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52 poker cards (half red and half black). Every time you draw a card, if you draw a red card, then you win one dollar. If you draw black card, then you lose one dollar.

Say you start with $n$ dollars, if you can stop/quit this game anytime you want, how much are you willing to pay this game?

What would be your optimal play strategy?


Attempt:

I think this is very similar to the gambler's ruin problem; We can let each state denote the amount of money that the gambler has. However, the transistion probability would be dependent/varied from state to state.

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  • $\begingroup$ Do you put back the card into the 52 card piles after u draw it? $\endgroup$ Commented Mar 26, 2017 at 5:43
  • $\begingroup$ if I can put it back, then it will become gambler's ruin problem. Let us assume that's the case. Even with this assumption, I still have trouble to figure out "how much you are willing to pay for this game", if we can stop anytime we want. $\endgroup$
    – wrek
    Commented Mar 26, 2017 at 5:45
  • $\begingroup$ what does this have to do with markov chains/processes ? $\endgroup$
    – mercio
    Commented Mar 26, 2017 at 8:38
  • $\begingroup$ stopping when $R < B - \sqrt(B)$ seems to be pretty close to the optimal strategy. $\endgroup$
    – mercio
    Commented Mar 26, 2017 at 9:02
  • $\begingroup$ "Let us assume that's the case" ?? The question is interesting when one does not put back the cards. $\endgroup$
    – Did
    Commented Mar 27, 2017 at 7:17

1 Answer 1

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Hint: First consider a deck containing only 6 cards, 3 +ve & 3 -ve (where + is for red and -ve is for black). Then generalize it for 52 cards.

I think you should take a lot at this page 52 Cards Win a Dollar.

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    $\begingroup$ Maybe this should be a comment. $\endgroup$
    – mlc
    Commented Mar 26, 2017 at 7:27

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