MJ: First Base: Better Chance of Catching High Cards?

[MJ]

Ok here is a newbie question hence I post it on the beginner's board. Why wouldn't first base have a better chance of getting the high cards during high counts then 3rd base??

Let me give you a simple example where this MIGHT be the case. Lets say I'm playing a 5 spot table with 5 players in a single deck game and there are only 5 cards left to be played(not likely but its just an example for simplicity sake). There are 4 high cards and 1 low card left. Now in this scenerio where would you rather be sitting? 3rd base which will receive the last card seems like it has a lesser chance of getting 1 of the 4 high cards because the 4 players to his right will most likely catch them.

On the other hand, is this a simple probability case where each player has a 4/5 chance of catching a high card regardless of where they are sitting? Can somebody present some type of simple mathematical example to demonstrate why 1st base does not have a better chance of receiving high cards as opposed to 3rd base in high count situations? Some will assert that the chance of catching high cards DIMINSHES as the dealer deals from 1st base to 3rd base in high counts. Thanks to anybody to who can respond.

-MJ

[KidDangerous]

> Ok here is a newbie question hence I post it
> on the beginner's board. Why wouldn't first
> base have a better chance of getting the
> high cards during high counts then 3rd
> base??

> Let me give you a simple example where this
> MIGHT be the case. Lets say I'm playing a 5
> spot table with 5 players in a single deck
> game and there are only 5 cards left to be
> played(not likely but its just an example
> for simplicity sake). There are 4 high cards
> and 1 low card left. Now in this scenerio
> where would you rather be sitting? 3rd base
> which will receive the last card seems like
> it has a lesser chance of getting 1 of the 4
> high cards because the 4 players to his
> right will most likely catch them.

> On the other hand, is this a simple
> probability case where each player has a 4/5
> chance of catching a high card regardless of
> where they are sitting? Can somebody present
> some type of simple mathematical example to
> demonstrate why 1st base does not have a
> better chance of receiving high cards as
> opposed to 3rd base in high count
> situations? Some will assert that the chance
> of catching high cards DIMINSHES as the
> dealer deals from 1st base to 3rd base in
> high counts. Thanks to anybody to who can
> respond.

> -MJ

I can't provide any math answers. I have heard people argue that 3rd base is better than 1st and that first is better than third. I think it really is a matter of personal preference vs. math probability which seat is more out to get the cards. ( I may be wrong on this if so, I welcome a correction) Now there othere reasons outside of likelyhood of getting hi cards that can be argued as to why 1 seat is better than another. As far as getting cards I think one seat has just a good a chance as another, as does the dealer.

I think most of the "worry" can be eliminated by not playing at a table with more than 2 other players and shoot for as much heads up play as possible. This also gives you more hands per hour anyway. More play = more money.

In short there are a few factors involved in ones choosing of a seat, but I don't think probability of getting hi cards is one of them.

I will add tho that it has been argued that sitting at 3rd base does give you the benefit of having the most information available to make your play decision with.

Kid

[Parker]

> On the other hand, is this a simple
> probability case where each player has a 4/5
> chance of catching a high card regardless of
> where they are sitting?

Yes, it is. Our counting system gives us no information regarding the order of the cards. Each spot is equally likely to receive the high (or low) card.

[Brick]

Nobody knows the order of the last 5 cards,they are dealt randomly. I dont think any math is required to explain why odds are the same for all players. An example question will probably help:

If you take 4 face cards and 1 small card from a deck and randomly shuffle them after each round. Do all five players have equal odds of getting a face card?

[Ouchez]

> Ok here is a newbie question hence I post it
> on the beginner's board. Why wouldn't first
> base have a better chance of getting the
> high cards during high counts then 3rd
> base??

> Let me give you a simple example where this
> MIGHT be the case. Lets say I'm playing a 5
> spot table with 5 players in a single deck
> game and there are only 5 cards left to be
> played(not likely but its just an example
> for simplicity sake). There are 4 high cards
> and 1 low card left. Now in this scenerio
> where would you rather be sitting? 3rd base
> which will receive the last card seems like
> it has a lesser chance of getting 1 of the 4
> high cards because the 4 players to his
> right will most likely catch them.

Why would an AP play with so many other players???

> On the other hand, is this a simple
> probability case where each player has a 4/5
> chance of catching a high card regardless of
> where they are sitting? Can somebody present
> some type of simple mathematical example to
> demonstrate why 1st base does not have a
> better chance of receiving high cards as
> opposed to 3rd base in high count
> situations? Some will assert that the chance
> of catching high cards DIMINSHES as the
> dealer deals from 1st base to 3rd base in
> high counts. Thanks to anybody to who can
> respond.

> -MJ

:-)
Ouchez.
Always looking out for the Ploppyi.

