Define a game of blindfolded cards as a game in which two players $A$ and $B$ take turns to choose a number from a deck of cards $\{1,2,3,\cdots , n\}$ blindfolded such that if $A$ gets a number $\leq a$ wins and $B$ wins if the number he picks is $\leq b$. A win ends the game.

  • If $1\leq a,b\leq n$ is given and $A$ begins the game, then who has a greater chance of winning. If $a=b$, show that $A$ has an advantage.

  • How does $n$ relate to the advantage of the player? For which $a,b$ will there be an advantage for either of $A$ or $B$ or there will be a $50-50$ chance of winning.

I don't really know who will have a greater chance of winning. But I believe that the fact that $A$ begins the game is counts. So if $a<b$, I think $A$ has a lesser chance of winning and vice-versa. Again, I don't get why if $a=b$, then $A$ might have an advantage in the game. After all, both will have a probability of $\frac{a}{n}$ and then, both will have a similar probability I guess. Anyway since $A$ begins the game, perhaps he can win it in the first chance, so that might be called as an advantage, though I believe there's a stronger version to it.

  • $\begingroup$ Not sure the rules are clear. If $A$ draws a winning number, does that end the game? $\endgroup$ – lulu Jan 16 '18 at 13:01
  • $\begingroup$ @lulu If $A$ draws a number $\leq a$, yes, the game will end. But yes, if $B$ draws a number $\leq a$ doesn't imply a win on $A's$ side. $\endgroup$ – Mathejunior Jan 16 '18 at 13:03
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    $\begingroup$ "though I believe there's a stronger version to it." Nope, that's it. Consider if they had drawn cards simultaneously, and they both get a low enough card to win. This would morally be a draw (and if it had been a draw, the game would be fair). However, according to the rules $A$ is the one who wins. And he has this advantage every turn until the game ends. $\endgroup$ – Arthur Jan 16 '18 at 13:06
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    $\begingroup$ @Mathbg I'm saying that if we change the game so that they draw simultaneously, and in if they both get a low enough card, $A$ alone wins, then it's mathematically the same game (I admit it was a bit unclear). It's a lot easier to see $A$'s advantage and that this advantage comes solely from him going first. $\endgroup$ – Arthur Jan 16 '18 at 13:09
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    $\begingroup$ It is clear why $a=b$ implies the first player has an advantage, but if you want to compute that advantage you should clarify whether drawing is with or without replacement. $\endgroup$ – hardmath Jan 16 '18 at 13:11

Note: in this computation I will assume that the choices are made with replacement, though this is not currently specified in the original problem.

To illustrate the computation, here is how to calculate $A's$ chance when $a=b$.

First, note that this is really just a bet on a biased coin. Since each player wins with probability $p=\frac {a}n$ they are effectively betting on a coin that comes up $H$ with probability $p$.

Let $P_A$ be the probability that $A$ wins (So $B$ wins with probability $1-P_A$). If $A$ wins the first toss the game is over, if $A$ fails on the first try, $A$ effectively moves into the second position. Thus $$P_A=p\times 1 + (1-p)\times(1-P_A)\implies \boxed {P_A=\frac 1{2-p}}$$

Note that this is always $≥\frac 12$ and equal only in the degenerate case $p=0$, so $A$ always has an advantage.

  • $\begingroup$ Note that this derivation assumes that that the cards are drawn without replacement. (Whether or not the cards are replaced between draws was not specified in the question.) $\endgroup$ – Michael Seifert Jan 16 '18 at 13:13
  • $\begingroup$ @MichaelSeifert Agreed. I'll edit to reflect that. $\endgroup$ – lulu Jan 16 '18 at 13:13
  • $\begingroup$ @MichaelSeifert Note, to be clear: I am assuming replacement. That is, I assume each choice is identical. I note that the OP has now commented that this is what was intended. $\endgroup$ – lulu Jan 16 '18 at 13:20
  • $\begingroup$ Yes, you're right — I meant "with replacement". That'll teach me to post before the coffee kicks in. $\endgroup$ – Michael Seifert Jan 16 '18 at 13:21
  • $\begingroup$ @MichaelSeifert Yeah, well I read your comment wrong for precisely the same reason, so we're even. $\endgroup$ – lulu Jan 16 '18 at 13:24

If cards are drawn with replacement, then A's chance of drawing a winning card in a particular try is $a/n = \alpha$, say, and B's chance is $b/n = \beta$. The probability that A wins after exactly $n$ failures by each player is $(1- \alpha)^n(1 - \beta)^n\alpha$. Hence A's chance overall is $$\sum_{n=0}^{\infty}(1 - \alpha)^n(1 - \beta)^n\alpha = \alpha/(\alpha + \beta - \alpha\beta).$$ Similarly B's chance overall is $(\beta - \alpha\beta)/(\alpha + \beta - \alpha\beta)$. Hence A has an advantage iff $\alpha > \beta - \alpha\beta$, which is equivalent to $n(b - a) < ab$. If $a \ge b$ then A has an advantage for all $n$. If $a < b$ then A has an advantage iff $n < ab/(b - a)$. (If the cards are not replaced it's probably harder.)

  • $\begingroup$ What do you mean by replacement? $\endgroup$ – Mathejunior Jan 16 '18 at 13:43
  • $\begingroup$ It means that when a card is drawn, and isn't a winning card, it is put back into the pack. $\endgroup$ – Michael Behrend Jan 16 '18 at 15:45
  • $\begingroup$ Oh, fine. Well, cards once drawn are put back definitely as the game ends only when the winning card is drawn. You can consider the same thing as an n-sided die for a clearer image of what I'm actually referring to $\endgroup$ – Mathejunior Jan 16 '18 at 15:48

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