# Math Help - How do I add successive diminishing probabilities for a single event?

1. ## How do I add successive diminishing probabilities for a single event?

Hi everyone. I play cards with friends, and I ran into this problem:

Say I have a deck of 52 cards. Now I add one joker to the deck. Now I pull the cards one at a time, checking the card each time. What are the odds that I draw the joker after n number of draws?

Logic tells me that it has to add up to 1 after 53 draws, but I don't know the formula. I tried a few things, but none of them made sense (like adding to one after 20 or so draws). My next intuition says something like, for the first 3 draws:

$\frac{1}{51}${ $1-[\frac{1}{52}(1-\frac{1}{53})]$}

But I could be way off my rocker here. I could test this formula, but I don't mean to be petulant when I say that there seems to be no easy way to run through this formula...53 iterations is really complicated. If this is the solution, or even if it is not, is there an easier way to do the math?

Just to be clear, I am not looking necessarily for the 53rd iteration, in case anyone was thinking of doing something tedious. I'm just interested to know the general formula, and perhaps a simpler way of calculating it if the one I gave is correct (hopefully it's not!)

Incidentally, what if there were 2 jokers in the deck? How would the probability for the 2nd joker change after drawing the first one? I am totally lost on that one!

Thanks for any tips/advice! Links of course are also appreciated. I tried searching for this particular question, but everything I put it sends me to simpler probability problems...or at least I don't think they apply to mine. Thanks for reading!

Edit When I tried a different approach, imagining a deck of 2 cards with 1 joker, I realized that my formula wouldn't work. In light of that, the only thing that I can think of is that your probability to draw the card remains constant and equal to the initial probability, despite having removed cards from the deck. In other words, after 10 draws in a 53 card deck, my probability would be 10/53 to have drawn the card, even though on that 10th draw my probability is only 1/43. Is this correct?

But that would still leave a bit of a dillema in that if I were to add up the individual probability for each of the first 10 draws, I would not have 10/53 by the end. So I guess back to square one.

For example, if I had 3 cards in a deck, one of which is a joker, then I have a 2/3 chance to draw the joker after 2 draws; however, on the second draw, I must have a 1/3 chance to draw it, even though only two cards are remaining. This is because I could not add 1/3 from the first draw, then 1/2 from the second draw to reproduce the 2/3 total. This seems like an extended example of that game show with the 3 curtains, but my logic is faltering at that point where I have 2 cards in my hand and only a 1/3 chance to pick the joker instead of 1/2. That doesn't make sense.

Anyone with some advice for the correct direction? Thanks!

2. ## Re: How do I add successive diminishing probabilities for a single event?

The answer is a bit counter intuitive at first but on reflection becomes clear.

The answer is that the probability of selecting your joker after k draws is $\frac{1}{53}~\forall k \in 1,2,\dots 53$

Think about it this way. Is the joker any different from any other individual card? No. Thus it's just as likely to find any selected card at the kth draw as it is to find the joker. If all the cards have equal probabilities then they must all have probability $\frac{1}{53}$ at each draw.

The sum of all these probabilities is just $53 \cdot \frac{1}{53}=1$

3. ## Re: How do I add successive diminishing probabilities for a single event?

Originally Posted by romsek
The answer is a bit counter intuitive at first but on reflection becomes clear.

The answer is that the probability of selecting your joker after k draws is $\frac{1}{53}~\forall k \in 1,2,\dots 53$

Think about it this way. Is the joker any different from any other individual card? No. Thus it's just as likely to find any selected card at the kth draw as it is to find the joker. If all the cards have equal probabilities then they must all have probability $\frac{1}{53}$ at each draw.

The sum of all these probabilities is just $53 \cdot \frac{1}{53}=1$
Hi there! Thanks for that answer. One point I'd like to clarify:

You are saying that if I have 3 cards, one of which is a joker, then I have a 1/3 chance to draw the joker on all the draws? So a $\frac{1}{3}$ chance on the first draw, and then when I only have two cards left, I still have a $\frac{1}{3}$ chance to draw the joker?

My reasoning is that I know that I have a $\frac{2}{3}$ chance after 2 draws to have drawn the joker, but if I were to approach the analysis step by step, then I should be able to sum the probability of each step individually to equal this $\frac{2}{3}$ chance.

With that statement, the implication is that if I don't know the initial state, I can say nothing about the probability? For example, if I arrived late to a deck drawing with n cards searching for 1 joker, then I would not know how many cards were initially there, and I could make no statement as to what the probability the joker may be drawn, even though I could count how many cards lay before me and know only one joker is there. I know this is wrong

That does seem very counter-intuitive, since we make statistical analyses in nature all the time, even though we are often not there at the beginning of an event.

Edit: Ah, I guess I have figured a formula that would work for this. It's less ambiguous to take an example with 4 cards, one of which is a joker. We have a $\frac{3}{4}$ chance to have drawn the joker after 3 draws. But using this formula, we can relate the probability at each draw such that it remains consistent with a 3/4 total probability after the 3rd draw.

The pattern would follow 1/4, then 1/3, then 1/2, then 1 for consecutive probabilities to draw the joker (call these individually the "present probability"), which are related to the current total chance by:

(sum of previous total chance) + (present probability) * (1 - sum of previous total chance) = current total chance

Then we would have:

1st draw

$(0)+\frac{1}{4}(1-0) = 0.25$

2nd draw

$\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4}) = 0.50$

3rd draw

$[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4}) + \frac{1}{2}(1-[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4})]] = 0.50 + \frac{1}{2}(1-0.50) = 0.75$

4th draw

$[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4})]+\frac{1}{2}[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4})]+1[1-[[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4})]+\frac{1}{2}[\frac{1}{4}+\frac{1}{3}(1-\frac{1}{4})]] =$

$= 0.75 + 1(1-0.75) = 1$

Am I understanding this right? I feel like I just did something very simple in a really weird way...

Thanks for any feedback from someone who bothered to wade through all this