That Computer Bob is one smart son of a gun!
I'm having difficulty understanding why this would be a card selection without replacement problem. If the question was what is the average of the next card after a 10 counter was drawn I agree the average would be lower but the question is what is the point total of the flop. If the first card is a 10 it does lower the average of the remaining cards but we already have 10 points in the house?
We are not burning any cards so all cards are in play....no?
Darts: 1-1-0 0 units
An easy way to see "without replacement" is to describe "with replacement":
The problem statement "with replacement" would be "first deal the hand, instead of a flop which is done "without placing cards back in the deck", deal a card, write its value down, put it back in the deck, deal another card, sum its value with the first value and write that down, put that card back in the deck and deal a third card, and add its value to the previous sum."
The probabilities are different for the two cases. For example, in your example, the chance for a sum of 30, what I call P30, is to get 3 ten values.
With replacement: P30 = (16/52)*(16/52)*(16/52)
Without replacement: P30 = (16/52)*(15/51)*(14/50)
Note also that we are assuming that the bet is made before anyone can look at their two hole cards (e.g. someone seeing pocket Aces would skew their calculation of the probabilities toward a higher breakeven point).. Whether a card is burned doesn't matter as long as no one sees the card.
I ran a longer simulation of 1,000,000 flops and got the average total = 19.607. P20 = 6.648%. 21 is the most frequent total at 8.042%
I admit most of the explanation is still too much for me to wrap my head around, so I'll just take your word for it computer bob. thanks for everything, will make holdem more fun for degenerates like me.
"It's not an addiction, it's a commitment"
Knowing your hole cards improve your chances.
Is this a bet you make prior to the deal? If not it could compromise the integrity of the hand.
9th Place in Dmans 2014 Contest
"You gotta figure some races are rigged.There's not much you can do about it as a bettor. You just hope you got money on the one that's supposed to win."
I was wrong about RAS 2013 season :(
delcodonny, all bets made before hole cards are dealt. Was tired of just betting the usual red/black and suits.
HigherPOVI admit most of the explanation is still too much for me to wrap my head around, so I'll just take your word for it computer bob. thanks for everything, will make hold'em more fun for degenerates like me.
One more try, since from your other posts you know some probability theory.
The key element to understand is that if I know the probabilities for each sum from 3 to 30 (called a discrete probability distribution), then I know everything about the solution.
The exact solution requires enumerating all of the possible combinations of cards for each sum. In the post above, I computed P30 which is easy (worked out its = 3360/132600 = 2.534%). Since the simulation gives only a approximate answer, I wrote a quick program to compute the exact probabilities.
For completeness, here they is the complete exact distribution (average total is 19.601538):
3 0.018%4 0.109%5 0.217%6 0.416%7 0.615%8 0.905%9 1.213%10 1.593%11 1.991%12 2.805%13 3.729%14 4.344%15 5.086%16 5.502%17 6.045%18 6.281%19 6.624%20 6.661%21 8.018%22 6.986%23 6.262%24 5.339%25 4.706%26 3.873%27 3.348%28 2.606%29 2.172%30 2.534%
Okay I can understand now if using the example P30. But isn't that a special case because you can only draw from a pool of the same 16 face cards to achieve a sum of 30. But how would you write that formula for say P20. With P20 couldn't any card can be the 1st card and still have the total ending up being 20. Would you need a different equation to explain P20. Thats the part that I cant get. It looks like you'd need advanced math to explain right?
HigherPOVOkay I can understand now if using the example P30. But isn't that a special case because you can only draw from a pool of the same 16 face cards to achieve a sum of 30. But how would you write that formula for say P20. With P20 couldn't any card can be the 1st card and still have the total ending up being 20. Would you need a different equation to explain P20. Thats the part that I cant get. It looks like you'd need advanced math to explain right?
Yes, of course, each P value has a different equation. Let's look at the other end to get a flavor for what is required.
In a similar fashion to P30, P3 is given only by A,A,A so P3 = (4/52)*(3/51)*(2/50) = 0.018%
Looking at P4 we see we have combinations AA2, A2A, and 2AA that add to 4. The denominator D is always 52*51*50= 132600 so let's just refer to the numerator, always dividing by 132600 (D) to get P4. The numerator is 4*3*4 + 4*4*3+ 4*4*3 = 3*48=144. P4 = 0.109%
For P5, we have the cases A22 (2A2 and 22A) as well as AA3 (A3A and 3AA). Numerators in this case are all 4*4*3 because we have one singleton and a doubleton so overall = 6*48=288. P5=0.217%
Slightly more complicated is P6's numerator, made up of AA4, A23, 222 which are 4*3*4 (3 combos), 4*4*4 (6 combos), and 4*3*2 (1 combo) or 3*48+6*64+1*24 = 552. P6 =0.416%
Now let's see what P20 would look like. We know the first few terms look like A,9,10; A,10,9; 9,10,A; etc. and 2,8,10; 2,10,8; 10,8,2; etc. and 2,9,9; 9,2,9; etc. Let's write the numerator as W(A,T(19))+W(2,T(18))+...+W(10,T(10)), where the W function indicates combinations of the first card with the T function indicating the total of last two cards.
P20 = (summation [W(I,T(20-i)] from I = 1 to 10) / D
we can see that there are a lot of terms involved in computing P20. If we had plenty of time, we could tediously write them all down. This is why I wrote a program ( a bit tricky because after the first and second cards are counted they need to be remembered) to automate the computation.
Hopefully this gives a flavor of how to attack problems of this type.
If anyone is still following this thread, take a shot at computing P7 (show your work, its pretty easy).
I JUST OPENED THIS THREAD, PRETTY INTERESTING THOUGHT, BUT WOULDNT THE OBVIOUS BET ON PRE-FLOP ALWAYS BE UNDER 19.599.... I WOULD IMAGINE IF VEGAS SET THE O/U FOR A PRE-FLOP (RANDOM) IT WOULD BE JUST UNDER THAT VALUE. JUST A THOUGHT.. GOOD CONVERSATION.. AS FOR ME TO TAKE A SHOT AT SHOWING MY WORK IN COMPUTATING THIS P7 THEORY--I AM GIVING IN TO THE BETTER KNOWLEDGE OF ComptrBob...... awesome man!