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If I replace the wrong card back into the pack, the odds of picking Ace Hearts remains one in 52, right?
what are the odds of me picking Ace of Hearts on two picks combined?
replacing?
and not replacing wrong cards?
I thought the answer would be 1 in 26 replacing and about 1 in 25.7 (in other words, better chance) keeping wrong pick out of deck but I'm being told on another forum that I'm speaking out of the wrong part of my anatomy.
are there any maths bods on here? :-)
can someone
You mean twice in a row?
1/52 X 1/52 if you put the card back in and 1/52 X 1/51 if you leave the card out after the first try (provided it's not the Ace of Hearts)
Not a math guy so take it with a grain of salt, and I assume that there's one Ace of Cards for pack (I don't play cards ;) )
if you put the card back, second pick is 1 in 52 again. that's 2 in 52 = 1 in 26 = a 3.84% chance of getting that ace within two draws.
if you don't put the wrong card back, your second pick is 1 in 51. with that rule, you have a 3.88% chance of getting the ace in the first or second draw.
Continue that to its conclusion: if you draw a third card, you have a 5.88% chance of having the ace.
fourth card: 7.92%
fifth card: 10.0%
sixth card: 12.1%
by the time you get to 33 draws, your odds are over 100%. If you don't have the ace by then, you're entering into "bad luck" territory.
at the fifty-first draw, your odds of getting the ace are 1 in 2. If you still pick the wrong one, you are losing against 451.88% odds.
prove it with javascript:
<script>
k=0;
for (i=0;i<52;i++){
j = 1/(52-i);
k=k+j;
document.write("<BR>"+i+" / "+j+" / "+k);
}
</script>
Putting the cards back after every draw, your odds are numberofdraws/52. Statistically you will get that ace once for every 52 times you draw.
1/52 + 1/52 + 1/52 ... = 1
at the fifty-first draw, your odds of getting the ace are 1 in 2. If you still pick the wrong one, you are losing against 451.88% odds.
clarification:
your odds of picking the ace on that last round are 1 in 2. By then you have already picked the wrong one 50 times; the odds against doing that are 1 in 451880.