MJ: Do other players affect your hand at the table?

[MJ]

Suppose a BS player is sitting at second base with a civilian at first base. It is a single deck game dealt to the very bottom (just for sake of simplicity).

The civilian is dealt a hard 20 and is contemplating splitting 10s. The BS player was dealt an 8,3 and decides to double down hoping to catch a 10. He pleads with the civilian not to split 10s for fear he might hurt his chances of drawing a 10 on his double.

At this very moment, the composition of the remaining deck is 10 cards (8 high, 2 low). Assuming the civilian stands on his 20, the BS player has an 80% chance of catching a 10 for a total of 21.

Now, what are the BS player's chances of drawing a 10 assuming the civilian splits his 10s and can only take one card on each 10 with no re-splitting allowed?

According to my calculations, the probability of the BS player drawing a 10 to go with his 8,3 is 74.4% assuming the civilian splits his 10s. So, the civilian has hurt the other player's chances of drawing a 10. Is this correct? I thought that the chances of catching high cards or low cards are not affected by how others play their hands.

Thanks for any responses.
MJ

[Mr. X]

Assuming your 8/2 ratio is correct and you have no extra info, the odds of the counter getting a 10 AT THE INSTANT BEFORE THE 10 SPLITTER HAS TO MAKE HIS DECISION is 80%.
Anytime anyone asks an "always" or "never" question, I instinctively try to think of exceptions. So, "CAN" the civilian have an effect on the odds? Sure...if he splits and resplits and hits (which I think your hypothesis excludes), theoretically he could eat up 9 cards, and since the last card of the deck is traditionally not dealt, the deck would be reshuffled, and you'd be facing a very negative deck, count so negative as to justify passing on the double. Obviously, this is mostly theoretical... I haven't run into a deck that had to be reshuffled due to lack of cards since maybe the early 90's.
The vast, vast majority of the time, what other players do doesn't matter. But not 100% of the time. A few months ago, I pleaded with a friendly but drunk Mexican not to split his 10's...not because of superstition, but because the dealer had already partially pulled the next card out of the shoe, and I got a peek at it...a 10, and I had a doubled 11 I wanted that 10 to go to!! I talked him out of it, and won an important dd21 lol.
There are other situations where players actions have a non-random effect on the table, but the ones I'm thinking of involve rarely used advantage play techniques that I don't want to get into on public boards.
Bottom line- the answer to "Do other players affect your hand at the table?" 99.999% of the time, the answer is no. But not 100% ;>

[MJ]

> Bottom line- the answer to "Do other players
> affect your hand at the table?" 99.999% of the
> time, the answer is no. But not 100% ;>

So is this really just a rare exception? I am not an outcome oriented person. I don't believe in voodo like taking the dealers bust card and that nonsense. However, the math (if correct) demonstrates that another player that precedes you can indeed hurt your chances of drawing a high card.

I thought that the probability of drawing a high (or low) card in the deck for a third baseman is equivalent to that of the first baseman and all players in-between regardless of how anybody plays their hands. Something must be off here.

Don? Norm? Is the math flawed? 80% > 74.4%.

MJ

[Fred Renzey]

> The math (if correct) demonstrates that
> another player that precedes you can indeed hurt your
> chances of drawing a high card. Something must be off here.

.snip: The generic chances never change. To simplify the explanation, let's say the other player wanted to just double his Ace/9 rather than split 10's (this way, we eliminate the compounding combinations of drawing Hi/Hi, Lo/Lo or Hi/Lo cards to the split 10's).

Of course, if the Ace/9 stands, we know the 11's chances of catching a Hi are 8 out of 10 -- or 80%.

Now when the Ace/9 doubles, 8 times out of 10, he will take a Hi. Those 8 times, the 11 will have only a 7 out of 9 chance (77.77%) to catch a Hi. But the other 2 times out of 10, the Ace/9 will take a Lo. Those 2 times, the 11 will have an 8 out of 9 chance (88.88%) to catch a Hi.

So when the Ace/9 doubles, 80% of the time the 11's chances to catch a Hi will be 77.77% -- and -- 20% of the time, they will be 88.88%.

