arithmetically challenged people

[William Sommerwerck]

I find it interesting that almost everyone who "agrees" with me is quite
wrong.

They are interpreting the problem and its explanation in terms of what they
would like the situation to be, rather than looking at it from a strictly
mathematical basis.

 

[William Sommerwerck]

"Don Pearce" <spam@spam.com> wrote in message
news:4dce456c.283347127@news.eternal-september.org...

On Sat, 14 May 2011 17:39:22 +1000, "Trevor" <trevor@home.net> wrote:


"Don Pearce" <spam@spam.com> wrote in message
news:4dcd9c92.240121013@news.eternal-september.org...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not,
ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics. And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

Here's the simplest-possible correct explanation...

2/3 of the time, your initial pick is wrong. The host will then show you the
"goat" prize (the other being the good prize). Ergo, switching will get you
the good prize 2/3 of the time. 1/3 of the time you'll lose the good prize.
This is obviously better than sticking with the initial choice (which is
right only 1/3 of the time).

How much simpler does it need to be, to be comprehensible?

 

[Arny Krueger]

"spamtrap1888" <spamtrap1888@gmail.com> wrote in message
news:dd7c10fd-38c7-43e2-9db6-5c4e25f4a946@35g2000prp.googlegroups.com

Declaring that there is no connection between the two
situations is the source of the poster's error. Monty
Hall knew if the player was correct or not, and so the
player's choice of the door in the first round
influenced the selection of the goat door. The graphic
helps you understand that there are still three scenarios
once a goat door has been revealed.

I get it now.

 

[Don Pearce]

On Sat, 14 May 2011 03:46:26 -0700, "William Sommerwerck"
<grizzledgeezer@comcast.net> wrote:

"Don Pearce" <spam@spam.com> wrote in message
news:4dce456c.283347127@news.eternal-september.org...
On Sat, 14 May 2011 17:39:22 +1000, "Trevor" <trevor@home.net> wrote:


"Don Pearce" <spam@spam.com> wrote in message
news:4dcd9c92.240121013@news.eternal-september.org...
No, you are in fact choosing one door (your first choice) or BOTH the
other doors - the choice if you swap. The revealed goat is one of the
two-door choice, so you have twice the chance of winning the car if
you swap.

What garbage, there are only now 2 doors whether you swap or not,
ignoring
the TV host likely manipulation, which CANNOT be determined as a simple
statistic.
(although could probably be measured from a large number of such TV game
shows. I am unaware of any such actual measurement however)

Trevor.

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics. And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.


Here's the simplest-possible correct explanation...

2/3 of the time, your initial pick is wrong. The host will then show you the
"goat" prize (the other being the good prize). Ergo, switching will get you
the good prize 2/3 of the time. 1/3 of the time you'll lose the good prize.
This is obviously better than sticking with the initial choice (which is
right only 1/3 of the time).

How much simpler does it need to be, to be comprehensible?

Some will never get it, no matter how it is explained.

d

 

[Arny Krueger]

"Trevor" <trevor@home.net> wrote in message
news:4dce3407$0$22473$afc38c87@news.optusnet.com.au

"spamtrap1888" <spamtrap1888@gmail.com> wrote in message
news:5f3fe288-fefb-4981-a73c-78c107ea2a97@z15g2000prn.googlegroups.com...
The player picks a door and has a 1/3 chance of being
right. This chance does not change when a losing door is
revealed, so the only remaining choice gives you a 2/3
chance.

Which totally ignores the fact that the only reason the
first door is opened is because the host already knows it
is incorrect. This is NOT a purely statistical game of
chance, the host can manipulate the odds either way, and
regularly do.

There are two iron rules that dictate which door the host opens.

(1) There can't be a car behind it

(2) It can't be the door the contestant picked.

If the contestant picks a door with no car, then there is only one other
door the host can choose. The host has no choice and can't affect the
outcome.

