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# if you pick 26 cards from 52 cards... watch

1. with replacement, what is the probability that it has exactly 13 reds and 13 black cards?
I got 52 C 26 * (1/2)^52,
is this even correct?
2. (Original post by Iconic_panda)
with replacement, what is the probability that it has exactly 13 reds and 13 black cards?
I got 52 C 26 * (1/2)^52,
is this even correct?
It looks to me as though what you've modeled there is, that from 52 picks with replacement after each pick, you will end up with exactly 26 reds and 26 blacks.
3. (Original post by Iconic_panda)
with replacement, what is the probability that it has exactly 13 reds and 13 black cards?
I got 52 C 26 * (1/2)^52,
is this even correct?
If you're doing with replacement, the cards wouldn't actually exist in your hand, so its a theoretical exercise like rolling a two sided dice with sides R&B?
If that's right the 52 is not important as you could consider a deck with just two cards (one red and one black) with replacement and you'd have an equivalent problem? The same chance of drawing a red or black and you just record 26 picks.
So there are
2^26
ordered hands in total (RBRR....). Can you find the number with exactly 13 of one type (and hence 13 of the other type)?
4. to be honest the question didnt specify whether or not it is being replaced or not. but I think i understand your approach if it is replaced thanks!
(Original post by mqb2766)
If you're doing with replacement, the cards wouldn't actually exist in your hand, so its a theoretical exercise like rolling a two sided dice with sides R&B?
If that's right the 52 is not important as you could consider a deck with just two cards (one red and one black) with replacement and you'd have an equivalent problem? The same chance of drawing a red or black and you just record 26 picks.
So there are
2^26
ordered hands in total (RBRR....). Can you find the number with exactly 13 of one type (and hence 13 of the other type)?
5. (Original post by Iconic_panda)
to be honest the question didnt specify whether or not it is being replaced or not. but I think i understand your approach if it is replaced thanks!
I'd be surprised if the "with replacement" was correct.

Without replacement
26C13 * 26C13 / 52C26
is probably right
https://math.stackexchange.com/quest...rom-deck-of-52
6. ah right that makes sense too, thanks!
(Original post by mqb2766)
I'd be surprised if the "with replacement" was correct.

Without replacement
26C13 * [/font][/size]is probably right
https://math.stackexchange.com/quest...rom-deck-of-52

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