Dice calculation of p(7), two different values?

  • #1
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Summary:

Six sided fair dice (two) labeled standard 1-6; two methods of calculating p(7) get different p?

Main Question or Discussion Point

Scenario 1]
The dice are rolled out of your sight and you are then asked the p(7), before you are allowed to look at the results.

Calculate p(7) by looking at table of 36 possibilities and counting the number of sum cells totals that equal 7, there are six of them, so

p(7) = 6/36 = 1/6

Scenario 2]
Same as in 1 except before you're requested for your calculation of p(7) you are informed that one of the faces shows a six.

You calculate p(7) by looking at the table of 36 possibilities and counting the number of possibilities that include six (the cells comprised of row 6 and column 6) so you count 11 of these cells whose row or column value is 6, and notice that 2 of them indicate 7 as their cell sums so

p(7) = 2/11 > 1/6

A closer look reveals that if you had been informed that one of the faces showed a 5, 4, 3, 2, or 1, you would find the same number of cells containing that row-column number (eleven of them), and the same number of cells totaling 7, (two of them).
So it would not have mattered if you were informed of one of the faces or not; you could have just imagined that it must be some number and used this method in the first scenario... what is wrong with this method? If the total cells selected came to 12 instead of 11, the p(7) would be right... Some reason the row-column intersection cell needs to be counted twice?
 

Answers and Replies

  • #2
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The first approach calculates ##P(a+b=7)## and the second calculates ##P(a+b=7|a=6\lor b=6)##. They are not the same quantity
 
  • #3
anuttarasammyak
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We hear that one shows 6, it does not matter. We think of 6 possible shows of the other dice including 1 winning show so the probability remains 1/6. Such information of one dice show does not effect the probability for this case of a+b=7. When you hear one dice show stop using 6*6 matrix and look at 6*1 column reduced from matrix by one dice opening.
 
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  • #4
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Yes... the second may be more generally P(a+b=7 |a V b where row a = col b)

If the second were P(a+b=7 |all a + all b where row a = col b) then knowing or assuming that a face is n, does the possibility (n,n) needed to be counted independently for row n and col n so that it is counted twice? That would make a+b=12, so 2/12 the right quantity for p(7), but I don't see how selecting possibilities could allow a double count of (n,n). But if this is done, and the value of p(7) is correct, is this really p or something else (artifact, nonsense)?
 
  • #5
anuttarasammyak
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If you would like to consider the 11 cells you quote, how about multiplying 1/2 to the 10 cells except (6,6) for probability calculation ? Cell (2,6) as well as (6,2) contains the cases that 2 is shown and 6 is shown half and half though we are focusing the case 6 is shown. We are sure (6,6) is surely in our focus.
 
  • #6
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I'm not following your meaning of multiplying by 1/2. The two cells of interest are the two that show 7 where the row and column labels match the known or assumed face. For a face up of 6, those two would be (1,6) and (6,1)... for all face up n=1-6 the two cells that show 7 will be on the same diagonal in the chart.
 
  • #7
anuttarasammyak
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Which of (n,6) or (6,n) cell you assign as the case of opened dice shows six ?
Which of (n,6) or (6,n) cell you assign as the case of opened dice shows n ?
(shown, not shown) or (not shown, shown) or any others ?
I am not sure about your setting so count these cells and halved the possibility ( 1/2 of open shows 6 not n ) to meet the both cases.
 
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  • #8
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bahamagreen said:
. . . before you're requested for your calculation of p(7) you are informed that one of the faces shows a six.

You calculate p(7) by looking at the table of 36 possibilities and counting the number of possibilities that include six (the cells comprised of row 6 and column 6) so you count 11 of these cells whose row or column value is 6, and notice that 2 of them indicate 7 as their cell sums so p(7) = 2/11 > 1/6
It appears to me that you are incorrectly counting the possibilities for the second die as being only 5 instead of correctly recognizing that the actual number of possibilities for the second die is 6, no matter whether you've seen or been told what the first die shows or not.

