Question to the smart people:

If one is using the kangaroo method, searching for 32 bit distinguished points, and searching in the 120 bit range, how many x coordinates (x coordinate is what the program finds of a given public key, which in kangaroo terminology means "distance traveled") would have a 32 bit distinguished point?

There are 2^120 possibilities, but we know not every private key in the 120 bit range has a public key x coordinate that has a DP of 32. Is there a good guesstimate or is it just anyone's guess?

I should clarify, how many tame points in a 120 bit range would/could have distinguished points of 32 bits?

Probably a good as a guess as any. One also has to remember there are leading DPs example: 00001234 and trailing DPs, example: 12340000. There are also in between DPs, example: 12000034. I've seen all three used in different Kangaroo programs. Not sure which is best/faster to search for. If they are of the in between type, they have to line up. Example, 12000034 won't work with 10000234. All interesting.
I went through a small range 1-300FFF and didn't find any 32 DPs. I was closely looking at trailing, but I don't think there were any leading or in between either.

Anywho, if we use your 2^28, that would mean 2^28 for each type of DP, leading, trailing, and in between. So that would mean 2^29.5 (minimum, because the in between ones can be in between anywhere.)

No, that is not accurate. A distinguished point could be anywhere. It's merely how many "zeros" are in a row (consecutive) in a given public key x coord.  It shouldn't matter how large the range or private key is. It will just take a lot longer to find a 32 bit DP, in any range. I found several 24 bit DP in that 2^24 range, just no 32 bit DPs.  I'm not looking at this to find a key, I am just curious how many trailing or leading 32 bit DPs could be in a range, in particular, the 2^120 range. I thought there might have been a good calculator for it but maybe it is all just mathematical luck, where DPs fall.

 is there any kangaroo version for ethereum?

Not to my knowledge...But there is a Bitcrack version for ethereum. Developer was selling it though.

Ok, I misread your first post.

Now, from the 8,589,934,592 32 bit DPs, I wonder how many are trailing, leading, and in between? If it's an equal spread or one occurs more than the other.

Been looking at different numbers; so how many 36bit Distinguished Points in a 72 bit range?

Applying what you said earlier, (2^36-72) = 2^-36. Then use 2^72-36 = 36 or 68,719,476,736 36bit DPs in a 72 bit range?

So you don't half the search space first then subtract the DP?

As you see in the images above the person that solved #110 and #115.

Example with #120 using 32 bit Dp: Operations = 2^(120/2+1) = 2^61.

61 - 32 = 29

32 + 29 = 61 When you merged both work files they both should be 2^29 each, then they should equal to 61 or close to it.

So 32 bit Dps for #120 is 536,870,912