First Right Side Up Jenny Souvenir Sheet Auction For $45,000

Wonder what the probability is of getting at least one Uvert in 3000 sheets. How many should you expect to purchase before your odds reach the 50% level? I'm too far removed from my probability & math classes to do the computation ... although this would seem to represent an example of a hypergeometric distribution. Due to the small probability of success, I assume that some approximation to hypergeometric would be used to do the actual computation ... perhaps binomial or normal.

My take on the probability is at the beginning, if no sheets were purchased yet, take the total number of sheets printed, divide by the number of "uninverts" printed, and divide by two, and you're at the number of sheets you need to purchase to have 50% probability of getting one "uninvert".

So, if the number of originally printed "regular" sheets (inverts) is "R", and the number of originally printed "error" sheets (uninverts) is "E", the number of sheets you would need to purchase to have 50% chance to get 1 "uninvert" would be:

50% probability of getting one = (R + E)/(E x 2)

SO, if R = 2,000,000 and E = 100, then then number of sheets is 10000.5 sheets, or 10001, since 10000 sheets would be slightly less than 50%!

On the other hand, if R + E = 2,000,000, then the number of sheets is
10000 sheets.

To answer JLLebert's question, if one purchased 3000 sheets at the beginning of the offering, they would have had a 15% probability of getting one sheet. NOT a good bet!

Once some sheets have been purchased, and one or more of the "uninverts" have been discovered, the equation changes and becomes more complex.

OK, to follow up on what the probability is once one or more sheets have been sold, here we go:

Once some sheets have been purchased, and one or more of the "uninverts" have been discovered, the equation changes and becomes more complex, so may be better to change definitions.

Let:

T = total number of originally printed sheets (including both "regular" and "error" sheets) = R + E

T' = number of sheets sold (this means the number of sheets still not sold = T – T')

E = total number of error (uninvert) sheets issued

E' = number of error sheets sold

P = Phantom number of error sheets sold, but not discovered or announced yet

Probability of getting one sheet at the beginning, before any sheets were sold was = (R + E)/(E x 2).

Once one or more sheets are sold it is:
(T - T') / (E – E' - P)(2)

So, in the case of T=2,000,000, T'=1,000,000, E=100, E'=25, P=25, the number of sheets needed to be purchased to have a 50% probability of getting one sheet is: 10000.

You probably notice the BIG problem here of "investing" in sheets once the gate is open – you DON'T know what the value of "P" is! P could be 75, so there could be really NONE still unsold.

Cheers,

Dave

The type of calculations by orstampman assumes that there is a single pool of all Jenny sheets, and the Uvert sheets are randomly distributed in this pool. In reality, however, there are hundreds of separate pools (all the post offices receiving allotments of Jenny sheets, plus USPS's SFS, ebay, and web stores) and no information on how the Uvert sheets are distributed among them. Thus, sheets sold and Uvert sheets discovered in one pool may have absolutely no impact on the probability of finding an Uvert sheet in another pool.

The bottom line is that it is impossible to calculate the probability of finding an Uvert sheet in a specific pool even if one has figures relating to the aggregate of all pools.

Definitely agree with the additional "unknown" factor of pools or uneven distributions. My calculation assumes a completely random distribution.

SO, given that this was shown to be a bad investment/bet already, this makes it even worse...

I think whether this is a bad investment depends very much on one's perspective.

For example, when you buy a lottery ticket and do not win, you lose the entire cost of the ticket. However, when you buy a Jenny pane and it is not an Uvert, you still have the stamps. Even if you do not use them as postage, you could still recover at least 70-80% of their face value in the secondary market--probably more if you hold on to them for a number of years after they go out of print.

Additionally, because of its widespread print anomalies, any given Jenny pane has a decent probability of containing some sort of profitable varieties (flattened wheels, truncated wings, color bleeds, etc. etc.).

I am not saying that this is necessarily a good investment, but the inherent cost of the investment is actually much lower than at first blush.

Good point. If one, in the process of buying many of these sheets, gets at least one of the "uninvert" sheets, then yes, it may be at least closer to a break-even "investment", if not profitable. Plus a great opportunity to find repeated varieties.

Bankruptcollector - srailkb is correct.

Rider - color bleeds on these are nothing - they all have them to various degrees. And ebay has been flooded with "flat tires". They might have sold for a premium at first, but I doubt it any more.

Attention: Collectors with unlimited budgets! Do you want to buy a right-side-up Jenny Souvenir Sheet outside of auction? It will only set you back $60,000 (+ tax) at this link, but is claimed to be a PSE graded XF-Superb (95) specimen:

http://www.hgitner.com/shop/1931-da...perb-nh.html

The posted picture is of insufficient resolution to discern whether it is completely free of any of the widespread printing anomalies that plagued the normal inverted Jenny pane.

a share of a bridge in NY is only 1/2 that (potholes included)

While this is out of the reach of most collectors, it will appeal to those who can afford it. Something that most will never have.

And I wonder if most of the collectos who gripe about this were suddenly in a position to afford an example, would they buy it?


Glenn Estus