Maarchk
12-27-2004, 04:16 PM
I was doing some magic tricks for the young ones over Christmas, and i was trying to think of what the odds were of pulling multiple cards "at random".
For instance, if i pull say 5 cards out, and i am looking for the one ace of spades. My chances are (1/52) + (1/51) + (1/50) + (1/49) + (1/48). Or aproximately 10.0080 % chance.
Now lets say i shuffle and start again, but i want to draw out the ace of spades and ace of diamonds. And i get my 2 percentages... the one from above plus the same calculations again? or would it be the same equation but the deck size would be -1 to incorporate the 2 card draw? Making my second draw out of a 51 deck, the odds would be about a 10.213 % chance on drawing the 2nd card out if one card is missing.
So i would then multiply .100080 X .10213 to get the odds of pulling both cards within the 5 cards i asked i draw from? and thus i get a 1.02208 % chance? I think this is correct but i am not positive. I almost want to say you could just sqaure the original percentage but i do not know which one is more accurate. If you are a stats person and happen to know, i would love to find out. But its not too big of a deal. I'll give you a cookie or a dead xmas tree if you want for your prize. :)
For instance, if i pull say 5 cards out, and i am looking for the one ace of spades. My chances are (1/52) + (1/51) + (1/50) + (1/49) + (1/48). Or aproximately 10.0080 % chance.
Now lets say i shuffle and start again, but i want to draw out the ace of spades and ace of diamonds. And i get my 2 percentages... the one from above plus the same calculations again? or would it be the same equation but the deck size would be -1 to incorporate the 2 card draw? Making my second draw out of a 51 deck, the odds would be about a 10.213 % chance on drawing the 2nd card out if one card is missing.
So i would then multiply .100080 X .10213 to get the odds of pulling both cards within the 5 cards i asked i draw from? and thus i get a 1.02208 % chance? I think this is correct but i am not positive. I almost want to say you could just sqaure the original percentage but i do not know which one is more accurate. If you are a stats person and happen to know, i would love to find out. But its not too big of a deal. I'll give you a cookie or a dead xmas tree if you want for your prize. :)