If the contestant chooses a door with a car, then the host can choose either
of two two doors that have no car, but which one he chooses doesn't seem to
affect the outcome. His effect on the odds comes from the fact that he
revealed one of the two doors with no car behind it.

How can the host manipulate the odds?

 

[William Sommerwerck]

"Arny Krueger" <arnyk@hotpop.com> wrote in message
newssadnZt_UpF1_FPQnZ2dnUVZ_v-dnZ2d@giganews.com...

"Trevor" <trevor@home.net> wrote in message
news:4dce3407$0$22473$afc38c87@news.optusnet.com.au
"spamtrap1888" <spamtrap1888@gmail.com> wrote in message

news:5f3fe288-fefb-4981-a73c-78c107ea2a97@z15g2000prn.googlegroups.com...
The player picks a door and has a 1/3 chance of being
right. This chance does not change when a losing door is
revealed, so the only remaining choice gives you a 2/3
chance.

Which totally ignores the fact that the only reason the
first door is opened is because the host already knows it
is incorrect. This is NOT a purely statistical game of
chance, the host can manipulate the odds either way, and
regularly do.

There are two iron rules that dictate which door the host opens.

(1) There can't be a car behind it

(2) It can't be the door the contestant picked.

If the contestant picks a door with no car, then there is only one other
door the host can choose. The host has no choice and can't affect the
outcome.

If the contestant chooses a door with a car, then the host can choose
either
of two two doors that have no car, but which one he chooses doesn't seem
to
affect the outcome. His effect on the odds comes from the fact that he
revealed one of the two doors with no car behind it.

How can the host manipulate the odds?

Correct. The host has no effect on the odds.

Part of the confusion occurs because people confuse permutations and
combinations. In the situation where the contestant has chosen the good
prize, the two bad prizes form a combination, not a permutation.

 

[David]

"Trevor" wrote in message
news:4dce2d98$0$2443$afc38c87@news.optusnet.com.au...


"David" <someone@somewhere.com> wrote in message
news:iqk5jt$jn1$1@dont-email.me...

"William Sommerwerck" wrote in message
news:iqk4b0$cop$1@dont-email.me...

But by not switching doors, you are ignoring the new
information that the
prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right.
By
switching
doors, you are including the new information that the prize
has to be
behind
one of the other two doors, and your new chance of winning is
50%.

No, it doesn't. Your new chance of winning is 2/3.

***
This is similar to another puzzle. A couple has two children.
What is the probability that the second is a boy? The couple
then volunteers that they are not both girls. Now what is the
probability the second is a boy?

The first case is 1/2. The second case is 2/3.

Wrong, on a purely statistical basis the first case is 50:50, BB,
BG, GB, or
GG. Two out of four meet the criteria.
The second case is 50:50 Boy or Girl, One out of two meets the
criteria.

<snip>
Trevor.

The second case is: BB, BG, GB. The couple told you the GG case
does not exist.
Get it now? The goat problem has similar probability outcome
changes.

David

 

[Bill Graham]

William Sommerwerck wrote:

But by not switching doors, you are ignoring the new information
that the prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching doors, you are including the new information that the
prize has to be behind one of the other two doors, and your new
chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

No. You now know that the prize is not behind door $3, so your chance of
winning in the, "second game" is 50-50. But you had to buy yourself this
chance at the second game. You did this by switching doors.

 

[William Sommerwerck]

"Bill Graham" <weg9@comcast.net> wrote in message
news:C8GdnTxkFsgeZ1PQnZ2dnUVZ5jidnZ2d@giganews.com...

William Sommerwerck wrote:
But by not switching doors, you are ignoring the new information
that the prize has to be behind one of the other two doors....

No, it doesn't. That's not correct.

You are sticking with
your original guess that had only a 1/3 chance of being right. By
switching doors, you are including the new information that the
prize has to be behind one of the other two doors, and your new
chance of winning is 50%.

