There’s a extremely simple solitaire game where you draw the first card and if it’s an ace, you lose. Then the second card, and if it’s a two, you lose. Keep going, naming cards in order (ace, two, three…ten, jack, queen, king) and if the card you name is the card you draw, you lose.
What’s the probability of winning this game?
One blind spot I’ve always had is when you have to calculate the probability of something happening vs. calculating 1-P(something doesn’t happen).
Here’s how I would do the calculation here:
P(losing = drawing the card you name) = 1/13. P(not losing) = 12/13. P(not losing 52 times in a row) = (12/13)^52 = ~1.557%, which is your chance of winning.
Is that right? Do I have to take into account all the cards that didn’t cause me to lose along the way?
Imagine if you played with a 2 card deck, eg A, 2. The first card drawn would entirely determine your chance of winning, because either both would match or neither. I think in general the fact that cards are discarded and not available to be drawn again means that you can’t treat the probability of losing on each turn as independent.
While we don’t know in advance what the first card is, we do know it’s not an Ace in the winning scenario, and it might have been a two. That means your odds are slightly better than 1/13 for the second card.
How this counts in the final calculation of odds I haven’t figured out yet.
This calculation is appropriate if you are replacing and re-shuffling the cards drawn, i.e. you are doing 52 separate draws from a full deck each time. Without looking at the link someone else posted as an answer, my gut tells me the actual odds of winning are going to be higher.
Since the card draws are not independent, numerical methods work better. After a couple millions trials, the crude simulation I whipped up is returning a winning percentage of about 1.62%
Aside on a similar question: Suppose you instead have a deck numbered from 1 to 52. You might have one “losing card”, that is, a card that’s in the position it’s “supposed to be in”. You might have more. You might not have any. But on average, the number of cards in their original position will be exactly 1. And this is, in fact, true no matter how many cards you have in your deck.
Just created a monte carlo simulation. After 25,000 iterations, I’m getting 1.562% which is much closer to the OP’s 1.557% than it is to the 1.623% from the link. I’m thinking the math geeks got some slight logic problem.
And, that’s the problem with Monte Carlo. Ran a few more iterations; now up to 100k, and I’m at 1.616%, which is now much closer to the math geeks than the OP.