Tacos,
I am currently trying to develop an overall formula to describe the thread @ hand. Problem is, do I want to use … Apples, Oranges, Bananas, Grapes, or something else.
You might know some of the cards, none of the cards, or all of the cards. Does that now mean I need 3 formulas? Is there a grand unified formula that takes into account all 3 of those possibilities ? Does it even matter ?
I have so far tried to come @ this problem from the standpoint :
- I know none of the cards, but I know the card in question is still avilable.
- How does having or not having burn cards, effect the implied adj. odds.
Problem is to have a complete answer, I must have a formula that describes ALL possibilities, but if some variables are ignored, the formula still works just fine.
My premise all along was :
Having a bigger pool of cards to draw from ( assuming the card in question has NOT already been used/dealt out/burned ) there is a increase in odds that any 1 specific card can be the next card. Mainly because if the card in question is already gone, your odds are , well… 0% chance, no different than drawing dead.
To me, that was a DUHHHHH (yes), just like saying if you spread all cards face down on a table and can draw 1 or 5 cards, does drawing 5 cards increase your odds of drawing 1 specific card, which I hope we all agree, is a Duhhhh (yes).
Because of some of the replies to my posts, kept adding back in “known cards” in some way, I finally decided to add a part to my formula to describe accurately, or account for , my assumption that you don’t know if the card is “gone”. Lets look now @ what the possibilities are…
- you know none of the cards.
- you know your cards, and community cards.
- you know the dead cards, including burn cards ( if there are any )
- some combination of the above #2 & #3 options.
As you move from 1 to 2, then 2 to 3, and finally 3 to 4… your formula gets more complicated, so thats why originally I chose option #1, with the only caviot being… the card in question has NOT been used/dealt out/burned. That is a form of “knowing” , so then I need’d a way to express that in the formula… because if the card in question is gone and your odds are 0%, any small chg in the odds , from there on, end up still being 0%. Other posters, kept adding back in some form of “knowing” , or ignoring I said… “assuming the card in question is still available”.
Basic odds here, if you are not a card-mechanic or a magician, and you spread all cards on a table face down ( yes randomized )… are 1:52 or 1.923%, that you can correctly draw 1 specific card, and since its the 1st card… your odds are 100% that card is still available. Therefore you effectively have a 100% chance of having a 1.923% chance.
( or 1 times .01923 = a 1.923% chance ) From there, this thread in my opinion seemed to devolve into a plethera of different answers. From the 2nd card on, you have 2 possibilties. Either it is gone and your chance is 0%, OR it is not gone, and depending on a few variables you chance is something other than a 0%.
To that end I tried to use the dice example, where success depended on rolling two Dice ( independantly ) and then combining those odds to a overall adjusted odds for success, by multiply’n them together. ( if you have (2) 6-sided dice, you need a 1-2 on the 1st one, and a 6 on the 2nd one (( so 2:6 times 1:6 )) therefore an adjusted odds of 1:18 (( 1:3 x 1:6 ))) … We both agreed on, 1:18 is correct, in that example. ( the 45 min talk with an un-named source.)
I spoke with someone in realtime here online, I won’t say whom, but after 45 minutes we could not agree on the correct way to … describe a generic way to combine 0% and something else, AND how that effected any other basic odds there are. I tried to tackle them 1 @ a time, but that didn’t seem to work well… I became frustrated, he had to get back to his day job, so the conversation ended without a resolution we both could agree apon. I thank’d him for the time he spent, and we went our separate ways.
I spoke to 2 ppl in Real Life, who only heard the orig question, not our discussions, who both told me “Yes” , having or not having Burn Cards does effect the odds.
I don’t use Odds-Calculators, but those rely on someone knowing thier cards, the community cards/lack thereof, and the range of thier opponents cards. I have never heard of an odds calculator including “burn cards/no burn cards”… but they must have taken into account somehow “cards removed from play, ie- the burn cards”. For the simple fact that on some level, if the question is concerning the River, they MUST account for ALL the cards that came before and are no longer in the “pool” of cards that are available “on the river”.
So, lets start by spreading all 52 cards on a table, face down ( assuming no funny buisness ). You can pick 1 card, thus your odds are 1:52 or 1.923% of correctly picking 1 “named card”.
Now I take away 3 cards, I do NOT look @ them. As described above, the “named card” is either gone, or its not. IF its gone, your chance is 0%, if its not gone ( 100% ) your chance is 1:49 or 2.041% of drawing that “named card”.
If we “assume” the card is not gone ( 100% ), your chance has increased from 1.923% to 2.041%. ( I bet we don’t all agree ). So, in order to “not assume” we must develop a mathmatical way to corretly assign a value between 0% & 100%, to eliminate that assumption. Then multiply that % chance, with whatever the odds are going forward, just like I did in my Dice example. ( I really think we don’t all agree ) Thus achieving a “adjusted” odds for success.
I really don’t think we need to go any further here. If the chances are NOT 0%, then it should be true your odds have increased from 1.923% to 2.041%, and the only other number we need to find and then multiply with, is what are the exact odds that those 3 cards are NOT the “named card”.
So, in my above Dice example it would look like ( ? % times 1:6 or (16.666667%)), yet this is exactly where everyone seems to to diverge and have different answers or claim I’m wrong, or its impossible to do it that way.
If I do the simple math (49/52) to represent the odds of the 49 cards left, thus containing the “named card” , multiplied by (1/49) to represent the odds of picking 1 of the remaining cards… therefore we get (49/52)(1/49)= .01812 … or 1.812% chance of picking the named card.
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1.812% is Less than 1.923% , that by removing 3 cards ( representing the Burn Cards ) and accounting for unknown-unknown (the card might have been removed) the adjusted implied odds have gone Down in reguard to someone picking the 1st card, and only 1 card.
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But, if we say those 3 cards are known-unknown ( the card has Not been removed ), the adjusted implied odds have gone UP , from 1.923% to 2.041%, in reguard to someone picking the 1st card, and only 1 card.
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Obviously if the 3 cards are known-unknown ( and 1 of them is the named card ) the adjusted implied odds are 0%, in reguard to someone picking the 1st card, and only 1 card.
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Obviously when I say “picking the 1st card” I am refferring to the 1st card picked from any/all remaining cards on the table, and it being the “named card”.
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Obviously I am using a term often used in the military… known-known , known-unknown, and an unknown-unknown. ( like the threat country is XXXX and the threat is from YYYY … so we can know both, just 1, or neither )
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Having burn cards do change the implied odds, when compared to the orig odds of having 52 cards, which was 1.923 %
My original hypothetical, mirror’d #2 and corroborated #6. We don’t really know any of the cards, but we did know that the target card was still available. Then due to no burn cards or with burn cards, what are the effective adjusted implied odds of drawing a named card, and is there a difference between the two, if so is it a increase or decrease.
I think we can ALL AGREE, that ( picking up cards 1 by 1 )… knowing you have NOT picked the target card, you odds steadially increase on each successive card.
I hope I have shown that by eliminating burn cards, thus increasing the pool of cards to draw from, and knowing the target card is still available… the effective adjusted implied odds, go UP of picking the named card.
This is a very narrow, semantically specific, hypothetical situation. Any rebuttle, can only change the outcome ( show how I am incorrect ), but not change the hypothetical itself. I welcome and hope actually, that if I am still wrong, someone sets me straight so I do NOT make this mistake in the future.
Sassy