Deck of cards without replacement

[jtkerr]

I have a standard 52 card deck of cards. I am trying to draw the ace of spades. I draw cards one at a time. I do not replace each incorrect card. I am about to draw my 50th card having drawn 49 consecutive cards without ever drawing the ace of spades. There are three cards left. I know that I have a 33% chance of drawing the ace of spades now. My question:

What are the odds against me being in this situation - having drawn 49 consecutive incorrect cards without replacement?

 

[Jameson]

I have a standard 52 card deck of cards. I am trying to draw the ace of spades. I draw cards one at a time. I do not replace each incorrect card. I am about to draw my 50th card having drawn 49 consecutive cards without ever drawing the ace of spades. There are three cards left. I know that I have a 33% chance of drawing the ace of spades now. My question:

What are the odds against me being in this situation - having drawn 49 consecutive incorrect cards without replacement?

Hi jtkerr,

Welcome to MHB! :)

So you are searching for 1 specific card out of 52 cards without replacement. For the 1st round there are 51 cards that aren't your card out of 52. So the probability of not drawing the As on round 1 is 51/52. Now for round 2 there are 51 cards left. The probability of not drawing the As is now 50/51. You can repeat this process all the way until there are 3 cards left and if you multiply all of that out you should get your answer.

Intuitively this should be very small, as drawing 49 out of 52 cards should get your specific card most of the time. What do you get as an answer when you try this approach?

 

[HOI]

There are, to begin with, 52 cards, 51 of which are "incorrect" (not the ace of spades). The probability of drawing an "incorrect" card is \(\displaystyle \frac{51}{52}\). If you do draw an "incorrect" card, there are 51 cards left, 50 of which are "incorrect". The probability of drawing an "incorrect" card now is \(\displaystyle \frac{50}{51}\). On the 49th such draw there will be 52- 48= 4 cards left, 3 of which are "incorrect". The probability of drawing an "incorrect" card on the 49th draw is \(\displaystyle \frac{3}{4}\).

The probability of drawing 49 consecutive "incorrect" cards is
\(\displaystyle \frac{51}{52}\frac{50}{51}\frac{49}{50}\cdot\cdot\cdot \frac{4}{5}\frac{3}{4}\).

You can see that the numerator in each fraction will cancel the denominator in the next fraction leaving \(\displaystyle \frac{3}{52}\).

 

[Ilya123]

Hi guys, can someone explain, how can I find out the probability and payout of hand, and than how to calculate the expected return of each hand? Let's say I've got the pair of Aces. There are 16215 possible combinations. Is it possible to count it somehow without using any computer or software?
In case of royal flush, there are only 47 possible outcomes. 27 times we won’t hit anything. 8 times we will hit Jacks or Better, 3 times a straight, 8 times a flush and one time a Royal flush.
I get it till here... How do I get that ,,expected value of the hand?'' If I don't know if someone will raise the bets or not...
Can someone explain it to me?

poker strategy source I've used.