This I think is based on the incorrect notion that cell (6,6} has already been counted once for either the row or the column, so that counting it for both the row and the column would be counting the same cell twice, which would seem to be incorrect; however, the dice roll independently of each other, so which of its 6 possibilities one die shows has no bearing on the fact of there always being 6 possibilities for the other die, or on which of those 6 it shows.

Instead of counting the 6 cells of a column or the 6 cells of a row and then counting only the other 5 cells in the row or column at the end of the column or row just counted, you must remember that a pair of dice is an ordered pair, so that e.g. (4, 3) is a different cell from (3, 4), but both pairs sum to 7, it takes 2 cells in the table to show both of them.

The 6 cases in which the 2 numbers match are the 6 different even-number sums, and each such matching pair is represented in a single cell, along the diagonal from upper left to lower right (or lower right to upper left). Every row and every column has exactly 1 such matching pair cell. You are throwing 2 dice, each of which has 6 faces, so you must count the entire column for one die, and the entire row for the other.

When you count all 36 cells, you count each cell exactly once, but then when you count the 6 cells that contain 7s, the 6/36=1/6 is clearly discernible: (1,6), (6,1) (3, 4), (4,3) (2, 5), and (5, 2) all sum to 7.

Regardless of which number is disclosed for the first die, the chance that the other number rolled will sum with it to produce 7 is 1/6.

Representing the 6 faces of one die as 6 rows, and the 6 faces of the other as 6 columns, yields a table with 6x6=36 cells, the lower left to upper right (or upper right to lower left) diagonal of which has a 7 in each cell:

\begin{array}{|c|c|c|c|c|c|}
\hline & \mathbf 1 & \mathbf 2 & \mathbf 3 & \mathbf 4 & \mathbf 5 & \mathbf 6 \\
\hline \mathbf 1 & 2 & 3 & 4 & 5 & 6 & \mathbf {\color{red} 7} \\
\hline \mathbf 2 & 3 & 4 & 5 & 6 & \mathbf {\color{red} 7} & 8 \\
\hline \mathbf 3 & 4 & 5 & 6 & \mathbf {\color{red} 7} & 8 & 9\\
\hline \mathbf 4 & 5 & 6 & \mathbf {\color{red} 7} & 8 & 9 & A \\
\hline \mathbf 5 & 6 & \mathbf {\color{red} 7} & 8 & 9 & A & B \\
\hline \mathbf 6 & \mathbf {\color{red}7} & 8 & 9 & A & B & C \\
\hline
\end{array}(I used hexadecimal notation to keep single-digit cell contents.)

It can be readily seen that whichever row or column is selected, there are 6 values in that row or column, exactly 1 of which is a 7, so the chance for any row or column that the cell in it that is indexed by the value on the face of the other die contains a 7 is 1/6.
 
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  • #9
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Yes... the second may be more generally P(a+b=7 |a V b where row a = col b)

If the second were P(a+b=7 |all a + all b where row a = col b) then knowing or assuming that a face is n, does the possibility (n,n) needed to be counted independently for row n and col n so that it is counted twice? That would make a+b=12, so 2/12 the right quantity for p(7), but I don't see how selecting possibilities could allow a double count of (n,n). But if this is done, and the value of p(7) is correct, is this really p or something else (artifact, nonsense)?
Sorry, your description here is confusing. I should have explained my notation better. In my notation we have two dice. ##a## is the random variable giving the value of the first dice and ##b## is the second. So one possible outcome of a trial is ##a=5 \land b=3## which is a distinct outcome from ##a=3 \land b=5##.

The notation ##P(7)## is confusing because ##P(.)## operates on random variables and 7 is not a random variable. So it is better to be clear that you mean ##P(a+b=7)##.

In this notation ##a+b=7## is the event that the total on the two dice is 7. ##a=6## is the event that the first dice is 6. ##b=6## is the event that the second dice is 6. ##a=6 \lor b=6## is the event that one (or both) of the dice is 6. ##a=b## is the event that the dice are the same.

