I play a variation of Gin Rummy with members of my family and we slightly adjusted the game rules and significantly tweaked the scoring. One of the players suggested a new rule to mark each occasion a player couldn't meld his cards at all, before his opponent knocks (to partially eliminate the luck out of the game). Each time it happens, the unfortunate player would still lose the game, but would be given a special "bad-luck point" in case the current tournament would end up with a tied score. In that particular scenario, the player that has more "bad-luck points" would win the tournament.
I agreed with the suggestion, but wanted to alter such rule with a more fair adjustment: there should be an inverse situation that cancels a single "bad-luck point" - if a player has been dealt with all 3 jokers during a single game!
The only thing missing is to equalize the probabilities of:
a) a bad luck situation when a player can not meld his cards until the end of the game, and
b) a good luck situation when a player has been dealt all 3 jokers during the game
Currently I'm not sure if a) and b) from above have even remotely similar probabilities, since the probability of a) depends on each player's strategy (therefore I may have to simulate it in Python), but I'm pretty sure the b) part could be calculated:
QUESTION:
There are 3 players and a dealer shuffles 2 standard decks of cards with added 3 joker cards (107 cards in total). Player 1 cuts the deck and if he cuts a joker card, he could keep it. If the card(s) that follows the joker were also joker(s), he may keep them as well, as long as the joker cards are not interrupted with a non-joker card. If Player 1 didn't cut the joker, the card will be dismissed and the players won't be able to use it throughout the game.
The dealer then deals Y cards, one-by-one to each player one-after-another, so every player will have to hold the same amount of cards (Y could be any number between 1 and 35).
a) What is the probability for one player to have no joker cards among his Y cards?
b) What is the probability for one player to have exactly 1 joker among his Y cards?
c) What is the probability for one player to have exactly 2 jokers among his Y cards?
d) What is the probability for one player to have exactly 3 jokers among his Y cards?
My attempt:
1 non-joker card being dismissed from the deck of 107 cards:
d) There are $ A_1 = {106 \choose Y} $ ways of choosing Y cards for the first player, $ A_2 = {106-Y \choose Y} $ ways of choosing Y cards for the second player and $ A_3 = {106-2Y \choose Y} $ ways of choosing Y cards for the third player. There are $ {3 \choose 3} $ = 1 way to draw the 3 jokers for a single player and $ B = {103 \choose Y-3} $ ways to select the remaining cards for the first player. Thus the probability is: $ { B \over A_1 } $.
Are the probabilities slightly in favor for the first player?
Complicated questions. Given the posting, and background, it is plausible that the OP (i.e. original poster) isn't really trained for this type of situation.
Typically, when faced with such a Probability problem, and given the choice between attacking the problem as either a Probability of events problem, or a Combinatorics-numerator-denominator enumeration problem, I usually opt for the latter.
This particular problem is so complicated, that I am going to partially use the former strategy, and partially use the latter strategy. What is more, I am going to dispense with any attempt at elegance, and plow through the individual Cases.
At the end of my answer, I will explain what the reader needs to do, to arrive at an overall answer to each of the questions.
Let $P(k)$ denote the probability that Player-1 initially cuts $(k)$ jokers, before the subsequent distribution of $Y$ cards to each of the $(3)$ players. Here, $k$ must be some element in $\{0,1,2,3\}$.
Assume that after Player-1 cuts the $(k)$ jokers, the resulting status of the deck is expressed as
$$[A::B]$$
to represent that after the cut, but before the distribution of the $Y$ cards to each player, there are $A$ cards, remaining in the deck, of which $B$ of these cards are jokers.
$\underline{\text{Simplifying Assumptions}}$
It should be assumed that:
If the status of the deck, for a specific case is $$[A::B],$$ that you can not have that $3Y > A.$
If $Y$ cards are distributed, and $s > Y$, then the probability of one of the three players receiving $s$ jokers, after the initial cut is $(0)$.
If the initial cut by Player-1 is $k$ jokers, and $s > (3 - k)$, then the probability of one of the three players receiving $s$ jokers, after the initial cut is $(0)$. So, the supplied formulas assume that $~s \in \Bbb{Z},~$ with $0 \leq s \leq (3-k).$
No mention will be made of Player-3. In all cases, the probabilities that pertain to Player-3 will always be identical to the probabilities that pertain to Player-2.
$\underline{\text{Case} ~0 : k = 0}$
$$P(0) = \frac{104}{107} :: ~\text{Deck Status} = [106::3]. \tag1 $$
The probability that Player-2 is then dealt $s$ jokers is
$$\frac{\binom{3}{s}\binom{103}{Y-s}}{\binom{106}{Y}}. \tag2 $$
Here, (2) above also applies to Player-1, since, in Case 0, Player-1 started by cutting $(0)$ jokers.
