Probability of winning your hand in Skull King with a 12 of red

Question

The question came up during a game of Skull King.

We were playing the first round and each of the 6 players received a single card.

With the 12 of red I drew the conclusion that I should mention 0 (not winning my hand, as opposed to 1 which is winning). In this game the person wins if he plays the highest card in this round.

A deck of card consists of 66 cards.

Out of those 66 cards, 22 can beat mine (13 red, 1-13 black, 5 pirates, 2 mermaids, 1 skull king = 22 cards). It is also worth mentioning that I will have to start the round, which means the color i play (red) will be trump. For that reason any yellow and blue card won't win against mine.

What is the probability that I will lose this hand as intended?

2. Is there an easy way to calculate that?

For those who are interested, you can find the official rules under the following url.

Answer 1

Btw, if I feed what you have written as it is, Wolfram returns an impossible probability of $\approx 1.84$

Anyway, the easiest way to compute:

P(you don't win) = $1$- P(you win) $= 1 - \dfrac{\binom{43}5}{\binom{65}5} \approx 0.8835$

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ADDED

An alternative way to compute it is

P(you don't win) $= 1 - \left(\frac{43}{65}\cdot\frac{42}{64}\cdot\frac{41}{63}\cdot\frac{40}{62}\right) \approx 0.8835$

Answer 2

I'm not 100% sure what you mean as I don't know the game, but from your comment I'm assuming the question is the following (or can be considered as the following):

There is a deck of 66 cards, with the $i$th card labelled $i, i = 1,2,\ldots, 66$. Six random cards are drawn. I flip the first and see that it has the label $44$. What is the probability that it is NOT the highest?

At least that's the question I'll answer. To do this we'll calculate the probability it is the highest. In this case, there are $43$ other cards that you beat. There are $\binom{43}{5}$ different combinations of those cards that can be drawn, and $\binom{65}{5}$ combinations in total. Therefore the probability of winning is:

$\frac{\binom{43}{5}}{\binom{65}{5}} \sim 0.1166$.

Hence probability of you losing is approx $0.8834$. You seem to be computing this card-by-card (but also you are not doing this for all six cards anyway, you're missing a term, and it doesn't seem to be correct). The above is a much less messy way and in general:

$X$ - number of cards in the deck $Y$ - number of cards that your card beats $Z$ - number of other players

Then if each player draws $1$ card, the probability you don't have the highest is:

$1 - \frac{\binom{Y}{Z-1}}{\binom{X-1}{Z-1}}$

Let me know if any issues.