How many cards, until the first ace: variations, extensions, lachrymae, confidence, Dirichlets
Abstract
From a deck of cards, how many cards do I need to draw, until the first ace? I identify the distribution for this waiting time T, and its satisfyingly nice expected value E T=(N+1)/(n+1), with N the number of cards and n the number of aces; hence 53/5=10.6 for the standard setup. After having solved this Question One I go on to certain alternative solutions and extensions, involving e.g. Beta approximations. I also consider the distributions and means for the 2nd, the 3rd, the 4th occurrences of aces, with generalisations, where there is a Dirichlet distribution in wait for us, with further links to order statistics for the uniform. Furthermore, an apparatus is developed for obtaining estimators and full confidence distributions for applications where one knows the number n of aces, but not the deck size N; and correspondingly for inference about the unknown population size N when n is known. If you have 1000 people in a room, and need to interview 11 of them until you've found the first left-handed person, how may left-handed are there in the room -- here we need both an estimate and a clear measure of uncertainty.
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BibTeX
@misc{hjort2026many,
title = {{How many cards, until the first ace: variations, extensions, lachrymae, confidence, Dirichlets}},
author = {Nils Lid Hjort},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.29596},
eprint = {2609.29596},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.29596},
}
RIS
TY - UNPB
TI - How many cards, until the first ace: variations, extensions, lachrymae, confidence, Dirichlets
AU - Nils Lid Hjort
PY - 2026
DA - 2026/09/01
PB - arXiv
DO - 10.48550/arxiv.2609.29596
UR - https://arxiv.org/abs/2609.29596
AB - From a deck of cards, how many cards do I need to draw, until the first ace? I identify the distribution for this waiting time T, and its satisfyingly nice expected value E T=(N+1)/(n+1), with N the number of cards and n the number of aces; hence 53/5=10.6 for the standard setup. After having solved this Question One I go on to certain alternative solutions and extensions, involving e.g. Beta approximations. I also consider the distributions and means for the 2nd, the 3rd, the 4th occurrences of aces, with generalisations, where there is a Dirichlet distribution in wait for us, with further links to order statistics for the uniform. Furthermore, an apparatus is developed for obtaining estimators and full confidence distributions for applications where one knows the number n of aces, but not the deck size N; and correspondingly for inference about the unknown population size N when n is known. If you have 1000 people in a room, and need to interview 11 of them until you've found the first left-handed person, how may left-handed are there in the room -- here we need both an estimate and a clear measure of uncertainty.
ER -