Is Playing a Card Game with Ascending Order Wins a Good Bet?

[boggled]

Hello so here is the question:

You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?

I am thinking a few different ways:

A the easiest: its symetric so equal chances higher or lower so the chance to get 5 cards each one higher than the next is 1/1 * 1/2 * 1/2 * 1/2 = 1/8

or b. the average pick out of the cards is 49 so the next has to be 50 or higher the average of 50-97 is the second choice(that has a probability of 47/96).. the third has to be higher than that average second pick which is 73 average of 73-95 is 85 and so on

so the prob is 1*47/96*26/95*10/94

Thank you

 

[Ray Vickson]

Hello so here is the question:

You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?

I am thinking a few different ways:

A the easiest: its symetric so equal chances higher or lower so the chance to get 5 cards each one higher than the next is 1/1 * 1/2 * 1/2 * 1/2 = 1/8

or b. the average pick out of the cards is 49 so the next has to be 50 or higher the average of 50-97 is the second choice(that has a probability of 47/96).. the third has to be higher than that average second pick which is 73 average of 73-95 is 85 and so on

so the prob is 1*47/96*26/95*10/94

Thank you

Your expression above is 13/912 ≈ 0.01425, but I get something very different: I get p{win} = 4465/18624 ≈ 0.2397.

The point is that when the first number is small, there are lots of ways to choose the second number (to get a win), but when the first number is large the number of choices for the second number is much more limited.

RGV

 

[awkward]

Are the 97 cards all distinct? If so, the probability of drawing 4 in ascending order is simply 1/4!, approximately 0.04167, since all 4! orderings are equally likely.

 

[Ray Vickson]

Are the 97 cards all distinct? If so, the probability of drawing 4 in ascending order is simply 1/4!, approximately 0.04167, since all 4! orderings are equally likely.

No fair, you did it the easy way! My previous post was nonsense, as I had made the stupid mistake of forgetting some factors, etc. When I correct the results (but using my lengthy method) I end up with P{win} = 1/24, as you say.

RGV

 

[awkward]

I didn't necessarily do it the easy way the first time, heh heh.

 

[Ray Vickson]

I didn't necessarily do it the easy way the first time, heh heh.

There must be something a bit deeper going on here. I used
P_{\text{win}}= \sum_{i=1}^{94} \frac{1}{97} \sum_{j=i+1}^{95} \frac{1}{96}<br /> \sum_{k=j+1}^{96} \frac{1}{95} \sum_{l=k+1}^{97}\frac{1}{94} = \frac{1}{24} = \frac{1}{4!}.
Of course, 1/4! is the volume of the region
R = \{(x_1,x_2,x_3,x_4) : 0 \leq x_1 \leq x_2 \leq x_3 \leq x_4 \leq 1 \}.
The volume of R can be computed by nested integrations; somehow, we get the same value by replacing the integrations by summations, and with slightly changing denominators (1/97, 1/96, 1/95, ... ). Surely this cannot be coincidental.

RGV