I am not exactly sure what you are trying to do in the post above. It may be some notational confusion due to my not being explicit above. In particular, I don’t know what you mean by “all a + all b where row a = col b“. The “row” is tripping me up. Do you mean something different from “row a=col b” and just ##a=b##
 
  • #10
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Knowing a 6 was showing, I meant counting all the row-column intersections for which 6 occurs within the face sums' ordered pairs, e.g., (1,6) (2,6), (3,6)...(6,3), (6,2), (6.1). Counting each intersection once I get eleven. The two die are independent so I should have counted (6,6) twice, or just counted all the 6s in the table.
 
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  • #11
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The two die are independent so I should have counted (6,6) twice, or just counted all the 6s in the table.
I am pretty sure that is not correct. Because the two dice are independent every cell has equal probability. So I cannot think why you would want to double the probability of one cell.

Whenever you are in doubt just do a Monte Carlo simulation. I wrote a Mathematica script that rolled a million pairs of dice.

Of the million rolls 166359 summed to 7 so ##P(a+b=7)=\frac{166359}{1000000}\approx \frac{1}{6}##.

Of the million rolls 305716 rolls contained a 6 so ##P(a=6\lor b=6)=\frac{305716}{1000000}\approx \frac{11}{36}##.

Finally, of the rolls that contained a 6 there were 55440 that summed to 7 so ##P(a+b=7|a=6 \lor b=6)=\frac{55440}{305761}\approx \frac{2}{11}##

So indeed the calculations in the OP are correct. Those two methods are calculating fundamentally different probabilities. They are not two ways of calculating the same probability.
 
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  • #12
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I am pretty sure that is not correct. Because the two dice are independent every cell has equal probability. So I cannot think why you would want to double the probability of one cell.

Whenever you are in doubt just do a Monte Carlo simulation. I wrote a Mathematica script that rolled a million pairs of dice. Of those 166359 summed to 7 so ##P(a+b=7)=\frac{166359}{1000000}\approx \frac{1}{6}##. Of the million rolls 305716 rolls contained a 6 so ##P(a=6\lor b=6)=\frac{305716}{1000000}\approx \frac{11}{36}##. Finally, of the rolls that contained a 6 there were 55440 that summed to 7 so ##P(a+b=7|a=6 \lor b=6)=\frac{55440}{305761}\approx \frac{2}{11}##

So indeed the calculations in the OP are correct. Those two methods are calculating fundamentally different probabilities. They are not two ways of calculating the same probability.
The second of those 2 does not reasonably simulate the roll of 2 dice; only the first one does ##-## the second counts a 6-containing cell only once even when that cell contains 2 6s ##-## of the 36 cells in the table, 6 contain a 7, so the probability is 6/36=1/6. You count a row of 6 cells for 1 die, and a column of 6 cells for the other die. The cell at which the row and the column intersect is counted once qua row cell and once qua column cell, because there are all 6 possibilities for each row, and all 6 possibilities for each column, whence the 6x6=36 cells.
 
  • #13
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The second of those 2 does not reasonably simulate the roll of 2 dice; only the first one does
The second of what 2?

I used exactly the same million dice rolls for all of my counts. There is only one set of a million rolls.
 
  • #14
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The second of what 2?
The second of the 2 Monte Carlo simulations, which counts only 11 cells instead of counting all 6 values in the row for one die and all 6 values in the column for the other die.
 
  • #15
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The second of the 2 Monte Carlo simulations, which counts only 11 cells instead of counting all 6 values in the row for one die and all 6 values in the column for the other die.
There was only one Monte Carlo simulation. All of the counts came from the same simulation of a million rolls. The point of the (one) simulation is to show that being told that one of the numbers is a 6 does in fact change the probability that the sum is 7.
 
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  • #16
anuttarasammyak
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I would like to confirm what Dale taught.

You did many times of two dice throws. I gathered and investigated the throw results and prepared three kind of data set

#1 I selected all the data showing at least one 6.
#2 I selected all the data of dice A show 6.
#3 I selected all the data of reading dice A or B with random choice and show 6.

