How much difference does it make?

[m1illion]

When playing live, cards are burned to prevent cheating. Another consequence is that cards are removed from play that will improve your hand. We have all seen a flop come 444 or [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img] [img]/images/graemlins/diamond.gif[/img],etc, so it is not unreasonable to say that the burn pile can have these same contents at times.
Online, however, there is no burn of any kind. It follows then that you are drawing to the entire deck. Therefore, miracle suck outs occur at a greater frequency than in live play. The question is, how much more often? How great is the effect on cash games vs. tourneys?

Sorry, if this has been covered, I couldn't find it.

[waffle]

Burn cards make 0 difference in the rate of 'miracle suck outs'. The game will play exactly the same, burn card or no.

[m1illion]

[ QUOTE ]
Burn cards make 0 difference in the rate of 'miracle suck outs'. The game will play exactly the same, burn card or no.

[/ QUOTE ]
It really can't.

[SamIAm]

[ QUOTE ]
[ QUOTE ]
Burn cards make 0 difference in the rate of 'miracle suck outs'. The game will play exactly the same, burn card or no.

[/ QUOTE ]
It really can't.

[/ QUOTE ]
I don't know what you're trying to say.

Waffle is right, of course; removing a random card doesn't make the next card any less random.
-Sam

[m1illion]

Heads up
dealer gives you two cards
dealer gives me two cards
dealer burns three cards
dealer flops three cards
dealer burns a card
52-2-2-3-3-1= 41 possible turn cards

computer deals you two card
computer deals me two cards
computer flops three cards
52-2-2-3=45 possible turn cards

[SamIAm]

I can't tell if you're kidding.

Assuming this isn't just a troll, let's say we're each dealt AA, and the flop is 2[img]/images/graemlins/spade.gif[/img]3[img]/images/graemlins/spade.gif[/img]4[img]/images/graemlins/spade.gif[/img]. By your calculation above, what are the 4 cards that are possible in the 2nd example that aren't possible in the first?

I suppose you could say "you can't be dealt the burn cards", but we don't know what those are. And that's the point. Any card is equally likely, and all 45 are possible.
-Sam

P.S. Dealers don't burn 3 cards before the flop.

[m1illion]

[ QUOTE ]
I suppose you could say "you can't be dealt the burn cards",

[/ QUOTE ]

This is exactly what I am saying.
The dealer holds in his hand a finite set of cards. I don't care what they are, I am interested in their relationship to the finite set of cards that the computer is dealing from, which is a larger set.

[cantona]

#Absolutely no difference whatsoever.

All unseen cards are equal in terms of probability, just as the cards in the other players hands are. To you they are just unseen.

For every time the burn card would have helped you , it would have hurt you exactly the same amount of times. I suppose thats the easy way to test it at home with a deck of cards, without doing the maths etc.

[Winenose]

Interesting..

Lets say this is the situation(heads-up):
Player A has a made hand (pair / whatever)
Player B has a drawing hand (straight / flush)

When dealer burns a card before dealing the flop then there is 1/48 chance that one of your outs is already gone.

Does not happen on internet.

I understand that the burned card is completely random, but it does seem to reduce the number of possible outs.

I guess I need a longer explanation to understand why it doesn't matter.

[Kerth]

I'm sorry that this ended up being long and boring, but hopefully this'll be of some help.

This is an example where the play is heads-up, no cards are burned before this situation and now we're finding out whether or not player A should want to burn a card before dealing the river.

Player A: A[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/heart.gif[/img]
Player B: 8[img]/images/graemlins/spade.gif[/img]8[img]/images/graemlins/diamond.gif[/img]

Board: 8[img]/images/graemlins/heart.gif[/img]8[img]/images/graemlins/club.gif[/img]Q[img]/images/graemlins/heart.gif[/img]J[img]/images/graemlins/heart.gif[/img]

8 cards are used, 44 are left in the deck. Player A has just one out, the ten of hearts, so the probability that he will win is 1/44. Now let's see how that probability changes if we decide to burn one card before dealing the river.

The probability that the burn card is the Th is 1/44. After burning the T[img]/images/graemlins/heart.gif[/img], there are 43 cards left in the deck, none of which help the Player A to win, so the probability that he will win is 0/43 = 0. One time in 44 burning causes the Player A's situation to be hopeless.

The probability that the burn card is NOT the T[img]/images/graemlins/heart.gif[/img] is 43/44. After burning the non-T[img]/images/graemlins/heart.gif[/img] card, there are 43 cards left in the deck, 1 of which turns the Player A into a winner, so then the probability that he will win is 1/43. 43 times out of 44 burning causes Player A to win a little bit more often than if there was no burning. Often enough to overcome the rare catastrophe of burning the T[img]/images/graemlins/heart.gif[/img]? Let's find out:

If you're familiar with EV calculations, the following should be self-explanatory. In case you're not: Calculating the expected value happens by multiplying the different possible values with their probabilites and then adding those together. Here we calculate the expected probability of Player A winning when a card is burned before dealing the river:

1/44 x 0 + 43/44 x 1/43 = 0 + 43/1892 = 1/44

Indeed, the rare catastrophe of burning the winning card is offset exactly by the small help Player A gets most of the time from burning, so it doesn't matter whether you burn a card or not. The probability of Player A winning stays the same: 1/44

It doesn't matter if you burn half the deck. It doesn't even matter if everything but the bottom card is burned. It changes nothing.

You get one random card. That's what happens whether you burn or not. Why would the Th reside more or less likely on the top of the deck? You could fan out the deck and pick the river yourself, it would still make no difference. You would
get one random card. You could shuffle the remaining deck before dealing the river and the situation would remain the same. I suspect that the confusion of burning's effect is caused by the perceived loss of possibilities that the burning supposedly causes. But that's just silly. If you deal the top card, then all the cards below that are "burned", they're not used, it's as if they weren't in the deck at all. The point is that you will get exactly ONE random card anyway, it doesn't matter if it's the first, second, third, or thirty first card, because the deck is randomized. Cards have no will, they don't care where they end up and have no means to alter their position. If you have one out before the river, the card you're hoping to get won't suddenly decide to move to the top of the deck to be burned away and piss you off. I keep repeating myself, but here it is once again: YOU GET ONE RANDOM CARD. You could feed half of the deck to your dog and it wouldn't change the probabilities as long as no one knows what those cards
were.

Burning is done to prevent cheating and has nothing to do with sucking out.

I chose this situation because of its simplicity. The points I make are true for different situations too (2 cards to come, preflop, more outs, whatever), and you could do the redundant math yourself if you wish to be sure. Hopefully you don't need to though, as you should be able to understand why it's redundant in the first place.

Everything I've said of course assumes a well-randomized deck. If the dealer wanted to deal in an unconventional way (e.g. from the bottom), it's probably still best to be suspicious in case you're being cheated. [img]/images/graemlins/smile.gif[/img]