5's and 10's

[Ulysses]

I did re-read what you said, and here's what you said in two posts:

So, I guess what I'm saying is that 5s and Ts have more relative inherent value than 2s, 3s, 4s, Js, Qs and Ks, because they make more straights

5, 6, 7, 8, 9, and Ts have more power. As such, both 5s and Ts have more power than the avg. card (since the avg. card includes the power of 2s, 3s, 4s, Js, Qs, and Ks).

My point is that even if your statements relate to the specific case of what your second card is given a random first card I intuitively would lean towards the group containing Js,Qs, and Ks, instead of leaning towards the group that makes more two card straights.

You also say:

the 'value jump' (i.e. increase in value) from a 4 in your hand to a 5 is a greater jump than from a 5 to a 6. Similarly, the jump from a 9 to a Ten is less than from a Ten to a Jack.

Now you're saying jumping from a Ten to a Jack is an increase in value. If you're correct on that count (and I think you are), then doesn't that go against your two-card straight power theory and groupings?

[Josh W]

you really pick and choose what you cut and paste...

yes, a jack is more powerful than a ten. we all know this.

but, and i've honestly said this about 6 times now, the increase in value from a jack to a ten is less than the increase in power from a ten to a nine.

similarly, the increase in power from a 6 to a 5 is less than from a 5 to a 4.

the bigger card has more power, more value. however, the value you gain by having a bigger card (i.e. a jack) is mitigated somewhat by having fewer straight possibilities.

i've said this about 94 times now. if you still don't get it, either i'm explaining it poorly, or you are understanding it poorly. regardless, it's a fact, and my voice is wearing out from repeating myself so much.

josh

[Ulysses]

if you still don't get it, either i'm explaining it poorly, or you are understanding it poorly. regardless, it's a fact, and my voice is wearing out from repeating myself so much.

Probably one or the other, though I'm not convinced your position re: their values is a fact. I suspect we're both reading different things into these statements. No biggie. I think I get part of what you're saying now, though - given two hands of x9 and x10, you'd rather change the 9 to a 10 than change the 10 to a J.

I cut and paste parts of your answer that I felt related to the initial question - is there some special inherent value to 5's and 10's. With these two statements,

So, I guess what I'm saying is that 5s and Ts have more relative inherent value than 2s, 3s, 4s, Js, Qs and Ks, because they make more straights

5, 6, 7, 8, 9, and Ts have more power. As such, both 5s and Ts have more power than the avg. card (since the avg. card includes the power of 2s, 3s, 4s, Js, Qs, and Ks).

I thought you were making a case that 5s and 10s did indeed have more inherent value (as a random second card) than the other cards you listed, which is what I disagree with. If you're not saying that, then I guess we don't disagree. [img]/forums/images/icons/smile.gif[/img]

[Bozeman]

Let Rx be the average winning percentage for all hands xy where y is any other card.

R2<R3<R4<R5<R6<R7....<RQ<RK&l t;RA. (One could argue theoretically against this, if one viewed straights as more important than pairs, etc., but practically there is no argument)

However (R5-R4)>(R6-R5), and (RJ-RT)<(RT-R9).

That is the point, based on the greater number of straights that 5,6,7,8,9,T can make.

Clear yet?

Craig

[Ulysses]

Thanks, Craig. I actually got that part, but it just didn't seem so obviously correct to me. I mentioned in my post that if that part is correct, given the opportunity to exchange x9, y10 one would rather take x10, y10 than x9, yJ. To me that feels like a tossup, but I'd probably take x9,yJ. Whatever. I suspect Josh is very correct when he says SUM(xJ)-SUM(x10) is very close to SUM(x10)-SUM(x9). (where SUM(xJ) is the sum of EVs for all hands containing a Jack)

[Ulysses]

I used this table. Heads-up EVs

So, I added up the "Pot Equity" column for:

Hands w/ a 9: 1265.1
Hands w/ a T: 1310.6
Hands w/ a J: 1353.4

diff(T,9) = 45.5
diff(J,T) = 42.8

So, it's very close, but it looks like (heads-up, at least) Josh is right!

I suspect this would have been a much shorter thread anytime but a holiday weekend. [img]/forums/images/icons/smile.gif[/img]

[MSchmahl]

In case you're interested, I calclulated these from Heads-Up equity vs. random opponent's hand.

R(A) = 61.1%
R(K) = 57.8%
R(Q) = 55.4%
R(J) = 53.3%
R(T) = 51.6%
R(9) = 49.7%
R(8) = 48.2%
R(7) = 46.9%
R(6) = 45.9%
R(5) = 45.0%
R(4) = 43.8%
R(3) = 42.6%
R(2) = 41.4%

i.e.: If you dealt two hold'em hands face down and looked at exactly one card, R(x) would be the probability of that hand winning, where x is the rank of the card you looked at.

R(5)-R(4) = 1.2% > R(6)-R(5) = 0.9% is true
R(J)-R(T) = 1.7% < R(T)-R(9) = 1.8% is true

I don't know if these numbers are very meaningful, but it was an interesting way to waste half an hour.

[Bozeman]

Hmm, I wonder if the differences are meaningful:

R(A) = 61.1%
R(K) = 57.8% d12=3.3
R(Q) = 55.4% d11=2.4
R(J) = 53.3% d10=2.1
R(T) = 51.6% d9=1.9
R(9) = 49.7% d8=1.9
R(8) = 48.2% d7=1.5
R(7) = 46.9% d6=1.3
R(6) = 45.9% d5=1.0
R(5) = 45.0% d4=0.9
R(4) = 43.8% d3=1.2
R(3) = 42.6% d2=1.2
R(2) = 41.4% d1=1.2

Notice that d1=d2=d3, but for higher cards the differences increase, with the noted exceptions around 5 and T. (Note: in the absence of hig card value, d5 should be less than d1, since d1 involves an increase in the number of straights allowed.) Presumably high card value increases with the number of undercards, apparently non-linearly.

In a real game, I would guess that the weighting toward big cards is even more extreme.

Craig

[Trefo]

No......but play that against me and raise preflop.

[Josh W]

AH!! The value of intuition [img]/forums/images/icons/smile.gif[/img]

Thanks all for doing the number crunching...

Josh