[Zenfighter]

Number of players = 5

Sample = {h,h,h,h,l}

pl = prob. of low
ph = prob of high = 1 - pl

 

 

Players		 

 

1)     pl = 1/5 		                 ph = 4/5 

 

2)     pl = 4/5 * 1/4 = 1/5	                 ph = 4/5 

 

3)     pl = 4/5 * 3/4 * 1/3 = 1/5	         ph = 4/5 

 

4)     pl = 4/5 * 3/4 * 2/3 * 1/2 = 1/5          ph = 4/5 

 

5)     pl = 4/5 * 3/4 * 2/3 * 1/2  * 1/1 = 1/5	 ph = 4/5 

 

 

Thus, all players have the same probability to get a high card.

Hope this helps, too.

Zenfighter

[MJ]

Well this example certainly proves it. I know just a little finite math. Why do you multiply by 1/4 for player #2?

-MJ

> Number of players = 5
> Sample = {h,h,h,h,l}
> pl = prob. of low
> ph = prob of high = 1 - pl
>
> Players
> 1) pl = 1/5 ph = 4/5
> 2) pl = 4/5 * 1/4 = 1/5
> ph = 4/5
> 3) pl = 4/5 * 3/4 * 1/3 = 1/5
> ph = 4/5
> 4) pl = 4/5 * 3/4 * 2/3 * 1/2 = 1/5
> ph = 4/5
> 5) pl = 4/5 * 3/4 * 2/3 * 1/2 * 1/1 =
> 1/5 ph = 4/5
>
>
> Thus, all players have the same probability
> to get a high card.
> Hope this helps, too.
> Zenfighter
>

[MJ]

[Don Schlesinger]

> Well this example certainly proves it. I
> know just a little finite math. Why do you
> multiply by 1/4 for player #2?

2) pl = 4/5 * 1/4 = 1/5

This type of calculation is referred to as a conditional probability. First, there is a condition, or stipulation, that has to take place before the actual probability you're trying to figure out; then you do a direct probability calculation. Think of it as a two-step process to get your final answer.

So, in this case, for the second player to have a shot at drawing the low card, it first has to be true that the first player has drawn a high card, because if the first player gets the low card right off the bat, the second player has no chance at getting it. So, the 4/5 represents the probability that the first player does, indeed, draw the high card, thus permitting the "experiment" to continue.

Once the first player draws the high card (Ph = 4/5), there are now 4 cards left for the second player, and only one of them is the low card. Thus, his probability of his now drawing the low card (Pl) is 1/4. Both of these events have to take place, one after the other, for the second player to draw the low card, so we multiply the two probabilities, getting 4/5 * 1/4 = 1/5.

Clear?

Don

[MJ]

Thanks for clearing that up Don. Many thanks to Zen for performing the continuous probablity calculation. The math is not as complex as I thought. I feel much better when playing at 3rd base.

What started this whole thread was I actually played in a 5 spot game at third base. I had to wait for what seemed like an eternity for the count to become favorable. When the count did become favorable the 4 players before me kept getting all the high cards and I was getting low cards. Bad luck I guess!

-MJ

> 2) pl = 4/5 * 1/4 = 1/5

> This type of calculation is referred to as a
> conditional probability. First, there is a
> condition, or stipulation, that has to take
> place before the actual probability you're
> trying to figure out; then you do a direct
> probability calculation. Think of it as a
> two-step process to get your final answer.

> So, in this case, for the second player to
> have a shot at drawing the low card, it
> first has to be true that the first player
> has drawn a high card, because if the first
> player gets the low card right off the bat,
> the second player has no chance at getting
> it. So, the 4/5 represents the probability
> that the first player does, indeed, draw the
> high card, thus permitting the
> "experiment" to continue.

> Once the first player draws the high card
> (Ph = 4/5), there are now 4 cards left for
> the second player, and only one of them is
> the low card. Thus, his probability of his
> now drawing the low card (Pl) is 1/4. Both
> of these events have to take place, one
> after the other, for the second player to
> draw the low card, so we multiply the two
> probabilities, getting 4/5 * 1/4 = 1/5.

> Clear?

> Don

[Don Schlesinger]

> When the count did become favorable the 4
> players before me kept getting all the high
> cards and I was getting low cards. Bad luck
> I guess!

Yes, unfortunately, that's all it really was.

Sometimes we ask this question in a slightly different way, and people can get confused. Suppose you put one $100 bill in a hat and nine $1 bills in with it. There are ten people, and each gets one pick out of the hat, but you don't look at the bill you've selected until everyone has drawn. Would you feel "advantaged" going first? "Disadvantaged" going last? Fact is, it makes absolutely no difference when you pick. All ten people have an equal chance (1/10) of picking the $100 bill, before the game begins. The order in which they draw is totally irrelevant.

Don

[KidDangerous]

Does this same math illustration apply to the scenario of just dropping the 5 cards in a hat and then deciding to draw first or last?
Do you still have a 1 in 5 chance this way?

Kid

[Zenfighter]