Online Algorigthm Debate - Burn Cards

[Shorty35]

OK - So I still have not been able to convince him. Maybe he is making a subtle point that I am not picking up on - here is one of our most recent exchanges

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THE WHOLE POINT IS THAT IT IS RANDOM!!!!!!!!!! It can be random that the card is on the bottom of the deck, or random that it can hit the flop/turn/river. It can be random that it is in the burn, or random that it is in a mucked hand. To remove any RANDOM cards from 2 Identical decks, will change the TOTAL POSSIBLE OUTCOMES between the two decks.

So do this Rich. Take one Ace from the deck and throw it out, rip it, shred it, whatever. Hold two aces in your hand. Run two decks, one that holds 1 ace and the other that holds two. over 100, 200, or 300 thousand deals.The more deals the better. I guarantee you that the POSSIBLILITY of hitting 3 or 4 of a kind is far greater than the deck that only holds 1 ace.

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And my response
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I agree with what you are saying in your example.

BUT the flaws in your "proof" are (1) that you are FIXING the shredded card as an A and (2) for purposes of the 200 deal eaxmple, you are assuming that the sam card is shredded every time (the burn card is random and different on each deal). Use your same hypothetical but instead of an A, tear up a random card. And tear up a different random card each deal. Now the results will be exactly the same as if you had a full deck each time



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And here is his argument restated in his own words

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My good friend and cousin in Law has a great dispute with the following. First, this is based on Texas Hold em between online play and real casino play. In a casino, 1 card is burned on the flop (3 community board cards), a burn on the turn card (4th community board card) and the River (5th) causing 3 cards to be taken out of the deck. Online, there is no reason for them to burn any cards, for there is no risk of anyone marking or seeing the card.

Now, based on a full table or ten players, 20 cards are dealt, leaving 32 cards remaining. Of those 32 cards, 5 have been dealt to the board, and three burned. This leaves 24 cards that are useless on the outcome of the hands, and their position in the deck is random, just as they could have been randomly placed on the top of the deck, or in position to have an effect.

My belief is the following, based on statistical combinations and including randomness of the deal. If there are two Identical decks, one having burned 3 cards and one having not, different outcomes COULD occur based on randomness. It is true that the impact of the burn cards could be identical as if that card had remained on the bottom of the deck or in a hand that has been mucked or discarded, being a non event. But I stand completely firm that in a deck that uses no burn cards, there is a definite POSSIBILITY that card could be in position within the deck and have a definite impact on someone's hand.

I am saying that if I hold AK of spades, there is a definite possibility that 1,2, or 3 spades COULD have been burned and that 1,2,or 3 spades COULD not have been burned. In an online deck, it is definite that 1,2,or 3 spades have NOT been burned, thus increasing the POSSIBILITY of me hitting a flush with an online deck versus that of a deck using burn cards. I realize that nobody knows whether the 3 spades will have an effect or not, as they could have been on the bottom of the deck, or in someone's hand that has been folded.

But the fact remains that this entire game is based on implied odds. I firmly believe that if a software program could run both type of decks for a given number of deals, say 300 thousand, the following would occur: The deck that doesn't use 3 burn cards COULD provide more positive outcomes than the other deck. This number COULD increase as the number of deals increased, as it is random. This outcome COULD also change by fewer remaining players, say 6 versus 10.

The reason I capitalize "possible" and "could" is because my friend here refuses to acknowledge their meaning

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Can anyone help me close this out?

[The Goober]

Well, if the burn or lack of burn cards has an effect on the probability of a player making a hand, then it must be possible to predict what that effect might be. This is step one - if he is simply saying that it must have *some* effect, but can't come up with what the effect is, then his argument must be flawed (if the effect is completely unpredictable, then clearly it can't change the probabilities).

Now, from this last quote, it sounds like he is saying that if he is deal A[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img] he has a slightly higher chance of making his flush if there are no burn cards - intuitively this almost makes sense, because he figures that none of his spades will be killed as burn cards. But now suppose there are 4 players in the hand, one has two [img]/images/graemlins/spade.gif[/img]s, one has two [img]/images/graemlins/diamond.gif[/img]s, one has two [img]/images/graemlins/club.gif[/img]s and one has two [img]/images/graemlins/heart.gif[/img]s. By his argument, *everybody* has a better chance of making a flush. Clearly, this isn't possible since if I have a higher chance of catching my clubs, you must have a lower chance of catching your hearts.

But this could go on and on. Start asking him how the results would change under lots of circumstances that we all know won't change the probabilities - you already mentioned burning two cards instead one. What if we burned from the top, but dealt off the bottom of the deck? What if we shuffled the deck again after dealing the hole cards? What if we shuffled the deck after every single cards that was dealt? What if we actually dealt the turn and river first (putting the cards to side to reveal later)? You get the idea - none of these would change the probabilities, and by asking him to describe to you the effect that each would have, he should find himself forced to pare down his logic until he sees that it just isn't consistent.

[elitegimp]

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OK - So I still have not been able to convince him. Maybe he is making a subtle point that I am not picking up on - here is one of our most recent exchanges
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My good friend and cousin in Law has a great dispute with the following. First, this is based on Texas Hold em between online play and real casino play. In a casino, 1 card is burned on the flop (3 community board cards), a burn on the turn card (4th community board card) and the River (5th) causing 3 cards to be taken out of the deck. Online, there is no reason for them to burn any cards, for there is no risk of anyone marking or seeing the card.

