# The expected number of a given digit on a dollar bill? **URL:** \ **Category:** Factual Questions\ **Created:** [December 23, 2002, 8:01pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580 "2002-12-23T20:01:27Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![js\_africanus](https://avatars.discourse-cdn.com/v4/letter/j/87869e/32.png) [@js\_africanus](https://boards.straightdope.com/u/js_africanus)\ **Post date:** [December 23, 2002, 8:01pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/1 "2002-12-23T20:01:27Z")
Howdy, A while ago I was discussing liar’s poker with a friend. It made me wonder what is the expected number of occurrences of a given digit in a dollar bill’s serial number. For example, suppose I have a bill with three sevens and I’m playing against 4 people. How many sevens should I expect between the five of us? I tried to figure it once and got twenty, which is clearly wrong since there are only eight digits in a serial number. I recently tried again by figuring that the odds of a seven being in any spot is 1/10 times the number of spots, 8, which gives 0.8 sevens expected on a dollar bill. So, in my previous example, I should expect my three plus 0.8 \* 4 = 3.2 which gives 6.2 sevens expected between the five of us. Is my figuring correct? ---
**Author:** ![Jpeg\_Jones](https://avatars.discourse-cdn.com/v4/letter/j/4bbf92/32.png) [@Jpeg\_Jones](https://boards.straightdope.com/u/Jpeg_Jones)\ **Post date:** [December 23, 2002, 8:31pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/2 "2002-12-23T20:31:28Z")
Close. Since you’ve already counted your three 7s, you just want to know how many 7s there are likely to be in your three opponents’ dollar bills. So, 0.8 x 3 = 2.4 Plus your existing 3 = 5.4 sevens between the four of you. ---
**Author:** ![Jpeg\_Jones](https://avatars.discourse-cdn.com/v4/letter/j/4bbf92/32.png) [@Jpeg\_Jones](https://boards.straightdope.com/u/Jpeg_Jones)\ **Post date:** [December 23, 2002, 8:35pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/3 "2002-12-23T20:35:02Z")
Oh. Wait. I see that there **are** 5 people total involved, so you were right the first time. ---
**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\ **Post date:** [December 23, 2002, 8:40pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/4 "2002-12-23T20:40:01Z")
The math would be right if the numbers were randomly assigned, but they aren’t. [http://www.sit.wisc.edu/~dmoffitt/serials/numb.html](http://www.sit.wisc.edu/~dmoffitt/serials/numb.html) To figure the expected value, one would need to know the series, denomination and make some assumption on the circulation at that point in time(the numbers then in circulation) of those bills. ---
**Author:** ![bcullman](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bcullman/32/8207_2.png) [@bcullman](https://boards.straightdope.com/u/bcullman)\ **Post date:** [December 23, 2002, 8:50pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/5 "2002-12-23T20:50:03Z")
assuming the digits in a serial number are random (and they are not, so this is where someone else will have to pick up the ball), yo should expect that if there are 5 one dollar bills (one for each person) and there are 8 digits in a serial number there are a total 40 digits we are inspecting. Out of a 10 possible digits, you sould see 7 on average 4 times total. On average that is slightly less than one 7 per dollar bill. The fact that you have a bill with three 7’s on it does not enter into the equation. ---
**Author:** ![Jpeg\_Jones](https://avatars.discourse-cdn.com/v4/letter/j/4bbf92/32.png) [@Jpeg\_Jones](https://boards.straightdope.com/u/Jpeg_Jones)\ **Post date:** [December 23, 2002, 8:58pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/6 "2002-12-23T20:58:47Z")
**bcullman** , you would be right if we didn’t know any of the 5 serial numbers and simply wanted to know our odds. But **js\_africanus** _does_ know that there are at least 3, because he (?) has them. I guess some clarification of the rules of Liar’s Poker might be warranted. ---
**Author:** ![js\_africanus](https://avatars.discourse-cdn.com/v4/letter/j/87869e/32.png) [@js\_africanus](https://boards.straightdope.com/u/js_africanus)\ **Post date:** [December 23, 2002, 9:39pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/7 "2002-12-23T21:39:12Z")
> [@](#): > > \*Originally posted by Jpeg Jones \* > **…he (?)…** Yup. > [@](#): > > \*\*I guess some clarification of the rules of Liar’s Poker might be warranted. \*\* It has something to do with claiming to have x number of whatever digit you choose between the players. So if I had a bill with three sevens, and I was playing with four other people, I could “safely” say that we have 6 sevens between us. Then you, for example, may bet that we have 7 nines between us. At some point somebody calls bullshit, or something like that, and a winner is determined. I don’t know much of the game beyond that. > [@](#): > > _Originally posted by Aahala_ > **The math would be right if the numbers were randomly assigned, but they aren’t.** Good eye–I hadn’t thought of that. I guess I had assumed that bills are distributed randomly. That may be a tenuous assumption, especially since the sort of people who bet money on such a game might keep an unusual bill in her wallet for just such an occasion. ---
**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\ **Post date:** [December 23, 2002, 9:51pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/8 "2002-12-23T21:51:47Z")
Assuming the site I gave above is correct(and my understanding of it), then the right five digits of $1 bills in circulation is(almost) random. That doesn’t leave much room for playing the game described–higher numbers of 7 on five bills is so unlikely, so you might want each person to have two or more bills. ---
**Author:** ![js\_africanus](https://avatars.discourse-cdn.com/v4/letter/j/87869e/32.png) [@js\_africanus](https://boards.straightdope.com/u/js_africanus)\ **Post date:** [December 23, 2002, 11:29pm UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/9 "2002-12-23T23:29:22Z")
> [@](#): > > \*Originally posted by aahala \* > \*\*That doesn’t leave much room for playing the game described–higher numbers of 7 on five bills is so unlikely, so you might want each person to have two or more bills. \*\* Indeed. Or it makes the informed player that much more formidable. Thanks for mentioning that site again, I forgot to check it out. If 96,000,000 is the maximum, then that _does_ change the calculation, doesn’t it? Hmm. That’s going to be bothering me now. It seems like a trickier problem than I had first imagined. Is Benford’s Law an issue in calculating the odds here, since the numbering is sequential? Just for fun, check out this serial number I have on a $2: G55955555A. ---
**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\ **Post date:** [December 24, 2002, 2:20am UTC](https://boards.straightdope.com/t/the-expected-number-of-a-given-digit-on-a-dollar-bill/144580/10 "2002-12-24T02:20:18Z")
From the currency site, I take it the entire 6.4 million series is printed and released nearly at once, then a new block, so the last five digits would all be used. The front digits would be subject to that Benford’s Law, and an additional factor. Paper money wears out rather quickly, so the early blocks would not have as many in circulation. I was aware of Benford’s Law effect but not its name, thanks for the link.