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

[KidDangerous]

I will preface by saying that I am not trying to beat a dead horse or prolong a question that has already been answered. Just trying to clear up my hang up on the 1 in 5 chance when there are only 5 choices and 5 participants. I just can't see how you have a 1 in 5 chance if you dont have access to "all 5" when your chance for drawing comes up. I had posted the following earlier, but then took it down, because I wanted to make sure that math include the scenario I was talking about. If anyone feels that this is indeed beating a dead horse, I will just drop it and research independently. No problems there. I just want to be on the same page as everyone but at the same time know why. Thank you for any input and patience. With that in mind here is my post from earlier.
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Quote


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.


That being said, it seems to me that the only thing that the math from above showed, is that the 5th person is 100% guarantee to get the odd card.

Because for 2 to be apart of the illustration it has to be assumed that 1 didn't get it. For 3 to be a part of it, 2 couldn't have got it.
if 4 is gonna be in it, then 3 can't have got. If 5 is to stay alive, then 1,2,3, or 4 can't have the card.

If 1,2,3,4 didn't get it, (which they can't for the illustration to even exist, as shown in quote above.) then it is 100% guaranteed that 5 has the odd card. So if that is so, how is it 5 is still considered to have a 1 in 5 chance?

In addition, with the above information about assuming the first person doesn't get it for the illustration to be able to carry on. Then you would have to assume that 1 will never get it. If 1 can never get the odd card in order for the illustration to exist past 1, then how does 1 have a 1 in 5 chance of getting it?

What I'm asking has left the blackjack table position problem. I fathom that because the possiblities outnumber the participants.

But when the participants equal the number of possiblities then I don't see everyone having the same 1 in 5 chance. Now I saw where in the mix it was included "before the drawing begins".

I can agree with that in a way because you have a 1 in 5 chance of being the first person to draw, who is the only person truly drawing 1 out of 5. Anyone after the first person is not drawing 1 from 5. They are drawing 1 from whatevers left.

The only way I see everyone having a 1 in 5 chance, a "true" 1 in 5 chance, is that when you draw you "couldn't" look at your card, and anyone after you has the option to draw from whats left and from whats already been picked. Even then I see a couple problems maybe.

I'm no math person as stated in the past. It hurts me sometimes to read posts. I have to take rests. But I'm purely stating what I see from a layman point of view. I'm asking if you can see what I'm saying, and help me see where I'm going wrong, or have I misread again???

Yes? No? Maybe? HelP?

Kid

[Fred Renzey]

> 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??

Thanks to anybody to who can
> respond.

> -MJ

snip> MJ, here's a down-to-earth layman's proof of why it makes no difference where you sit regarding your question.
Suppose there are, as you say just five cards left, with four cards being 10's and the fifth card a 5. To simplify, let's say there are only two players, first base and third. First base will receive a 10 four times out of five on average. Those four particular times, third base will receive a 10 three times and the 5 once, again, on average.
First base however, will also receive the 5 one time out of five. That one time, third base must receive a 10.
So that's four 10's in five averaged tries for first base and four 10's in five averaged tries for third base as well. It just can't ever be anything else.
Notice that this was merely a verbal illustration of the conditional probability expression given by Zenfighter.

[Fred Renzey]

> I will preface by saying that I am not
> trying to beat a dead horse or prolong a
> question that has already been answered.

> Yes? No? Maybe? HelP?
> Kid

>snip: Kid -- the thing to remember is that sometimes third base's chances to receive a high card will go down by the time the deal comes around to him, and sometimes they will go up. If there are mostly high cards left, third base's chances for a high card will USUALLY go down a LITTLE -- and occasionally will go UP a LOT.
If for example, there are eight highs and two lows, first base is an 80% shot to get a high card. If he does get the high card, the next player's chances drop 2% to 78%. But if first base gets a low card, then the next player's chances for a high card shoot up 9% to 89%. The averaged sum of all subsequent players' chances for a high card is 80% -- the same as first base.

[Don Schlesinger]

> I will preface by saying that I am not
> trying to beat a dead horse or prolong a
> question that has already been answered.