You pick up one data blindly among these sets.
How much is the probability of picking up data that one dice show 1 ?

Ans. #1 2/11 #2,3 1/6

Summary:: Six sided fair dice (two) labeled standard 1-6; two methods of calculating p(7) get different p?

So it would not have mattered if you were informed of one of the faces or not; you could have just imagined that it must be some number and used this method in the first scenario... what is wrong with this method? If the total cells selected came to 12 instead of 11, the p(7) would be right... Some reason the row-column intersection cell needs to be counted twice?
How "informed one of the faces" differs in these three cases.
 
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  • #17
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There was only one Monte Carlo simulation. All of the counts came from the same simulation of a million rolls. The point of the (one) simulation is to show that being told that one of the numbers is a 6 does in fact change the probability that the sum is 7.
I acknowledge that you did only 1 simulation of 1 million rolls; however, you did 2 countings, and the second of them does not accurately reflect the rolling of 2 dice.

The values shown on the faces are independent of each other, so whether you observe one of them before observing the other, the probability of 7 remains 1/6.

Being told that one of the dice shows a 6, does not alter the 1/6 probability that the other shows a 1, so the probability of a 7 is still 1/6.

Regardless of what the first die shows, the probability that the value on the second will, with the value shown on the first, sum to 7, is always 1/6.

Being told one number before the other has no effect or bearing on the 1/6 probability of each number for each of the 2 dice.
 
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  • #18
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Regardless of what the first die shows, the probability that the value on the second will, with the value shown on the first, sum to 7, is always 1/6.

Being told one number before the other has no effect or bearing on the 1/6 probability of each number for each of the 2 dice.
Sure, but that is not what was described in the OP. What you are saying here is ##P(a+b=7|a=6)=\frac{1}{6}##

This is a correct statement but the condition from the OP is ##a= 6 \lor b=6## not ##a=6##. My simulation and my counts correctly reflect that
 
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  • #19
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Sure, but that is not what was described in the OP. What you are saying here is ##P(a+b=7|a=6)=\frac{1}{6}##

This is a correct statement but the condition from the OP is ##a= 6 \lor b=6## not ##a=6##
If you are told that at least one of the 2 dice shows a 6, a 5, a 4, a 3, a 2, or a 1, the probability that the other die shows a number that with the first sums to 7 remains 1/6. If you are told that exactly 1 of the dice shows a certain 1 of the 6 possibilities, you can eliminate that possibility for the other die, so that the probability for the other die of a number that with the disclosed number sums to 7 would be 1/5. That's not what was postulated in the OP ##-## it says that you are told that 1 of the dice shows a 6, but you are not told that exactly 1 of the dice is a 6, and if you are told only that at least 1 of them is a 6, it doesn't matter which one it is ##-## the other could still be any of the 6 possibilities, resulting in a probability for a sum of 7 that is 1/6.
 
  • #20
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If you are told that at least one ... If you are told that exactly 1
Neither of those wordings was used. The wording was
you are informed that one of the faces shows a six
That is pretty standard phrasing meaning ##a=6 \lor b=6## but in case there were any doubt on the meaning the OP said very clearly
the number of possibilities that include six (the cells comprised of row 6 and column 6) so you count 11 of these cells
So there can be no ambiguity that the conditional is ##a=6 \lor b=6## rather than ##a=6##

it says that you are told that 1 of the dice shows a 6, but you are not told that exactly 1 of the dice is a 6, and if you are told only that at least 1 of them is a 6, it doesn't matter which one it is −−- the other could still be any of the 6 possibilities, resulting in a probability for a sum of 7 that is 1/6
Here your words don’t match your math. Your words say ##a=6 \lor b=6## but ##P(a+b=7|a=6 \lor b=6)\ne 1/6##
 
