Ram draws a card randomly among cards numbered 1-23 and keeps it back. Then Sam draws a card among those left. What is the probability that Sam has drawn a card greater than Ram?
My Approach:
I could not solve this question.
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7 Answers
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I suppose that you have understood by now that you need not actually use brute force ?
Just by symmetry, Pr = 1/2.
If you can see that symmetry exists in a situation, use it !
answered Aug 25, 2015 at 3:51
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1$\begingroup$ It takes some experience to see that the situation is symmetric under interchanging the order of draws, though. $\endgroup$ Commented Aug 25, 2015 at 12:20
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$\begingroup$ Just by symmetry we can only conclude that P(greater)=P(lower). Which becomes 1/2 when we consider that there are no further possible outcomes, as you do implicitly. If the chosen card were returned and the 2nd drafter could also get it then there would be a third possible outcome; the situation would still be symmetrical but P(greater) would not be 1/2. $\endgroup$ Commented Aug 25, 2015 at 12:39
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$\begingroup$ @JoseAntonioDuraOlmos: The question states "Then Sam draws a card among those left" So there is no third outcome. $\endgroup$ Commented Aug 25, 2015 at 14:36
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1$\begingroup$ Indeed that is the case hence why I say you are implicitly considering that there are no further outcomes. But I thik that your already correct answer would be slightly improved if you explicitly mentioned that Pr=1/2 comes from symmetry and the fact that there are only two possible outcomes. You have my +1 either way. $\endgroup$ Commented Aug 25, 2015 at 14:42
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$\begingroup$ @JoseAntonioDuraOlmos: I get your point, would surely have explained it a bit more had not the previous (accepted) answer laid out a detailed explanation from which the symmetry would be discernible. I guess i'll let sleeping cemeteries lie. (Excuse the bad pun !) $\endgroup$ Commented Aug 25, 2015 at 15:09
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There is a one-to-one correspondence between the scenarios in which Sam draws the higher card and those in which Ram draws the higher card--this is given by exchanging the cards each has drawn. In this way we see that the number of ways that Sam draws a higher card is equal to the number of ways that Ram draws a higher card and constitutes exactly half of the possible outcomes (since in each scenario either Sam or Ram has a higher card). It follows that the probability you are looking for is $\frac{1}{2}$. Using this approach you don't need to 'calculate' any probabilities.
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If $R$ draws a $1$, then $S$ has a $\dfrac{22}{22}$ probability of winning.
If $R$ draws a $2$, then $S$ has a $\dfrac{21}{22}$ probability of winning.
$\vdots$
If $R$ draws a $23$, then $S$ has a $\dfrac{0}{22}$ probability of winning.
Summing all the probabilities and dividing by the number of events ($23$) gives the probability of $S$ winning as:
$$\dfrac{\dfrac{22\times23}{2}}{22\times23}=\dfrac{1}{2}$$
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$\begingroup$ you're using "greater" as "greater or equal" though :) $\endgroup$– AntCommented Aug 25, 2015 at 10:10
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$\begingroup$ @Ant Since all the cards are uniquely numbered and drawn without replacement, that doesn't matter. It's impossible for them to draw two equal cards. $\endgroup$– KyleCommented Aug 25, 2015 at 13:19
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Another approach: it is clear that drawing first conveys no advantage or disadvantage, as the setup just takes two cards at random and gives one to each player. Since there is no possibility of a draw, the only outcomes are victory for Sam and victory for Ram. By the symmetry we observed, these must have equal probability, so we can conclude that both probabilities are $\frac{1}{2}$.
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Brute Force If Ram draws card 1, what are the chances Sam draws higher? What if Ram draws 2? 3? 4? See the pattern?
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If Sam keeps his card then there are exactly $23\times 22$ ordered pairs, resulting from the draw. And exactly half of them have the first card greater than the second one, and another half of them have the first card less than the second one.
So the probability of Sam win is equal to the probability of Ram win – and both are $\frac 12$.
The result would be a bit less if you consider returning the card to the pool, because there is a third option: a draw, which is $1/23$ of all $23\times 23$ possible results. Then the probability that Sam's card is greater than that of Ram is $\frac 12 \times \frac {22}{23}$.
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There are 3 possibilities in such games:
- Sam wins.
- Ram wins.
- It's a draw.
Because each card only exists once, it cannot be a draw, the probability is 0. Next it's always a very good approach to try to quickly figure out who has a higher chance to win. You can later use this to do a sanity check of the calculated result.
Assuming the chance to draw each card is uniform (big assumption - you can influence humans by making some cards dirtier than others, or marking them with scratches, even if they don't know what the value of the marked card is) drawing second does not give a disadvantage. The game is the same as a neutral judge drawing 2 cards while not revealing them and Ram being allowed to pick which of the 2 card he wants. Therefore neither has a higher chance to win. Because neither has a higher chance to win, and we know the probability of the draw (0), we know that each has a probability of 1/2 to win. At that point we can skip the calculation.
The game becomes more interesting if the probability for each card is different, or even better, if the probability depends on both the card and the player - Sam likes cards with cleaner backs while Ram prefers the dirtier cards. In that case, brute force is the way that's most adaptable.
$p_{Ram} = \sum_{i=1}^n \left( p(i,Sam) \cdot \frac{\sum_{j=i+1}^n p(j,Ram)}{1-p(i,Ram)}\right)$
where $p(i,Sam)$ is the probability of Sam drawing the card with value i. In the simple case $p(x,y)$ is a constant $1/23$. This turns to:
$p_{Ram} = \sum_{i=1}^{23} \left( 1/23 \cdot \frac{\sum_{j=i+1}^{23} 1/23}{22/23}\right) = 1/23 \cdot \sum_{i=1}^{23} \frac{\sum_{j=i+1}^{23} 1}{22} = 1/23 \cdot \sum_{i=1}^{23} \frac{n-i}{22}$
$=1/23 \cdot \sum_{i=0}^{22} \frac{i}{22} = 1/23 \cdot \frac{22(22+1)/2}{22} = 1/2$