No, it doesn't. Your new chance of winning is 2/3.

No. You now know that the prize is not behind door $3, so your chance of
winning in the, "second game" is 50-50. But you had to buy yourself this
chance at the second game. You did this by switching doors.

I know it's unkind to tell people who agree with you that they're wrong,
but... you're wrong. You really need to think this through carefully.

 

[Carey Carlan]

spam@spam.com (Don Pearce) wrote in
news:4dcd5ee1.224328189@news.eternal-september.org:

On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulfjoe@hotmail.com
wrote:

spam@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
arnyk@hotpop.com> wrote:

"Bill Graham" <weg9@comcast.net> wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganews.com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a
chance to guess again, there is a 50% chance the car is behind door
#1 and a 50% chance the car if behind door #2. Change your choice
or not, you have a 50% chance of being right.


Lets make it ten doors. You pick one, and get a one in ten chance of
being right. That means that the chances are 90% that the car is
behind one of the 9 doors you did not pick. You know for certain
that at least eight of those doors conceal a goat, so when eight
goats are revealed, you have no new information. The chances are 90%
that the car is behind one of the nine - only now there is only one
remaining to open.

One vital fact here is that the person doing the revealing knows the
contents of the doors and chooses to reveal only goats. Had he been
guessing too, and just happened to reveal only goats, then yes, you
would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the situation.
How do the preceding 8 steps affect the final step?

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still 1/2.

 

[Don Pearce]

On Sun, 15 May 2011 12:14:44 GMT, Carey Carlan <gulfjoe@hotmail.com>
wrote:

spam@spam.com (Don Pearce) wrote in
news:4dcd5ee1.224328189@news.eternal-september.org:

On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulfjoe@hotmail.com
wrote:

spam@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
arnyk@hotpop.com> wrote:

"Bill Graham" <weg9@comcast.net> wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganews.com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a
chance to guess again, there is a 50% chance the car is behind door
#1 and a 50% chance the car if behind door #2. Change your choice
or not, you have a 50% chance of being right.


Lets make it ten doors. You pick one, and get a one in ten chance of
being right. That means that the chances are 90% that the car is
behind one of the 9 doors you did not pick. You know for certain
that at least eight of those doors conceal a goat, so when eight
goats are revealed, you have no new information. The chances are 90%
that the car is behind one of the nine - only now there is only one
remaining to open.

One vital fact here is that the person doing the revealing knows the
contents of the doors and chooses to reveal only goats. Had he been
guessing too, and just happened to reveal only goats, then yes, you
would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the situation.
How do the preceding 8 steps affect the final step?

That isn't the final situation. I will take this a step at a time.

There are three doors - one with a car, two with goats

I choose one. I have a 1 in 3 chance of being right

That means there is a 2 in 3 chance of the car being in the other two

I know for a fact that at least one of the other two is a goat.

That does not change the odds - it is still 2 in 3 that the car is in
one of those

The host shows me one of the two - one he knows to contain a goat.

This is not new information, I knew there was a goat there, I still
know there was a goat there.

The odds are still 2 in 3 that the car is in one of those two doors.

But now those 2 in 3 odds have been concentrated into the one
remaining door of the two, which I will open because that is better
than the 1 in 3 chance of it being my first choice.

d

 

[spamtrap1888]

On May 15, 5:14 am, Carey Carlan <gulf...@hotmail.com> wrote:

s...@spam.com (Don Pearce) wrote innews:4dcd5ee1.224328189@news.eternal-september.org:



On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulf...@hotmail.com
wrote:

s...@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
ar...@hotpop.com> wrote:

"Bill Graham" <w...@comcast.net> wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganews.com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous problem.
I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3  and are given a
chance to guess again, there is a 50% chance the car is behind door
#1 and a 50% chance the car if behind door #2.  Change your choice
or not, you have a 50% chance of being right.

