probabilitystatistics
In a gambling game, a woman is paid dollar 3 if she
draws a jack or a queen and dollar 5 if she draws a king or
an ace from an ordinary deck of 52 playing cards. If
she draws any other card, she loses. How much should
she pay to play if the game is fair?
i am new to probability and i am stuck on this question , couldn't find an answer with a good explanation ,please help ……..thanks in advance
Normal approximation:
Consider a single card draw from the deck:
E(X)=7
Var(X)=14
With the sum of 5 cards (drawn without replacement):
$E(X)=7 \times 5$
$\text{Var}(X)=14\times 5\times \frac{52-5}{52-1}$
(the last term being the 'finite population correction', which applies as much to the variance of the sum as it does to the variance of the mean).
The approach to normality in this situation is reasonably rapid.
This suggests the sum on 5 cards might be be roughly approximated by a normal distribution with mean $35$ and variance $70\times\frac{47}{51}\approx 64.5098$. Using a continuity correction this gives an approximate probability of totalling at least 40 on 5 cards of:
$$1-\Phi(\frac{39.5-35}{\sqrt{70\times\frac{_{47}}{^{51}}}})\approx 0.288$$
Simulation (in R) of ten million five-card draws indicates the probability is
about $0.293$ (with s.e. $\approx 1.4\times 10^{-4}$):
res=replicate(10000000,sum(sample(rep(1:13,4),5,replace=FALSE)));mean(res>=40)
[1] 0.2927447
As a check on the earlier calculation of the variance, the standard deviation of of those simulated sums was 8.0316; the previous calculation gives 8.0318.
Edit: Here's a comparison of the empirical cdf of the simulated data with the above normal approximation; they're pretty close:
More extensive simulations are consistent with the other two answers based on complete enumeration:
Here is a hint:
Given that you have drawn one ace and one king, you will need either to draw the remaining three aces, or draw the remaining three kings, from the remaining $50$ cards.
Can you continue from this point?
(Spoiler answer)
There are two winning ways to draw the remaining three cards, and ${50 \choose 3}$ total ways to draw the remaining three cards, so the probability is $2/{50 \choose 3}$.
Best Answer
To put you on track:
Let's say that she pays $l$ dollars to play.
Three things can happen and all with a certain probability: she wins $3$, she wins $5$ or she looses. Denoting the corresponding probabilities with $p_3$, $p_5$ and $p_l$ there is an expectation of: $$3p_3+5p_5-lp_l$$ A fair game means that this expectation equals $0$. Now start finding $l$.