My husband and I have been using ivf to try to get pregnant. We have 7 embryos frozen.

My research shows that at my age (38) each embryo (blastocyst actually) has a 40% chance of being euploid (having the correct number of chromosomes and therefore able to make a baby). Each euploid embryo has a 60% chance (for a variety of reasons) of actually being born. So each embryo has a 24% chance of becoming a baby. I rounded to 25% for the ease of the math.

My husband is worried that we'd have 7 children. I showed (I believe) how that would be statistically extremely unlikely. I believe it's .25 to the 7th power which would be 0.006% chance of them all making it. I also think that the chance of having at least one make it would be 1- (.75^7) = 86.65%. So, the chance that we'll have zero children would be 13.35%.

Those numbers look right and my gut math says that at a 25% chance for each one of 7 we'll have 1-2 children from those 7. But what exactly are the odds that we'll have 1, 2, 3, 4, 5, or 6 children. My college math is a little rusty. (To assuage any ethical concerns we will use all frozen embryos and we will be putting them back as single embryo transfers. There will be no chance of twins or higher.)

Assuming that the success rate of 0.25% is independent (i.e. if one baby is successful it does not affect in any way the probability that another is succesful) then your math is correct. The general formula for P(k successful) comes from the binomial probability formula:

P(k) = C(n,k) p^k (1-p)^{n-k}

Where C(n,k) is the combination of k out of n items:

C(n,k) = \frac {n!}{k!(n-k)!}

For your problem the math works out as:

P(0) = 13.3%
P(1) = 31.1%
P(2) = 31.1%
P(3) = 17.3%
P(4) = 5.8%
P(5) = 1.2%
P(6) = 0.1%
P(7) = 0.006%

So while you are correct that P(7) is very, very small, it would not be unusal to end with 0, 1, 2, 3, or even 4. Good luck!

Sounds wonderful! Let's hope it's not zero.

Thank you so much for your help.