I'm purely curious and google has failed me.
Based on the photo from Jolenealaska, I did some analysis:
Note: I'm no mathematician, so let me know if my math is terrible, and how to fix it!
I took the image and opened it in ImageJ, then as the image is lighter on the left than the right I adjusted the window level to even out the background. I then converted to binary (true black and white, not greyscale) and "watershed" the image to separate out the clusters of granules.
Following this, I used the "analyze particles" command with the parameters "size = 20 - infinity" to exclude any small crumbs that might skew our data. This resulted in a total of 429 particles counted, and gave me some more data about the area of each in pixels (as there's no scale on the image). Fortunately this tied in nicely with my hand counting of the particles. So:
If there are 20 *10^9 cells/gram, then there are 1 *10^9 per 0.05g
1*10^9/429 = 2331002 cells/granule on average.
For those interested, is this at all correct?:
The statistics as calculated in R using the
summary() command on the values generated in ImageJ are (in pixels):
Min. : 20.00
1st Qu.: 26.00
Median : 31.00
Mean : 36.27
3rd Qu.: 43.00
This tells us that there are a couple of outliers in the data, that are probably skewing the mean area a bit (and also that the watershedding didn't work fully). The data doesn't look quite like I would expect - I was expecting more of a bell-shaped curve, less heavily skewed to the left, but I did cut off data less than 20 pixels, so the left-side of the curve is a bit selection-biased anyway.
According to this answer a 7 g packet of yeast is 2.25 teaspoons, and a US teaspoon is 4.92892 ml (AKA cubic cm/cc, also thanks google for knowing the volume) - so
4.92892 ml * 2.25 ml = 11.09007 ml (or cubic cm). so for 1 g we have 11.09007/7 = 1.5843 cc/g 1.5843 cc * 0.05 g = 0.0792 cc 0.0792/429 granules = 0.000185 cc/granule
According to this paper (paywall?) the average volume of a baker's yeast growing at 30 C is 309 um^3 and there are 10^12 um^3 in a cc. I can't find the volume of a freeze-dried yeast cell, but I expect it'll be a bit smaller.
0.000185 cc = 184615400 um^3 So 184615400 / 309 = 597461 cells/granule.
The good news about this is that we are in the right ball-park for numbers with the top estimate being in the low millions and this being in the mid 100,000s. There will be some error in this estimate too, as not all the volume measured here is taken up by cells - some of it is air, so this number is possibly a bit high.
The granule size will vary from brand to brand. Even within a packet, granule size is not completely uniform. To get the estimate, you'd need to weigh out a sample (say 1g or 0.1g), count the number of granules in that sample, then do the math. The larger the sample you count, the better estimate you'll have.