Two people, (call them C and D), decide to play a card game for fun. They use an ordinary fair deck of $52$ cards, shuffled well before each hand is drawn, and randomly draw cards from it one a time without replacement, both using (sharing) the same drawn cards to determine who wins. A win is defined as follows:
C wins if he gets at least one of all $13$ ranks of the cards (regardless of suit as they can be mixed suit or even all the same suit) in a hand.
D wins if he gets either $5$ reds or $5$ blacks in a row (consecutive) for a particular hand. Each new hand starts with $0$ in a row so far so there is no "carryover" from a previous hand.
It is possible that C and D can both "win" on the same card draw so normally that would be a tie but a "twist" in the game is that ties are awarded to C but not just as a single win. Since ties are likely rare, C gets a triple win for ties. That is, if C and D bet even money and they "tied", C would then win $3$ to $1$ odds of whatever D bet him for that particular hand. So let's take an example run so there is no confusion. Suppose they both bet $1$ dollar per hand and the following happens:
D wins game $1$ so he is then up by $1$ dollar.
C wins game $2$ so they are both back to even money.
D wins the next $2$ games so he is then up $2$ dollars over C (C is down $2$ dollars).
The next game is a tie so C is awarded $3$ dollars so is then ahead by $1$ dollar.
Another way to think about it is to not think about money but just count up the number of wins. If there is a tie, C gets awarded $3$ wins for that hand.
So the question is who has the mathematical advantage here and by how much? For example, if it was a rainy day and they played this game for many hands, who would likely be ahead as far as net money gained as a result of playing this game?
Some interesting things to consider are:
- D can immediately win with only $5$ card draws while C requires $13$ minimum.
- It is possible that D will not win even if all the cards are drawn, never getting $5$ in a row of either color.
- A decision can take anywhere from $5$ to $49$ cards. $49$ is the max because imagine if $12$ of each rank (of all $4$ suits) have been chosen but D hasn't won yet for that hand, the next card will complete one of those set of ranks. For example, if the last $4$ cards in the deck are all Kings (K), the $49$th card will give the win to C (assuming D doesn't win or tie).
$$UPDATE$$ I ran a simulation of 1 billion decisions (ties included) and the results are as follows:
C won : $469,102,581$ times. (excluding triple wins for ties).
D won : $514,835,119$ times.
C,D tied : $16,062,300$ times. (C awarded triple win).
C won : $517,289,481$ times. (including triple wins for ties).
Advantage for C: about $0.48$%
Average # of cards drawn to make a decision is $20.579$.
So the triple win award for ties gives C a very slight edge over D but without that D has a decent advantage. So in theory, if they played this game for many hands, they would about break even. However, in the shortrun, someone could take a sizable lead. Sometime I may try about $10$ hands with actual cards and see what I get.
I would like to know how to set this problem up mathematically or if it is even possible. Perhaps we could first solve a simpler variation where we draw exactly $21$ random cards then check for a winner. Perhaps that will give us some insight into how to solve the more general question with a variable # of cards (from $5$ to $49$ is possible.).
Also, can someone tell me how to plot a graph on here because I have data for the # of wins of each # of cards drawn from $1$ to $52$. The numbers show some interesting patterns. Out of $1,000,000$ decisions, $5$ cards drawn accounts for the most wins at about $5$%. Next is very close between $23, 24,$ and $25$ cards which account for about $4.4$% each.