 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  In a bar (Posted on 2016-02-23) 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.

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution, with caveat | Comment 2 of 6 | 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 N-1 successive non-6's and then a 6, or N successive non-6's followed by the same situation again.

Call the probability that this person will throw the next 6, p.

p = ((5/6)^(N-1) / 6) + (5/6)^N * p
p(1-(5/6)^N) = ((5/6)^(N-1) / 6)
p = ((5/6)^(N-1) / 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)^(N-1)//6)//(1-(5//6)^N)
25    Q=1-P
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 A-Z
Dim crlf\$, who(3)

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 2016-02-23 14:03:10 Please log in:

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