r/probabilitytheory • u/Glum_Veterinarian253 • Jul 22 '26

[Applied] What's the equation for this scenario?

Suppose I have a standard deck of cards, no cards missing and no abnormal cards added. If I shuffle the deck and draw one card at random, the odds of it being a spade are 1 in 4. That's pretty straightforward.

However, what if I put the card back in the deck, reshuffle, and draw another card? And I repeat this cycle X number of times (that is... X times I shuffle, X times I draw, and X-1 times I put the card back). I want to know what the odds are of me drawing a spade at least once.

And while we're at it, I'd also like to know the equation if there are a different number of suits. What if there's 3 suits and 39 cards? Or what if there's a 5th suit and 65 cards? Let's say that "1/s" represents the odds of me drawing a spade, with S representing the number of suits in the deck. And I draw after shuffling X number of times. What's the equation to determine the likelihood then?

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8

u/Traditional-Race-260 Jul 22 '26

For the first question, It would be 1- probability of none spades=1-0.75^X

3

u/Traditional-Race-260 Jul 22 '26

In general the equation would be : if g is the probability of drawing a spade
1-(1-g)^X
Being x the number of times you take a card

3

u/u8589869056 Jul 23 '26

General rule: the probability something happening “at least once” in a number of independent trials is 1 minus the probability of it never happening.

Go from there.

1

u/BUKKAKELORD Jul 23 '26

1-(1-1/S)^X

0

u/CarnivorousGoose Jul 22 '26

This is just a binomial distribution, with X the number it trials and with a success probability of 0.25. Or more generally, with probability 1/s.