You have a normal deck of 52 playing cards You draw cards one by one (Cards drawn are not returned to the deck).
A red card pays you a dollar. A black one fines you a dollar.
You can stop any time you want.
a. What is the optimal stopping rule in terms of maximizing expected payoff?
b. What is the expected payoff following this optimal rule?
c. What amount in dollars (integer values only ) are you willing to pay for one session (i.e. playing as long as you wish, not exceeding the deck), using your strategy?
Source will be disclosed after the solution is published.
The below table shows the results of 10,000 trials each of setting an amount by which one's being ahead would cause one to stop playing.
stop prob of expected
point wins win value
1 9654 0.9654 .9654
2 8569 0.8569 1.7138
3 7181 0.7181 2.1543
4 5489 0.5489 2.1956
5 3873 0.3873 1.9365
6 2472 0.2472 1.4832
7 1531 0.1531 1.0717
8 862 0.0862 .6896
9 433 0.0433 .3897
10 220 0.022 .22
So it looks as if, when considering a constant value to be ahead, regardless of the stage of the game, you should stop when you're ahead by 4, giving an expected gain of about $2.20.
It might (likely) be possible to improve the expected value by changing the stop point as the game progresses.
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
For v = 1 To 10
Randomize Timer
wins = 0
For tr = 1 To 10000
DoEvents
red = 26: black = 26
posn = 0
For draw = 1 To 52
r = Rnd(1)
If r < black / (black + red) Then
black = black - 1
posn = posn - 1
Else
red = red - 1
posn = posn + 1
If posn >= v Then wins = wins + 1: Exit For
End If
Next
If tr = 10000 Then
Text1.Text = Text1.Text & Str(v) & " " & wins & " " & wins / tr & Str(wins / tr * v) & crlf
End If
Next tr
Next v
Text1.Text = Text1.Text & crlf & " done"
End Sub
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Posted by Charlie
on 2015-09-24 15:05:47 |
computer exploration