  • #21
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Neither of those wordings was used. The wording was [##\dots##you are informed that one of the faces shows a six.] That is pretty standard phrasing meaning ##a=6 \lor b=6##
That expression is true if ##a= 6##, if ##b=6##, and if both ##a=6## and ##b=6##.
but in case there were any doubt on the meaning the OP said very clearly So there can be no ambiguity that the conditional is ##a=6 \lor b=6## rather than ##a=6##
I didn't claim that what the OP said meant ##a=6##; I said that it didn't assert that exactly one of ##a \lor b## is true.
Here your words don’t match your math. Your words say ##a=6 \lor b=6## but ##P(a+b=7|a=6 \lor b=6)\ne 1/6##
To me it seems that you are interpreting ##a \lor b## as if it meant ##a \oplus b## ,
i.e.,
treating (##a## OR ##b##) as if it meant (##a## XOR ##b##).
 
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  • #22
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That expression is true if ##a= 6##, if ##b=6##, and if both ##a=6## and ##b=6##.
Correct. I agree.

I didn't claim that what the OP said meant ##a=6##; I said that it didn't assert that exactly one of ##a \lor b## is true.
Yes, I agree here also.

To me it seems that you are interpreting ##a \lor b## as if it meant ##a \oplus b## ,
i.e.,
treating (##a## OR ##b##) as if it meant (##a## XOR ##b##).
I am not sure why you think that. I am using ##\lor## and not ##\oplus## in all of my notation and in my Monte Carlo simulation.

##P(a+b=7|a=6\lor b=6)=2/11## with ##a=6 \lor b=6## including the event when both ##a=6## and ##b=6##.

If you believe that ##P(a+b=7|a=6\lor b=6)=1/6## then I encourage you to do a Monte Carlo simulation of your own.
 
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  • #23
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Correct. I agree.

Yes, I agree here also.

I am not sure why you think that. I am using ##\lor## and not ##\oplus## in all of my notation and in my Monte Carlo simulation.
To put it another way: it appears that you are using conditional probability for independent events, which results in 2/11 probability for 7, just as treating 'at least one' as if it meant the same as 'exactly one' would.
##P(a+b=7|a=6\lor b=6)=2/11## with ##a=6 \lor b=6## including the event when both ##a=6## and ##b=6##.

If you believe that ##P(a+b=7|a=6\lor b=6)=1/6## then I encourage you to do a Monte Carlo simulation of your own.
For each possibility for die ##a##, there are 6 possibilities for die ##b##, and for each of the 6 possibilities for die ##b##, there are 6 possibilities for die ##a##.

Being told that one of the dice shows a ##6## does not change the ##1/6## probability of each value for the other die; it tells you that if die ##a## shows a ##6##, die ##b## will have to show a ##1## for the numbers showing on the two dice to sum to ##7##, and if die ##b## shows a ##6,## die ##a## will have to show a ##1## for the sum of the numbers showing on the two dice to be ##7##; it has no bearing on the fact that when one of the two dice shows a ##6##, the ##6## possibilities for the other are equiprobable at ##1/6## each, including ##1 /6## chance of a ##1##, and consequently ##1/6## chance for a sum of ##7##.

Once we are told that ##(a=6 \lor b=6)## then ##P(a=6 \land b=6)## increases from ##1/36## to ##1/6##, and ##P(a=6 \lor b=6) = 1## but ##P(a=1 \lor b =1)## remains ##1/6##, just as it was before we were told anything, while after we are told that ##(a \lor b) = 6## then we know that ##P(a=1 \land b=1) =0##

Die ##a## does not lose its ##1/6## chance of showing a ##6## if die ##b## shows a ##6##, and die ##b## does not lose its ##1/6## chance of showing a ##6## if die ##a## shows a ##6##. After we are told that either '##a \lor b## shows a ##6##', or 'die ##a## shows a ##6##', or 'die ##b## shows a ##6##', the probability that ##a \lor b## shows a ##1## is ##1/6##; not ##2/11##.

