Because in other cases, where a stupid mistake won the others a hand, I’m sure the newbie wasn’t thanked.
I think a player could reason that taking a card could hurt the other players. But why? The player should always do what is best for him. He doesn’t care if the other players win or lose.
And the answer is no. The newbie should play his hand, he can’t affect the other players unless he has knowledge of what cards are coming.
Suppose there are only two cards left, one of which busts the dealer (although the player can’t know this) and one of which gives him blackjack. Let’s say they’re a ten and a five for the dealer has 16 example.
We don’t know the order of the two cards. So if the newbie draws a card he has a 50% chance of getting the 5, leaving the dealer a 100% chance of busting. However if the newbie doesn’t draw the card the dealer has a 100% chance of winning. Overall newbie taking cards == dealer has 50% chance of winning.
( See where we’re going ) Newbie not taking card, dealer has a 50% chance of drawing the bust card, hence 50% chance of winning.
So the newbies action had no effect on the overall game.
Applying the same principle to the general case (we’ll ignoring drawing multiple cards here).
It’s been a very long time since I’ve had to do this sort of stuff, so sorry in advance if I screw it up.
If the dealer has N cards that bust him out of M left then the following is true, I’ll give different events letter ( A -> D ) to make things clearer.
Dealer prob. to bust if newbie doesn’t draw … N out of M
Dealer prob. to bust newbie draws …
IF:
[A] … Newbie draws a dealer busting card … (N-1) out of (M-1)
** … Newbie draws a non-dealer busting card … N out of (M-1)
… Prob. of [C] newbie drawing a dealer busting card … N out of M
… Prob. of [D] newbie drawing a non-dealer busting card … (M-N) out of M
Since pairs ([A], **) and ([C], [D]) are mutally exclusive but the combination the dealers draw is independant of the newbies draw once we’ve taken the changed probablities into account.
Thus, I think, the dealers chance to bust if the newbie draws is
([C] * [A]) + ([D] * **)
= (N/M * (N-1)/(M-1)) + ((M-N)/M * (N/(M-1))
= (N*(N-1)) / (M*(M-1)) + ((N*(M-N)) / (M*(M-1)))
= (NN - N + NM - NN) / (M*(M-1))
= (NM - N) / (M*(M-1))
= N(M-1) / M(M-1)
= N/M
= Same as if the newbie hadn’t drawn a card.
Of course if you look at the card the newbie draw then you can know if it was a good or bad call, but the newbie can’t change the probability without knowing the cards in advance. So play yer own cards, count the ones ya can see if you want, but don’t worry about making/breaking others hands.
SD
The answer is no. No matter how bad anyone at the table plays, based on the rules of probability. everyone else at the table has the same chance of winning.
Wrong.
If the dealer is showing a 2-6, then he needs to draw cards to reach a total of 17 or higher.
What are the odds in a single deck (assuming only 1 player against the dealer) to make it “easier”) where the player is holding 12 or higher that the dealer will draw enough cards to gain 16-20 total more points in order to bust his 2-6 showing?
If this is answered, then we know the relative odds of a dealer busting. There are a multiple of factors to consider:
a) all of the permutations for a player hand to get up to 12 points or higher in order to justify standing
b) the permutations possible of undercards for the dealer who is showing a 2-6
c) the permutations of a dealer drawing enough cards to bust (16-20 more cards points including the undercard), while not drawing to a 17 through 21.
Anyone want to tackle that math?
CuriousCanuck, what does this have to do with anything?
Here is a Blackjack Odds Table but it doesn’t matter as far as this discussion is concerned.
Well, if you want to know the odds of a dealer busting when you stand on a 12 or higher, then this is a great place to start. If you want to know the odds of a dealer busting while holding a 6, then adjust my caveats appropriately.
When you have that math done, you can even look at the difference between having a 12 or higher and taking a card vs not to see how it affects the outcome of the dealer busting. It is all probability.
I suggest that the OP pose a question to the “newbies stole the dealer’s card” friend:
Let’s say that the dealer is showing a 6, and there are five players at the table. Should all five players refuse to take any hits, regardless of the sum of each of their hands? That is what he seems to be advocating.
