Breeding Question

[Tad Carlucci]

Here's this week's question.

You just got a box. It shows Pandi Ebony & Ivory with Purple Rain.

Mother is a Pandi Ebony & Ivory with Odyssey Kaleidoscope eyes. Her parents were Pandi Evony & Ivory with Water and Ocicat Lavender with Blush Quartz.

Father is a Pandi Ebony & Ivory with Purple Rain. His parents were Chateau Pink & White No 1 with Forest and your box's mother (Ebony & ivory with Odyssey Kaleidoscope).

Arrange the furs and eyes in order or most likely to least likely to be hidden in your box. Using Saga's charts, show the most dominant fur and eye which is recessive to all those listed and place it in your list.

Chateau Pink & White No 1
Ocicat Lavender
Pandi Ebony and Ivory

Blush Quartz
Forest
Genesis Water
Odyssey Kaleidoscope
Purple Rain

Extra credit for working out the numerical odds for each fur and eye in your answer.

[Malayaa Resident]

(06-09-2013 09:02 AM)Tad Carlucci Wrote:  Here's this week's question.

You just got a box. It shows Pandi Ebony & Ivory with Purple Rain.

Mother is a Pandi Ebony & Ivory with Odyssey Kaleidoscope eyes. Her parents were Pandi Evony & Ivory with Water and Ocicat Lavender with Blush Quartz.

Father is a Pandi Ebony & Ivory with Purple Rain. His parents were Chateau Pink & White No 1 with Forest and your box's mother (Ebony & ivory with Odyssey Kaleidoscope).

Arrange the furs and eyes in order or most likely to least likely to be hidden in your box. Using Saga's charts, show the most dominant fur and eye which is recessive to all those listed and place it in your list.

Chateau Pink & White No 1
Ocicat Lavender
Pandi Ebony and Ivory

Blush Quartz
Forest
Genesis Water
Odyssey Kaleidoscope
Purple Rain

Extra credit for working out the numerical odds for each fur and eye in your answer.

Hidden Fur:
Pink and White, Oci Lav, or Pandie again (equal chance, 33%)
Dominance Order most to least: Pandie, then some overlap of Pink and White which may mean still to be proven?, then Oci Lav

Hidden Eye:
Kaleido or Quartz, 50-50
Dominance Order most to least: Gen Water, Forest, Purple Rain, Kaleido, Blush Quartz

(not used to working out the odds, this should be interesting LOL)

[Liriel Garnet]

OK, here comes some fun LOL.

parents = shown/hidden
mom = E&I/OciLav+ OdyK/BlushQ+
dad = E&I/P&W?+ Purple Rain/OdyK or BlushQ+

TWO cases for the furs (since there is overlap between E&I and P&W in what has been proven):

First, IF P&W is recessive to E&I, then:

box furs: 1/3 E&I/E&I 1/3 E&I/P&W+ 1/3 E&I/OciLav+
so technically chances are 1/3 to hide E&I, 1/6 to hide P&W, 1/6 to hide whatever the P&W hid (which may also be P&W), 1/6 to hide Oci Lav, 1/6 to hide whatever the Lav hid (which may also be Lav)

furs likelihood to be hidden:
E&I (1/3 chance)
P&W/Oci Lav (equal at 1/6 each)
two unknowns (equal at 1/6 each)
Chat Cream & White 1 is most dominant that's recessive to all listed.

Second, if P&W is NOT recessive to E&I, then the only change is that dad's hidden MUST be what hid behind the P&W (which by definition means it cannot be more P&W since if it was dominant it would show), which changes the odds to:
1/3 to hide E&I, 1/3 to hide whatever the P&W hid, 1/6 to hide Oci Lav, 1/6 to hide whatever the Lav hid

furs likelihood to be hidden:
E&I/unknown hidden from behind the P&W (equal at 1/3 each)
Oci Lav/unknown hidden from behind the Oci Lav (equal at 1/6 each)
Chat Cream & White 1 is still the most dominant that's recessive to all listed.


Eyes:
Ody K/BlushQuartz+ (one for sure hides, equal chances for each)
which means technically, the chances are:
1/2 OdyK
1/4 Blush Quartz
1/4 Whatever is hidden behind the Blush Quartz (which may be more Blush Quartz)
the others on the list are not even in the running for the hidden.
Ody Bellini is the only eye known recessive to Blush Quartz, therefore it is the most dominant that's recessive to all listed.

[Tad Carlucci]

To correct Liriel's odds statements ...

Mom is:
Pandie Ebony and Ivory hiding Ocicat Lavender or better.
Odyssey Kaleidoscope hiding Blush Quartz or better.

Dad is:
Pandie Ebony and Ivory hiding Chateau Pink and White No. 1 or better
Purple Rain hiding Odyssey Kaleidoscope or better.

The box is:

Pandie Ebony and Ivory hiding either Pandie Ebony and Ivory, or Chateau Pink and White No. 1 or better

Purple Rain hiding either Odyssey Kaleidoscope, or Blush Quartz or better

There are three possibilities for the fur:

There is a 1-in-2 chance that the box is Pandie Ebony and Ivory hiding Pandie Ebony and Ivory.

There is a 1-in-4 chance that the box is Pandie Ebony and Ivory hiding Ocicat Lavender or better.

There is a 1-in-4 chance that the box is Pandie Ebony and Ivory hiding Chateau Pink and White No. 1 or better.

[Note, at this point, that there is a common fallacy that the odds for each of the three cases, above, are 1-in-3. This is impossible. This math error arises from misusing the Purnett Square when calculating the odds.]