The counting method that results in ##2/11## when one of the dice shows a ##6## would also result in ##2/11## each when we are told that one of the dice shows a ##5, 4, 3, 2,## or ##1##. That would make the possible outcomes sum to ##12/11## instead of ##1##, which is clearly incorrect.

Whether that results from treating independent events as conditional, or treating 'at least' as if it meant 'exactly' or treating 'OR' as if it meant 'XOR' does not change the fact that what shows on one die, regardless of us being told that one die shows ##n## such that ##(n=1 \oplus n=2 \oplus n=3 \oplus n=4 \oplus n-5 \oplus n=6)##, leaves 2 equiprobable sets of 6 ordered pairs for each of the 6 possible values of ##n##, e.g. when ##n=6##: ## [(6, 1), (6, 2), (6, 3), (6, 4), (6, 5,), (6, 6)]##, and ##[(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)]##. Of those 12, 2 sum to 7, meaning ##2/12=1/6## probability of ##7##.

It may be objected that##(6, 6)## is being counted twice, but that's not legitimately objectionable, because the ##1/6## chance for each number for one die is independent of the ##1/6## chance for each number for the other die, regardless of which is 'one' and which is 'the other', and, as long as what we say leaves at least one or the other of the dice unrestricted, also regardless of whatever we may say about what is showing on the two dice.

Being truthfully told that one of the dice shows a ##6 ## does not affect the ##1/6## probability that we could also be truthfully told that one of dice shows a ##1##, which means that us being told that one of the dice shows a ##6## does not change the ##1/6## probability that the sum of what is shown on the two dice is ##7##.
 
  • #24
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To put it another way: it appears that you are using conditional probability for independent events, which results in 2/11 probability for 7, just as treating 'at least one' as if it meant the same as 'exactly one' would.
This is not correct. ##P(a+b=7|a=6 \oplus b=6)=1/5 \ne 2/11## so I am not sure why it appears that way to you. Again, I encourage you to do the Monte Carlo simulation yourself.

Regarding the rest, I will go through your math after work. In the meantime here is my Mathematica code for the Monte Carlo simulation:

For simulating rolling two six-sided dice:
d6 := RandomInteger[{1, 6}]
d66 := {d6, d6}
For counting events:
contains6[{a_, b_}] := Or[a == 6 , b == 6]
containsOne6[{a_, b_}] := Xor[a == 6 , b == 6]
sumsTo7[{a_, b_}] := a + b == 7
For the Monte Carlo simulation:
nrolls = 1000000
rolls = Table[d66, {nrolls}];

rollss7 = Select[rolls, sumsTo7];
rollsw6 = Select[rolls, contains6];
rollswOne6 = Select[rolls, containsOne6];

rollsw6s7 = Select[rollsw6, sumsTo7];
rollswOne6s7 = Select[rollswOne6, sumsTo7];
 
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  • #25
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Once we are told that (a=6∨b=6)(a=6∨b=6)(a=6 \lor b=6) then P(a=6∧b=6)P(a=6∧b=6)P(a=6 \land b=6) increases from 1/361/361/36 to 1/61/61/6, and P(a=6∨b=6)=1P(a=6∨b=6)=1P(a=6 \lor b=6) = 1 but P(a=1∨b=1)P(a=1∨b=1)P(a=1 \lor b =1) remains 1/61/61/6, just as it was before we were told anything, while after we are told that (a∨b)=6(a∨b)=6(a \lor b) = 6 then we know that P(a=1∧b=1)=0P(a=1∧b=1)=0P(a=1 \land b=1) =0

I too am having trouble getting my head around this but I believe @Dale to be correct (aside from the fact that he usually is) .

What probability do you calculate for P(a≠6 AND b≠6)?? My math says 25/36. Therefore P(a=6 OR b=6)=11/36 exactly.
P(a=6 OR b=6|a+b=7)=1/3
P(a+b=7)=1/6
Bayes Theorem gives the result !!! First time I ever used it.....pretty good stuff.

P(a+b=7|a=6 OR b=6)=(1/3)(1/6)/(11/36)= 2/11
 
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