Lets make it ten doors. You pick one, and get a one in ten chance of
being right. That means that the chances are 90% that the car is
behind one of the 9 doors you did not pick. You know for certain
that at least eight of those doors conceal a goat, so when eight
goats are revealed, you have no new information. The chances are 90%
that the car is behind one of the nine - only now there is only one
remaining to open.

One vital fact here is that the person doing the revealing knows the
contents of the doors and chooses to reveal only goats. Had he been
guessing too, and just happened to reveal only goats, then yes, you
would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the situation.
How do the preceding 8 steps affect the final step?

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still 1/2.

The big difference: In the Monty Hall problem there is only one "coin
flip". Only one random choice is made -- the first choice of a door.
In the coin flip situation, there are five coin flips, five random
choices.

Now, in contrast, if the car and remaining goats were randomly
shuffled after each goat door was revealed, then the situation would
be different. But in the MHP problem the car does not move.

 

[Bill Graham]

spamtrap1888 wrote:

On May 15, 5:14 am, Carey Carlan <gulf...@hotmail.com> wrote:
s...@spam.com (Don Pearce) wrote
innews:4dcd5ee1.224328189@news.eternal-september.org:



On Fri, 13 May 2011 16:05:33 GMT, Carey Carlan <gulf...@hotmail.com
wrote:

s...@spam.com (Don Pearce) wrote in
news:4dcd3b49.215216046@news.eternal-september.org:

On Fri, 13 May 2011 08:09:11 -0400, "Arny Krueger"
ar...@hotpop.com> wrote:

"Bill Graham" <w...@comcast.net> wrote in message
news:t_ydnZKHN4u_QlHQnZ2dnUVZ5rWdnZ2d@giganews.com
Soundhaspriority wrote:
"Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind
the others, goats. You pick a door, say No. 1, and the
host, who knows what's behind the doors, opens another
door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your
advantage to switch your choice?" The above is a famous
problem. I've left out the
attribution to give you a few minutes (or forever, if
you want) to enjoy it. Bob Morein
(310) 237-6511

When you pick door #1 you only have a 1/3 chance of
winning. But after you see that there is a goat behind
door #3, your chance of winning is 1/2, so I would change
doors and pick door #2. But I don't really know
why....It's just gambler's instinct with me.

After you know there is a goat behind door #3 and are given a
chance to guess again, there is a 50% chance the car is behind
door #1 and a 50% chance the car if behind door #2. Change your
choice or not, you have a 50% chance of being right.

Lets make it ten doors. You pick one, and get a one in ten chance
of being right. That means that the chances are 90% that the car
is behind one of the 9 doors you did not pick. You know for
certain that at least eight of those doors conceal a goat, so
when eight goats are revealed, you have no new information. The
chances are 90% that the car is behind one of the nine - only now
there is only one remaining to open.

One vital fact here is that the person doing the revealing knows
the contents of the doors and chooses to reveal only goats. Had
he been guessing too, and just happened to reveal only goats,
then yes, you would be down to 50/50.

Alternate:

You walk in with 8 doors already open revealing 8 goats.
The car is behind one of the two remaining doors.
Convince me that your odds are not 50% to find the car.

Why? That isn't what happens. Read again and try to follow,
particularly the last part, which is the vital proviso.

Why? Because at the point of the final decision, that's the
situation.
How do the preceding 8 steps affect the final step?

Just as in flipping coins.
Getting 5 heads in a row is 1/32.
But getting the 5th head after already getting 4 is still 1/2.

The big difference: In the Monty Hall problem there is only one "coin
flip". Only one random choice is made -- the first choice of a door.
In the coin flip situation, there are five coin flips, five random
choices.

Now, in contrast, if the car and remaining goats were randomly
shuffled after each goat door was revealed, then the situation would
be different. But in the MHP problem the car does not move.

The really interesting thing is that, even if the car does not move,
conditional probability theory says the odds have changed, and you should
switch doors.