Er, really?! :-)

> Just trying to clear up my hang up on the 1
> in 5 chance when there are only 5 choices
> and 5 participants. I just can't see how you
> have a 1 in 5 chance if you don't have access
> to "all 5" when your chance for
> drawing comes up.

Really pretty simple, but we'll go over it again.

> I had posted the following
> earlier, but then took it down, because I
> wanted to make sure that math include the
> scenario I was talking about. If anyone
> feels that this is indeed beating a dead
> horse, I will just drop it and research
> independently.

We'll try to help.

> 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.

Sounds very wise to me! :-)

> That being said, it seems to me that the
> only thing that the math from above showed,
> is that the 5th person is 100% guarantee to
> get the odd card.

No, of course not. He's "guaranteed" to get the odd card IF AND ONLY IF all the people before him didn't get it. That surely isn't a certainty! So, for the first four people to NOT get the odd part requires a 4/5 prob. for the first person, a 3/4 prob. for the second, a 2/3 prob. for the third, and a 1/2 prob. for the second. Multiply the four fractions (be sure to cancel all over the place!), and eureka! you get your 1/5 prob. for the fifth guy.

> Because for 2 to be a part of the
> illustration it has to be assumed that 1
> didn't get it.

Right. But you have to factor in that prob. as part of your answer.

> For 3 to be a part of it, 2
> couldn't have got it.
> if 4 is gonna be in it, then 3 can't have
> got. If 5 is to stay alive, then 1,2,3, or 4
> can't have the card.

Exactly. See string of probs. above.

> If 1,2,3,4 didn't get it, (which they can't
> for the illustration to even exist, as shown
> in quote above.) then it is 100% guaranteed
> that 5 has the odd card. So if that is so,
> how is it 5 is still considered to have a 1
> in 5 chance?

See above.

> In addition, with the above information
> about assuming the first person doesn't get
> it for the illustration to be able to carry
> on. Then you would have to assume that 1
> will never get it. If 1 can never get the
> odd card in order for the illustration to
> exist past 1, then how does 1 have a 1 in 5
> chance of getting it?

He doesn't NEVER get the odd card. He gets it, right off the bat, one time in five. If he does, good for him; if he doesn't, the experiment continues.

> But when the participants equal the number
> of possiblities then I don't see everyone
> having the same 1 in 5 chance. Now I saw
> where in the mix it was included
> "before the drawing begins".

Obviously, you're talking about pre-deal probabilities. Once the cards are dealt, the probs. keep changing.

> I can agree with that in a way because you
> have a 1 in 5 chance of being the first
> person to draw, who is the only person truly
> drawing 1 out of 5.

You're looking at it the wrong way.

> Anyone after the first
> person is not drawing 1 from 5. They are
> drawing 1 from whatevers left.

That's the second part of the conditional prob. The first part involves multiplying by the prob. that the later person ever does get to draw in the first place.

> The only way I see everyone having a 1 in 5
> chance, a "true" 1 in 5 chance, is
> that when you draw you "couldn't"
> look at your card, and anyone after you has
> the option to draw from whats left and from
> whats already been picked.

Makes no difference.

> I'm no math person as stated in the past. It
> hurts me sometimes to read posts. I have to
> take rests. But I'm purely stating what I
> see from a layman point of view. I'm asking
> if you can see what I'm saying, and help me
> see where I'm going wrong, or have I misread
> again???

Hope I've helped.

Don

[KidDangerous]

[KidDangerous]

Thank you FR for the input. I understand and accept the seating position at a table. It is just the picking a straw theory that I can't.
I obviously misinterpreted something somewhere.
I think it falls in the "before the pick or draw" part that I got confused on. On another site they were trying to say that when you pick straws everyone picking has the same chance. Maybe I even read that wrong. I will double and triple check myself again. But I will just leave it be as far as a thread goes. Thanks again for offering your input.

Kid

[Battery]

> Quote
> 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.

> That being said, it seems to me that the
> only thing that the math from above showed,
> is that the 5th person is 100% guarantee to
> get the odd card.

Perhaps a very short explanation might work best:
The odd card has a one in five chance of going to each of the five positions.