Now, based on a full table or ten players, 20 cards are dealt, leaving 32 cards remaining. Of those 32 cards, 5 have been dealt to the board, and three burned. This leaves 24 cards that are useless on the outcome of the hands, and their position in the deck is random, just as they could have been randomly placed on the top of the deck, or in position to have an effect.

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Can anyone help me close this out?

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no, this leaves 27 cards useless to the outcome, because the burn cards don't do anything.

[kmvenne]

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I am saying that if I hold AK of spades, there is a definite possibility that 1,2, or 3 spades COULD have been burned and that 1,2,or 3 spades COULD not have been burned. In an online deck, it is definite that 1,2,or 3 spades have NOT been burned, thus increasing the POSSIBILITY of me hitting a flush with an online deck versus that of a deck using burn cards.

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Here is where his argument falls apart. If no live game burned cards are spades, then his chance of a spade flush is HIGHER live then it is online. If one, two, or three are burned, lower. The average number of burned spades will be 33/50th of a card. This number evens out the online and live game spade flush draws.

This quoted section is ripe for the picking. Go at it to your friend.

[SNOWBALL138]

you are correct. your friend is incorrect.

[carsten]

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I am saying that if I hold AK of spades, there is a definite possibility that 1,2, or 3 spades COULD have been burned and that 1,2,or 3 spades COULD not have been burned. In an online deck, it is definite that 1,2,or 3 spades have NOT been burned, thus increasing the POSSIBILITY of me hitting a flush with an online deck versus that of a deck using burn cards. I realize that nobody knows whether the 3 spades will have an effect or not, as they could have been on the bottom of the deck, or in someone's hand that has been folded.


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I can't tell if your friend just likes to be contrary or if he knows just enough about probability theory to be thoroughly confused. It seems like he doesn't know the difference between the probability of an event and actually observing an event.

It astounds me that your friend already brings up the argument against his own assertion, but fails to realize its significance: It is just as likely that the cards that would have been burned in live play are sitting at the bottom of the deck in online play.

In an earlier post in this thread I have provided an actual mathematical proof of the fact that the probability for any one specific card to be dealt is the same between burning and not burning. However, as long as your friend doesn't realize or understand this fact, the whole debate is as pointless as debating with a blind man whether the sky is blue.

Hope this helps,

Carsten.

[Yads]

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Anyone that still doubts the assertion that the probability after burning is still 1/29 should consider the following. Either the A [img]/images/graemlins/spade.gif[/img] is burned or it's not. If it is burned, the probability of it coming up as the next card is 0. If it is not burned, the probability of it coming up as the next card is 1/28. The probability for A [img]/images/graemlins/spade.gif[/img] being burned is 1/29, the probability of it not being burned is 28/29.

With all that, the overall probability of A [img]/images/graemlins/spade.gif[/img] being drawn after burning is 1/29 * 0 + 28/29 * 1/28, which simplifies to 1/29, q.e.d.

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The proof was already given. Just show him this. If he still doesn't get it then ask him how his odds of making the flush change if you burn all, but 2 cards.

[blank frank]

OK, your friend is a bone head. (sorry, but he is).

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My belief is the following, based on statistical combinations and including randomness of the deal. If there are two Identical decks, one having burned 3 cards and one having not, different outcomes COULD occur based on randomness.

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??? If you have two identical decks, except one has burned and the other hasn't, different outcomes will occur no matter what. If you deal different cards, you'll get different outcomes. The question isn't what will happen on one hand, the question is what will happen over all the hands. And over the course of all possible hands, every time burning cards changes the winning hand from X to Y, there is a corresponding case where not burning cards changes the winning hand from Y to X, balancing things out.

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But the fact remains that this entire game is based on implied odds. I firmly believe that if a software program could run both type of decks for a given number of deals, say 300 thousand, the following would occur: The deck that doesn't use 3 burn cards COULD provide more positive outcomes than the other deck. This number COULD increase as the number of deals increased, as it is random. This outcome COULD also change by fewer remaining players, say 6 versus 10.


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Having programmed more poker simulations than I expect your friend has, I would have to say 300K is a bit on the low side for firm results. Also, I will unequivocably say he is wrong. Burning cards will not change the results in a statistically significant manner.

Try this on him. Imagine a game where a deck is shuffled, and you win if you draw the ace of spades off the top of the deck. What are your odds of winning? He should say 1 in 52.

Now imagine a second game where the deck is shuffled, and you win if you draw the ace of spades off the bottom of the deck. What are your odds of winning? He should say 1 in 52.

Now imagine a third game where the deck is shuffled, and the top 51 cards are discarded face down (burned). He wins if the remaining card is the ace of spades.

Note that there is no difference between the second game and the third game. The chance of the ace of spades not being the bottom card in the second game is exactly the same as the chance of the ace of spades being burned in the third game. In fact, if you have two identically shuffled decks, the results will be the same every time.

If the second and third games are functionally the same, and the first game and the second game have the same odds of winning, then the first game and the third game have the same odds of winning.

If the first game and the third game have the same odds of winning, burning cards had no effect.

If that doesn't work, I suggest taking the mathematical proof previously give (which should have been sufficient), carving it onto a piece of wood, and beating him over the head with it until he learns the error of his ways. However, while I must emphasize that I am not a lawyer, I think that may be illegal in some states.

[valenzuela]

ure right.