 

[Trevor]

"Don Pearce" <spam@spam.com> wrote in message
news:4dce456c.283347127@news.eternal-september.org...

This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics.

Right!

And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

And understanding that games of pure chance have NO memory for any previous
actions. It's just like the old question, what are the odds of tossing a
coin 10 heads in a row? If you toss 9 heads in a row, what are the odds of
tossing a 10th?
(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

Trevor.

 

[Trevor]

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in message
news:iqlm7b$i37$1@dont-email.me...

I find it interesting that almost everyone who "agrees" with me is quite
wrong.

They are interpreting the problem and its explanation in terms of what
they
would like the situation to be, rather than looking at it from a strictly
mathematical basis.

Or interpreting it from a view of TV game show reality rather than a purely
statistical basis.

Trevor.

 

[Trevor]

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in message
news:iqlmhv$itv$1@dont-email.me...

How much simpler does it need to be, to be comprehensible?

Nobody said it wasn't comprehensible. But it simply ignores the fact that
game shows are NOT pure chance.


Trevor.

 

[Don Pearce]

On Mon, 16 May 2011 15:08:08 +1000, "Trevor" <trevor@home.net> wrote:

"Don Pearce" <spam@spam.com> wrote in message
news:4dce456c.283347127@news.eternal-september.org...
This is like pulling teeth. I'm not going to explain it any more.
Either you understand or you don't. It helps to have studied maths and
statistics.

Right!

And no, there isn't any manipulation. It is purely a
matter of understanding what is and isn't new information.

And understanding that games of pure chance have NO memory for any previous
actions. It's just like the old question, what are the odds of tossing a
coin 10 heads in a row? If you toss 9 heads in a row, what are the odds of
tossing a 10th?
(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

Trevor.

What has this to do with the question?

d

 

[Trevor]

"David" <someone@somewhere.com> wrote in message
news:iqm8lg$3bf$1@dont-email.me...

The second case is: BB, BG, GB. The couple told you the GG case does not
exist.

Oops you are quite correct, there are still 3 possibilities.

Trevor.

 

[Trevor]

"Don Pearce" <spam@spam.com> wrote in message
news:4dd0b4d6.442940877@news.eternal-september.org...

(First you MUST assume the coin is untampered with, you cannot assume the
same for a TV game show however!)

What has this to do with the question?

Nothing I guess for the specific case in question. As was pointed out there
is only one possible set of events if a switch is offered for a one of 3
game of chance. I admit to confusing this with other TV games where the host
can and does influence the outcome.

Trevor.

 

[Carey Carlan]

spam@spam.com (Don Pearce) wrote in news:4dd0c3d7.381246241
@news.eternal-september.org:

That isn't the final situation. I will take this a step at a time.

There are three doors - one with a car, two with goats

I choose one. I have a 1 in 3 chance of being right

That means there is a 2 in 3 chance of the car being in the other two

I know for a fact that at least one of the other two is a goat.

That does not change the odds - it is still 2 in 3 that the car is in
one of those

The host shows me one of the two - one he knows to contain a goat.

This is not new information, I knew there was a goat there, I still
know there was a goat there.

The odds are still 2 in 3 that the car is in one of those two doors.

Stop there.
No, I didn't know there was a goat THERE. I knew there was a goat
behind at least one of the door besides the one I chose, but I didn't
know which one. Now a variable is removed from the equation.

Revealing a goat behind a door doesn't change the odds?
Of course it does.
Otherwise, revealing the car behind a door also wouldn't change the
odds.

Once the host has revealed a goat, then there's an even chance that the
car is behind one of the two remaining doors--and I have no information
either way (unless you're counting the psychological factors) that the
door I chose is or is not the correct one.

Not trying to be argumentative, but I still don't see the logic.

But now those 2 in 3 odds have been concentrated into the one
remaining door of the two, which I will open because that is better
than the 1 in 3 chance of it being my first choice.

d