Timeline for Guessing the next card colour in a deck [closed]
Current License: CC BY-SA 3.0
7 events
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| Feb 13, 2012 at 10:54 | comment | added | Johan Wästlund | Perhaps the question was closed too quickly? Based on Douglas Zare's comment there is a clear answer: $E(r,b)$ can be computed as a single sum / recursion involving binomial coefficients (rather than over two variables). In particular it seems that $E(n+1,n+1) = E(n,n) + 1 + (1/2)\cdot (2\cdot 4 \cdots (2n))/(3\cdot 5 \cdots (2n+1))$, and that asymptotically, $E(n,n) = n + \sqrt{\pi n}/2 + o(\sqrt{n})$. | |
| Feb 13, 2012 at 7:04 | comment | added | Douglas Zare | There is a big simplification possible. If there are $k>0$ more red cards than black cards, you will make $k$ more correct than incorrect guesses by the time the deck is even again. So, the expected number of correct minus incorrect guesses is a sum of $1$ for each time the deck changes from even to uneven. That is the expected number of times the deck will be even before the end, or the sum of the probabilities that the deck will be even at time $0, 2, 4, ... 2n-2$. | |
| Feb 13, 2012 at 1:45 | comment | added | A Chuh | Sure, so there is no explicit formula for a situation like this? | |
| Feb 13, 2012 at 1:31 | history | closed |
Bill Johnson Chris Godsil S. Carnahan♦ |
too localized | |
| Feb 13, 2012 at 1:31 | comment | added | S. Carnahan♦ | You already set up a recursion, and the initial cases are not too hard to work out. Please feed this information to a computer algebra system. | |
| Feb 13, 2012 at 1:19 | comment | added | Joel David Hamkins | You can solve $E(r,b)$ by recursion. The base of the induction is when there is only one color left, and you are sure to be right. Then, your equations provide the value at $E(r,b)$ from smaller values. So the complete table of values can be exactly calculated. | |
| Feb 13, 2012 at 0:33 | history | asked | A Chuh | CC BY-SA 3.0 |