Possible Ambiguity
Under the assumption that Player-1 starts out by cutting $1$ or $2$ jokers, it is unspecified, whether the subsequent card that Player-1 cuts, which stops Player-1 from initially receiving any further jokers, is also discarded.
Consistent with the concept behind the specification when $k = 0$, I will assume that when $k = 1$ or $k=2$, that the subsequent non-joker card is discarded.
$\underline{\text{Case} ~1 : k = 1}$
$$P(1) = \frac{3}{107} \times \frac{104}{106} :: ~\text{Deck Status} = [105::2]. \tag3 $$
The probability that Player-2 is then dealt $s$ jokers is
$$\frac{\binom{2}{s}\binom{103}{Y-s}}{\binom{105}{Y}}. \tag4 $$
With Player-1 starting out by cutting $(1)$ joker, the expression in (4) above is also the probability that Player-1 is then dealt $s ~\color{red}{\text{additional}}~$ jokers.
$\underline{\text{Case} ~2 : k = 2}$
$$P(2) = \frac{3}{107} \times \frac{2}{106} \times \frac{104}{105} :: ~\text{Deck Status} = [104::1]. \tag5 $$
The probability that Player-2 is then dealt $s$ jokers is
$$\frac{\binom{1}{s}\binom{103}{Y-s}}{\binom{104}{Y}}. \tag6 $$
With Player-1 starting out by cutting $(2)$ jokers, the expression in (6) above is also the probability that Player-1 is then dealt $s ~\color{red}{\text{additional}}~$ jokers.
$\underline{\text{Case} ~3 : k = 3}$
$$P(3) = \frac{3}{107} \times \frac{2}{106} \times \frac{1}{105}. \tag7$$
When Player-1 starts out by cutting all $(3)$ jokers, it is irrelevant how many cards are then dealt out. That is, there are $(0)$ remaining jokers to distribute.
$\underline{\text{Putting it All Together}}$
$$P(0) = \frac{104}{107}.$$
$$P(1) = \frac{3}{107} \times \frac{104}{106}.$$
$$P(2) = \frac{3}{107} \times \frac{2}{106} \times \frac{104}{105}.$$
$$P(3) = \frac{3}{107} \times \frac{2}{106} \times \frac{1}{105}.$$
The probability of Player-2 receiving exactly $s$ jokers is
$$\left[P(0) \times \frac{\binom{3}{s}\binom{103}{Y-s}}{\binom{106}{Y}}\right] ~+~ \left[P(1) \times \frac{\binom{2}{s}\binom{103}{Y-s}}{\binom{105}{Y}}\right] ~+~ \left[P(2) \times \frac{\binom{1}{s}\binom{103}{Y-s}}{\binom{104}{Y}}\right]. \tag8 $$
For Player-2, in the particular case of $s=0$, to the expression in (8), you also have to add the explicit value of $P(3)$, which represents the probability that (for example) Player-2 received no jokers specifically because Player-1 initially cut all $(3)$ jokers.
For Player-1, the computation is more complicated, because the expression in (8) represents the probability of Player-1 receiving $~s ~\color{red}{\text{additional jokers}},~$ after the initial cut.
So, I will explicitly specify the Probability that Player-1 ends up with $s$ jokers, overall as follows:
For $s = 0~$:
$$P(0) \times \frac{\binom{3}{0}\binom{103}{Y}}{\binom{106}{Y}}.$$
For $s = 1~$:
$$\left[P(0) \times \frac{\binom{3}{1}\binom{103}{Y-1}}{\binom{106}{Y}}\right] + \left[P(1) \times \frac{\binom{2}{0}\binom{103}{Y}}{\binom{105}{Y}}\right].$$
For $s = 2~$:
$$\left[P(0) \times \frac{\binom{3}{2}\binom{103}{Y-2}}{\binom{106}{Y}}\right] + \left[P(1) \times \frac{\binom{2}{1}\binom{103}{Y-1}}{\binom{105}{Y}}\right] + \left[P(2) \times \frac{\binom{1}{0}\binom{103}{Y}}{\binom{104}{Y}}\right].$$
For $s = 3~$:
$$\left[P(0) \times \frac{\binom{3}{3}\binom{103}{Y-3}}{\binom{106}{Y}}\right] + \left[P(1) \times \frac{\binom{2}{2}\binom{103}{Y-2}}{\binom{105}{Y}}\right] + \left[P(2) \times \frac{\binom{1}{1}\binom{103}{Y-1}}{\binom{104}{Y}}\right] + \left[P(3)\right].$$