which I tried to make clear.

Also by beat me to it I meant supply a correct answer, not neccessarily correct my mistakes.

Not sure what you mean by use probabilities since I would assume any probability would have to include counting combinations or permutations of outs either through math or simulation. Perhaps you mean convert the counting to probabilities immediately?

Also I agree with you that once the rules about the drawing decisions are set the calculation should be straight forward without the need for computer simulation. The simulation might be less error prone haha.

Correcting the math mistake from before I get something like .45% which as I said is an over estimate because there are definitely draws included that the player would not take at the wheel.

If I use my method but restrict the sequences to only those starting with 24 then I get .1527% which is in line with Scotch's answer.

I will take Scoth's advice and avoid the sequence thing and see if I get an answer closer to .16 as I may be missing some cases there as I am still off by a decent amount.

I did rough calculations and got about .31%.

You have about a 2/13 chance the first card is a 2 or 4 on the first draw and then a 1/13 chance the second card makes it 65432, so about a 1.3% chance of 65432 on the first draw. Then 1/13 chance of a 7 on the 2nd draw for a 0.1% chance of making a wheel on the second draw. Assuming you pat an 8 on the 2nd draw, you then have about a 0.09 chance of making a wheel on the 3rd draw. So about a .019% chance of making a wheel by making 65432 on the first draw.

Now there is about a 47% chance you don't improve on the 1st draw. If so, and assuming you don't fold, you would again have a 1.3% chance of getting 65432 on the 2nd draw. So about a 0.6% chance. Then you catch a 7 on the 3rd draw about 1/13 of the time. So about a 0.05% chance of making a wheel that way.

Then you have a 2/13 * 10/13 = 20/169 = 12.2% chance of making a 2 or 4 on the 1st draw and not making 65432 or a hand you will pat. If you do that, you have a 1/13 chance of making 65432 on the 2nd draw, then a 1/13 chance of making a wheel on the 3rd draw. So 12.2% * (1/169) = 0.07%.

So I get 0.19% + 0.05% + 0.07% = 0.31%.

This is not considering dead cards, which would make it lower. If you were to consider the dead cards in the #2 hand, you would also have to estimate the dead cards that were discarded or from folded hands, which would tend to be high cards, so the effect of dead cards would not be great.

There are only a few specific things that need to happen so I guess what I meant is that you can just explicitly state them and them sum up the various odds

1.pulls 2 and 4 on 1st draw

2. Gets 4 and 7 and draws to 7543

3. Gets a 2 and other card is not a 4, 7, 8. By not including the 9 here we are making an assumption he breaks pat 9

4. Gets a 4 and other card is not a 2, 7 or 8

5. UI, no 2,4,7, or 8

What you guys are doing may be correct but everyone has their own method so it can be hard to follow

I missed that. You can get to 76543 and pitch the 6 as well as get to 65432. It should be about the same percentage of that happening, so it should double my calculations to about 0.6%.

I think I was more specific than others and would be interested in a critique or verification of my approach. I will try to figure out what the others are saying. I am sure that 0.16% is way too low though. I will work on other approaches to get a more definite answer.

I am only getting .23% using this method and exact numbers. This is not including the 76543 draw.

Also I am getting .16% and some change for your scenario 1 which is all Scotch and the other source where using for the D2's continuing range.

I did it very quickly so I may have made some mistakes but if I add in the 76543 draw for both the first and second draw then I get .32%.

If I add it in for just the first draw it's around .30%.

I wrote a program to do a simulation and got 0.74% based on 1,000,000 iterations and 7353 wheels.

I am going to run it with more iterations. Also, I just generated cards with replacement. I will have it consider dead cards, including those discarded by Arieh in the simulations, and those and both of their hands. I will also estimate the cards discarded by Arieh on the first draw and those discarded by the two players who folded. All of the discarded and folded cards would tend to be high cards. There may be a slightly higher percentage of wheels considering dead cards, because there aren't many 7s, 4s, and 2s dead except in Stockey's hand.

I got 0.62% analytically, so this result is fairly close.

I got 0.738% with 10,000,000 iterations. I didn't calculate the statistical margin of error, but this should be very close to the exact answer given the assumptions I used.

I was assuming Arieh folds made 9s. If he keeps them on later draws, then the percentage is lower.

Accounting for dead cards should make the percentage higher. Yockey had a 2, 4, and 7, but Arieh had none is his initial hand. The two folded hands should not have many of those cards. Therefore 2s, 4s, and 7s should be slightly live. Also, cards discarded on the second draw should not be the ones Arieh needs on the third draw.

I made a mistake. It was 0.717% with 10,000,000 iterations.

Also, it is impossible to make a #2 to chop. If you make a straight, you always ditch the 6 and keep the 5.

Intuitively, 0.16% is too low. You have 3 draws to make a straight and then catch a 7 or deuce depending on the straight. That isn't that impossible.

.16% aa stated before is assuming only that 23456 is drawn on the first draw and that is the only time the player continues.

Even your rough calc before which yielded .19 before was close to that.

Yeh, I assumed that he would always keep drawing and would pat 7s and 8s but not 9s. I practice Arieh probably assumes that a large part of Yockey's range is pat 9s. So Arieh probably usually continues, but also usually pats smooth 9s.

If he doesn't make 23456 on the first draw, Arieh needs to draw to 653 or to a straight 7543, 7653, or 6543. Realistically, assuming he only continues with 23456 on the first draw is too tight, but I don't think he draws to 6543. He could also fold unimproved after the first draw and he probably pats a smooth 9 at least after the 2nd draw.

So a realistic percentage is somewhere between the 0.16% discussed initially and the 0.71% I got.

Totally agree that it's probably too tight given the situation.

Strangely when I try to work it out using the conditions Scotch laid out I am only getting .33% but assuming I am making a mistake somewhere just can't figure out where I am missing such a large portion of cases.

Oh well it doesn't matter too much to me other than curiousity and a little bit of pride at this point haha.

Thanks for the work and discussion.

The vast majority of Yockey’s range is either 7 or an 8. He will have nines for sure but if you look at a list of made 9s you will see that many want to be played as a D1.

And when Arieh is holding a 9 himself it further reduces those combos