I am developing a card game with a friend and I want to calculate the probability of different win conditions being achieved (I am putting together an excel sheet).

I have trouble calculating the probability of drawing a four-card straight out of fifteen cards (which is the average amount of cards you see per game). These are the relevant parameters:

120 cards in the deck Numbered cards run from 1-10. Each number is on 6 cards in the deck. (60 in total) There are fifteen joker cards which can represent any number. The remaining 45 cards are blanks for the purposes of this calculation, i.e. they can be drawn but not used to complete a straight.

I've been watching different videos on how to do this, and tried out different ways of calculating it, but have somehow ended up thoroughly confused. Could one of you help me out?

Edit to answer questions that came up:

  • The 45 non-specified cards are 'action cards' that allow you to interact with your opponent's cards. For the purposes of this calculation, assume that they are blanks, i.e. can be drawn but can not be used to complete a straight.
  • You can use more than one joker, so four jokers would be an automatic straight. Three jokers does not make an automatic straight, because you could draw only action cards with the remaining twelve draws (i.e. blanks for the purposes of making a straight).
  • The intention is that the probability of getting to a straight will be high because there are many ways to mess with your opponent in the game. I am aware that this throws off the probabilities, but I'm just looking for baseline probabilities so we can balance the different win conditions.
  • 4
    $\begingroup$ The number don't seem to add up. If you have $6$ each of cards numbered $1-10$ that makes $60$ cards, and $15$ jokers makes $75$ cards. What are the other $45$ cards? $\endgroup$ – saulspatz Feb 13 at 17:28
  • $\begingroup$ You have to carefully specify the rules. Can you use more than one joker? Would three jokers guarantee a straight because you will have another card? I am guessing the chance will be high, but we need to know what the deck is. $\endgroup$ – Ross Millikan Feb 13 at 20:36

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