N people roll a die in turn, following a prearranged list. The first to get a 6 names the drink. The second to get a 6 drinks it. The third, pays! Then the next on the list rolls and so on till the closing time.
The above goes on for 120 minutes, averaging 5 rolls per minute (the time for naming, drinking, plus settling disputes and paying the bills was discounted).
How many times was it the same player that named the drink, consumed it but did not pay for it?
a. Provide your estimates for N=3, N=6 and N=12
b. Please explain the meaning of the results.
Rem: Verification of analytical results by simulation is welcome.
If it averaged 5 rolls per minute, there were 600 rolls.
That would then usually result in 100 6's, worthy of 33 1/3 drinks.
We know that someone will name each drink. We're not concerned with individuals here and it doesn't matter who in fact names the drink. But, whoever that is, what is the likelihood that the person who has just rolled a 6 will roll the next 6?
For that to happen, on the next N throws, there must be N1 successive non6's and then a 6, or N successive non6's followed by the same situation again.
Call the probability that this person will throw the next 6, p.
p = ((5/6)^(N1) / 6) + (5/6)^N * p
p(1(5/6)^N) = ((5/6)^(N1) / 6)
p = ((5/6)^(N1) / 6) / (1(5/6)^N)
Then, assuming the player has achieved the double 6's, he must fail to get the next 6. Call that probability q = 1  p. The probability asked for, as we're not concerned with which particular player this happens to, is just p*q.
But we're not asked for the probability, but rather the expected number of times this is to happen during the evening of drinking. That would be 100*p*q/3.
N expected number (rational and decimal approximation)
3 55000//8281 6.6417099384132351165
4 2275000//450241 5.0528494739483965253
5 83875000//21631801 3.8773932877803378461
6 2906875000//962922961 3.0188032872133371009
7 96971875000//40727679721 2.3809820658651314382
8 3153296875000//1661497798081 1.8978640107991723111
9 100702421875000//66008653934041 1.525594234592734918
10 3173660546875000//2570545871703601 1.2346251361667751073
11 99024513671875000//98576489633282761 1.0045449380502282657
12 3066102841796875000//3735103983097007521 0.8208882150730856022
5 open "inabar.txt" for output as #2
10 for N=3 to 12
20 P=((5//6)^(N1)//6)//(1(5//6)^N)
25 Q=1P
30 print #2,N,100*P*Q//3,100*P*Q/3
40 next
50 close #2
Simulation results:
3 6.5678
4 5.0283
5 3.8246
6 2.9555
7 2.3835
8 1.8623
9 1.4913
10 1.211
11 .9877
12 .7894
The simulation results are consistently lower than the calculated results, and this may be due to abandonment of partially completed sets at the 600th toss of the die, related to the clarification asked for by Steve in the previous post for this puzzle.
Simulation program:
DefDbl AZ
Dim crlf$, who(3)
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
ntrials = 10000
For n = 3 To 12
successct = 0
For tr = 1 To ntrials
ct6 = 0: tosser = 0
For toss = 1 To 600
DoEvents
tosser = tosser + 1: If tosser > n Then tosser = 1
rslt = 1 + Int(6 * Rnd(1))
If rslt = 6 Then
ct6 = ct6 + 1
who(ct6) = tosser
If ct6 = 3 Then
If who(1) = who(2) And who(2) <> who(3) Then successct = successct + 1
ct6 = 0
End If
End If
Next
Next tr
Text1.Text = Text1.Text & n & Str(successct / ntrials) & crlf
Next n
Text1.Text = Text1.Text & crlf & " done"
End Sub

Posted by Charlie
on 20160223 14:03:10 