According to Saga's Charts there is still a possibility that Chateau Pink and White No.1 is dominant to Pandi Ebony and Ivory. If so, Chateau Pink and White No. 1 would have odds of 0 to be hidden, and the box has a 1-in-2 chance (instead of a 1-in-4 chance) to be Pandi Ebony and Ivory hiding Ocicat Lavender or better. [Edit: in actual point of fact, the box being discussed has produced an offspring, today, when mated with a Siamese Flame hiding Balinese Cream Lynx or better, which was Chateau Pink and White No. 1 proving that Pandi Ebony and Ivory is, in fact, dominant to Chateau Pink and White No. 1 and the odds of it being so were 1-in-4.]

For the eyes:

There is a 1-in-2 chance the box is Purple Rain hiding Odyssey Kaleidoscope.

There is a 1-in-2 chance the box is Purple Rain hiding Blush Quartz or better.

Genesis Water and Forest were in the list and should appear here at 0 odds.
These two eyes are NOT possible to be hidden since they were eliminated
by prior breeding.

The most dominant fur known to be recessive to the known furs is Chateau Cream and White No. 1

The most dominant eye know to be recessive to the known eyes is Odyssey Bellini.

---

How NOT to count when using Purnett Squares ...

[Charles Courtois]

Actually, Liriel's analysis of the situation where Pandie Ebony & Ivory is dominant to Pink & White No. 1 is correct, whether derived from the Punnet square, or the less visual version of calculating the conditional probability.

A conditional probability is the probability of an event occurring when a given condition exists, when the event and condition are not independent. In this case, it is the probability of a certain fur hiding given the condition that Pandie Ebony & Ivory is showing. In this case the event & condition are clearly not independent, since removing the condition that the Pandie is showing would change the calculation ( the possibility of Mom passing Oci Lav or better & Dad passing Pink & White or better would have to be included).

The solution would look like this:
There is a 3/4 chance of Pandie E+I showing in a breeding of the 2 parents involved (every case except P&W +/ Oci Lav+ would result in the Pandie showing).

There is a 1/4 chance of each of the following events:

Pandie E & I hiding Pandie E + I,
Pandie E & I hiding Oci Lav + and
Pandie E & I hiding P & W +.

I'll show the calculation for the last case (the other 2 calculations would be the same):
P(P&W+ hiding, given the condition that Pandie E & I is showing)= P(Pandie E & I hiding P&W +)/ P(Pandie E & I is showing)= (1/4)/(3/4)= 1/3.

For another example of an application of conditional probability to genetics, see example 2 on this page: http://nitro.biosci.arizona.edu/courses/...ure16.html

[Liriel Garnet]

(06-10-2013 12:07 AM)Tad Carlucci Wrote:  [Edit: in actual point of fact, the box being discussed has produced an offspring, today, when mated with a Siamese Flame hiding Balinese Cream Lynx or better, which was Chateau Pink and White No. 1 proving that Pandi Ebony and Ivory is, in fact, dominant to Chateau Pink and White No. 1 and the odds of it being so were 1-in-4.]

Could you possibly send me a notecard in-world with the pedigree information for the box under discussion and the offspring where you pulled the P&W from behind the E&I so that I can update the charts? Thanks!

[Tad Carlucci]

You're correct. I did state the question in the conditional case. My bad, I grabbed the wrong formula.

I guess it shows you where my head is: I've been working on a different implementation of the genetics engine where the parent's genetics are not set until the last possible moment and grabbed the formula from that rather than for the specific question asked. /me sighs. Here I am always saying you need to be sure you answer the question as asked and failing to do so myself.

[Liriel Garnet]

(06-10-2013 07:29 AM)Charles Courtois Wrote:  Actually, Liriel's analysis of the situation where Pandie Ebony & Ivory is dominant to Pink & White No. 1 is correct, whether derived from the Punnet square, or the less visual version of calculating the conditional probability.

A conditional probability is the probability of an event occurring when a given condition exists, when the event and condition are not independent. In this case, it is the probability of a certain fur hiding given the condition that Pandie Ebony & Ivory is showing. In this case the event & condition are clearly not independent, since removing the condition that the Pandie is showing would change the calculation ( the possibility of Mom passing Oci Lav or better & Dad passing Pink & White or better would have to be included).

The solution would look like this:
There is a 3/4 chance of Pandie E+I showing in a breeding of the 2 parents involved (every case except P&W +/ Oci Lav+ would result in the Pandie showing).

There is a 1/4 chance of each of the following events:

Pandie E & I hiding Pandie E + I,
Pandie E & I hiding Oci Lav + and
Pandie E & I hiding P & W +.

I'll show the calculation for the last case (the other 2 calculations would be the same):
P(P&W+ hiding, given the condition that Pandie E & I is showing)= P(Pandie E & I hiding P&W +)/ P(Pandie E & I is showing)= (1/4)/(3/4)= 1/3.

For another example of an application of conditional probability to genetics, see example 2 on this page: http://nitro.biosci.arizona.edu/courses/...ure16.html

for those who like the representation to be visual, i find this handy:

mom = E&I/OciLav+ OdyK/BlushQ+
dad = E&I/P&W?+ Purple Rain/OdyK or BlushQ+

looking at only the furs, possibly box combos

E&I/E&I
E&I/P&W+
OciLav+/E&I
OciLav+/PW+

of those four possiblities, the first three are the only ones to show the E&I, so we've eliminated the fourth possibility. Of those remaining three, there's an equal chance of each, so 1/3 hides E&I, 1/3 hides P&W+, 1/3 hides OciLav+

[Tad Carlucci]

(06-10-2013 09:10 AM)Liriel Garnet Wrote:  Could you possibly send me a notecard in-world with the pedigree information for the box under discussion and the offspring where you pulled the P&W from behind the E&I so that I can update the charts? Thanks!

I rarely actually go in-world these days.