Problem:
Someone hands you a deck of cards which you thoroughly shuffle. Next, you start to deal them, face-up, counting the cards as you go. “One, Two, Three …”
The aim is to predict what the count will be when you encounter the second black Ace in the deck.
If you had to select one position before you started to deal, what number would you select that maximizes your chance of guessing the location of the second black Ace?
Someone hands you a deck of cards which you thoroughly shuffle. Next, you start to deal them, face-up, counting the cards as you go. “One, Two, Three …”
The aim is to predict what the count will be when you encounter the second black Ace in the deck.
If you had to select one position before you started to deal, what number would you select that maximizes your chance of guessing the location of the second black Ace?
last
ReplyDeleten^th position probability = 2*(n-1)/((52)(51)) , maximizes at n=52
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ReplyDeleteLast,
ReplyDeleteThe problem can be thought of picking the first Ace from the desk upside down.
Now:
Prob Of 1st card being first black ace I pick = 2/52
Prob Of 2nd card being first black ace I pick = (50/52)*(2/51)
Prob Of 3rd card being first black ace I pick = (50/52)*(49/51)*(2/50)
As you see.. this sequence is decreasing, with max prob at starting, so prob is max at first card (for upside down deck)
Hence our prob is max for 52nd card !
Let the 2nd black ace be picked in the n-th turn, n>=2
ReplyDeleteA: The first black ace is picked in n-1 turns
B: The second black ace is picked in the n-th turn
P(A) = 2 * 50C(n-2) / 52C(n-1) = 2*(53-n)*(n-1)/(52*51)
P(B) = P(B|A) * P(A) = 1/(52-(n-1)) * P(A) = 2(n-1)/(52*51)
P(B) is max for n = 52