Monthly Archives: May 2013

Calculating the odds

Breeding rabbits can be very expensive, in time, facilities and of course dollars.  While we hope that each litter can pay for itself, that is not always the case, so it is important to carefully choose each breeding to have the most likelihood of producing the bunnies that we are hoping to get.  Most breeders just pick two rabbits that have characteristics that they want and hope for the best.  However, if you keep good records and are willing to pull out a calculator, you might be able to get better odds.  Since color genetics are fairly well understood, let me give you an example using them.  If you have read up on color inheritance (and if not, you should check out which has an excellent introduction),  you know that blue is a recessive trait, where the bunny needs two copies of the recessive gene “d” (dd), in order to become blue.  One or more copies of the full strength gene, “D” (DD or Dd), will give you a black bunny.  So, let’s say that I want blue babies.  I have a doe and a buck that I know from looking at their pedigree each recessively carry “d.” I want to know whether breeding these two would be worthwhile.  Since each of these rabbits have two genes (a D and a d), I know I have a 50% (or .5) chance of the baby getting the “d” from each parent.

If you go back to your math lessons, you will remember P(A and B)=P(A)*P(B) or the Probability of two independent events occurring together = Probability of one event multiplied by the probability of the second event.  Therefore, to determine the chance that I have of a blue baby I multiply the Probability that the baby will get a dilute gene from the father, .5,  times the probability that it will get a dilute gene from the mother, .5, which gives me a .25 or 25% chance of a blue baby.  The following chart is an easy way to visualize this:

Inheritance Diagram 1

If my litter size is a typical 8, I should end up with two blue babies out of the match (.25*8=2).  I can use this information to decide whether or not two blue babies would be worth the trouble of raising a litter.

Now let’s change the situation slightly.  Say instead of a blue baby, what I really wanted was a lilac bunny.  From our color genetics we know that lilac is created when the bunny has the full dilute recessive or dd as well as the brown recessive, or bb.  Well, I’m in luck – my doe is chocolate so both her genes are bb and I know that the probability of her contributing a “b” is 100%.  The buck only carries b so the probability of him contributing a b instead of black(B) is 50%.  The probability that the baby will be brown, or P(brown for doe and brown from buck) is P(brown from doe)*P(brown from buck) = 1.0*.5=.5=50% of the babies will be brown.  But I want lilac instead of brown so I pull out the calculator again and calculate that the probability that the baby will have both dd and bb = probability of dd times the probability of bb = .25*.5=.125 or 12.5% chance.  That means that if I have a typical litter size of 8, on average I could expect one baby to be lilac (.125*8).  That is not a good chance – I might be better off checking with other breeders and buying my lilac baby rather that wasting breeding time and money on a full litter.  But let’s say that I am less picky – I would be happy with a blue or brown or lilac.  Back to statistics class, I know that P(A or B)=P(A)+P(B)-P(A and B).  That complicated equation simply states that I take the chance that I will get a blue baby add the chance I will get a brown baby and take out the chance I will get both at the same time so I do not count it twice.  You can see how that works with the circles below – in order to calculate the total “Red overlapped with Blue” you must add the Red circle with the Blue circle and remove the amount that the Blue circle overlaps with the Red circle because we already accounted for it.

Inheritance Diagram 2

So our Probability(brown or blue) = P(brown) + P(blue) – P(both, or lilac) = .5 +.25 – .125 = .625 or about 62%. So if I have a litter of 8, I know that 2 of them should show blue, 4 of them show brown but since one is lilac or “both blue and brown”, I remove one from the blue group and one from the brown group and just count it as lilac. Therefore one will be lilac, one will be blue and three will be brown, (5 out of 8 = 5/8=.625=62.5% chance of blue, brown or both/lilac). I think that I would be willing to try for a litter with a 63% chance of getting a baby that I want.

You can take it further: assuming that I kept good records, I know that my buck has been throwing good hips in the last 7 out of 16 babies, so the babies should have a 7/16=.44 or 44% chance of nice hips.  Therefore my chances of getting a blue or brown showing baby with good hips is P(brown or blue)*P(good hips) = .625*.44= 28%.  And let’s say the doe has a white toenail that she passed on to 6 of the kits in her first litter of 8 so the chance of a baby NOT inheriting a white toenail is 1.0-6/8 or 25%.  So the chance of getting a blue or brown showing baby, with good hips and matched toenails is .28*.25=.07 or only 7%.  Since I am not feeling lucky, maybe I will pass on that litter after all!

Statistics is only gives you the most likely outcome; just like you can flip a coin five times and get five heads, you can get lucky, but it does give you the knowledge to plan your breedings better.  Maybe you would have a better chance of getting what you want if you pick a different doe or buck.  There is not an unlimited market for rabbits and any tool that gets you where you want to be without creating extra mouths to feed is valuable.  The better records you keep, the more you can make statistics work for you.