If he (sensibly) disagrees with that proposition, then he can’t truly believe his nonsense idea that players hitting helps the dealer.
Every other player at the table could play with mathmatical precision and still get beat by the dealer. This seems like more of a question of the newbie’s understanding of basic Blackjack strategy and using his unsound play as an excuse for being beat at a hand of 21.
Good point Ravenman. A similar question for the OP:
Lets say a friend and I have a chance of winning something. There are four black balls in a hat and one white one. Whoever picks the white ball wins. Do I have a better chance of winning if I let him go first? Probability says that he will most likely pick a black ball leaving me with 4 to 1 odds, where if I picked first my odds would be 5 to 1.
But none of that affects the chances of other players at the table.
If I do not take a card, then the dealer busts if cards 1,2,3… remaining in the shoe bust his hand. If I do take a card, then he busts if cards 2,3,4… bust his hand. The probability is the same in either case.
It is true that taking a card affects the outcome of each particular hand, but taking a card is exactly as likely to make the dealer bust as it is stop him from busting.
Thanks for restating the OP.
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It may be a bad idea to take the card. That doesn’t mean it has a higher probability of adversely affecting anyone else. It has no effect on the probability of anyone’s elses results being better or worse.
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The goal of blackjack is not to help or hurt other players. If the player split aces and drew two 10’s (bust cards), you wouldn’t be mad at him, right? So why is it different if he does it with a 16? Because it’s a worse play for him? Why do you care about that, you’re suddenly concerned with his welfare? No, you don’t care about his hands, he doesn’t care about yours.
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In fact, him taking a card helps you if you are card counting – the more cards revealed, the more you know about the distribution of what’s left.
But mostly #1. It just plain doesn’t affect the expected outcome of the hand. It just doesn’t.
I hit send prematurely last post. I was drawing an anology to poker. Say I have a pair of aces. The other guy has 6-10 off suit and goes all in. I call, and he ends up converting a miracle inside straight. Would you say I made the wrong play? Of course not, the expected outcome was vastly in my favor. Regardless of the actual outcome on that particular hand, it was still the right play. Similarly, the actual outcome of what happens in any particular blackjack hand does not say anything to whether the play that resulted in that particular blackjack outcome was good.
The player who goes first has a 1/5 chance of winning.
The other player, in the remaing 4/5 of the time, has a 1/4 chance of winning. (4/5)*(1/4) = 1/5.
Doesn’t matter when you go, it’s the same odds. I hope that was your point X-ray! Probability says it doesn’t matter.
Of course, if it was me on the anchor, I would tell the guy betting $40 that I would stand pat on whatever I had… if he paid me $5.
Not his money, not his choice. Not his right to complain.
Yes, that was my point muttrox.
You’re playing at a $5 table in your scenario, right CuriousCanuck?
Uh…for the probability-impaired, can you xplain this?
I assumed that it didn’t finish until someone got the white ball… therefore your probabilities are only correct for the first round.
second round player A, who picked first, has a 1/3 * 3/5, or again 1/5 chance of winning with his second pick, as does player B at 1/2 * 2/5.
And if each of them has picked twice without getting the white ball, then logically A should get his turn once again, and have a 100% chance of winning at that point.
Total probabilities of a player A win 3/5 or 60%, 40% for player B.
No, each player only gets to pick one ball.
Yeah, but the dealer cuts the deck after the 3rd ball anyhow!!
lol… yeah, I see your point, but it’s all dependent exactly what the game is – I think it was supposed to be an illustration of how the probabilities cancel out.
Honestly, it’s a bit hard to do quickly. When you are determining the probability of an event that is composed of two *independent * events, you multiply their individual probabilities. Clear as mud?
Example 1: You flip a coin twice. The probability of getting two heads is 1/4. This is because the probability of each flip is 1/2, and they are independent events, so (1/2) * (1/2) = 1/4.
Example 2: On any given day, the odds of a traffic jam occuring on highway X is 20%. Your odds of taking highway X to work during that timeframe is 40%. To get the overall probability of you getting stuck in a traffic jam, you multiply (20%)(40%)=8% -or- (1/5)(2/5)=2/25 for the fraction people.
Hope this helps.