Stay with me -- we're almost there.
If I double my 11 eight times with a 77.77% chance to catch a Hi, that's 6.22 Hi's caught. If I double another 2 times with an 88.88% chance to catch a high, that's another 1.78 Hi's caught. All in all, it's 8 Hi's out of 10 attempts -- or an 80% rate.

Bluebook II contains a more concise explanation of this with a simpler example. But just keep in mind that somebody else's chances of taking a high card in front of you are always exactly offset by their chances to take a low card in front of you.

[Fred Renzey]

If the 6.22 and 1.78 Hi's don't ring true with you, then think of it this way.

Suppose the Ace/9 stands 45 times. Then you'll catch a Hi 36 times, or 80% of the time.
If the Ace/9 doubles 45 times, 36 of those times you'll catch a Hi 77.77% of the time for 28 Hi's caught.
The other 9 times you'll catch a Hi 8 times for a total of 36 Hi's in 45 attempts -- or 80%.

[MJ]

> .snip: The generic chances never change. To simplify
> the explanation, let's say the other player wanted to
> just double his Ace/9 rather than split 10's (this
> way, we eliminate the compounding combinations of
> drawing Hi/Hi, Lo/Lo or Hi/Lo cards to the split
> 10's).

Alright, if the probability of catching a high card is unchanged by the other player, then apparently I made an error somewhere in my analysis. I'll post how I came up with my probabilities later and maybe you or somebody else can point out where I screwed up.

MJ

[Don Schlesinger]

> Alright, if the probability of catching a high card is
> unchanged by the other player, then apparently I made
> an error somewhere in my analysis. I'll post how I
> came up with my probabilities later and maybe you or
> somebody else can point out where I screwed up.

Suppose you have a bag with ten balls numbered 1-10. Suppose you're trying to draw the one marked 10. Does it matter if you draw first, or if you wait and draw last?

Your example is no different.

Don

[Fred Renzey]

> Apparently I made an error somewhere in my analysis. I'll post how I came up with my probabilities later.
MJ

snip> What if there were just three cards left -- two 10's and a 5. If the Ace/9 stands, the 11 will catch a 10 two times out of three.
Now if the Ace/9 doubles, he will take a 10 two times out of three, leaving one 10 and one 5. Those two times then, the 11 will on average, catch a 10 once and the 5 once.
But that third time, the Ace/9 will take the 5. That third time, the 11 must catch a 10.
So after all three times, the 11 has caught a 10 two times out of three.

As Don points out, suppose the dealer takes all three cards out of the shoe (two 10's and a 5), and fans them out face-down on the table -- then offers that you pick your own double down card. Would you be any worse off choosing the last card (as if the first two were taken by the 10-splitter)?

Or -- what if the dealer reaches into the shoe and reverses the order of those three cards? Which card is most likely to be the 5 now? In fact, which card was most likely to be the 5 before?

[MJ]

> Alright, if the probability of catching a high card is
> unchanged by the other player, then apparently I made
> an error somewhere in my analysis. I'll post how I
> came up with my probabilities later and maybe you or
> somebody else can point out where I screwed up.

In summary: 10 cards total {8 tens, 2 fives}
tens = H, fives = L

Player at first base has 2 tens and is contemplating a split. If he splits, he may only take 1 card on each ten with no re-splitting permitted. Player at second base has 8,3 and doubles down. What are the second baseman's chances of drawing a ten based upon the above summary?

Solution: Assuming the first baseman stands, then clearly the second baseman has an 80% chance of drawing a high card.

If the first baseman splits 10s, then all the ways that the second baseman can draw a ten are accounted for below:

p(HHH) = 8/10 x 7/9 x 6/8 = 46.6%
p(HLH) = 8/10 x 2/9 x 7/8 = 15.6%
p(LHH) = 2/10 x 8/9 x 7/8 = 15.6%
p(LLH) = 2/10 x 1/9 x 8/8 = 2.2%

The sum of the above probabilities equals 80%, which is identical to that of the second baseman standing on his 10s.

Sorry I made a silly arithmetic error when I said 74.6%.

MJ