Help with probability please?

Question

Suppose Bob and Nick are playing a game. Bob rolls a die and continues to roll until he gets a 2. He keeps track of all the outcomes of the rolls. He calls the sum of the rolls X. Nick does the same thing as Bob, only he keeps rolling until he gets a 5 and he calls the sum of his rolls Y. At the end of the game, Bob pays Nick Y dollars and Nick pays Bob X dollars. Does one of the boys have an advantage over the other? Which one?

Answer 1

both will get what they themselves rolled. bob stops at 2 and nick at 5.

bob's average in a roll before he stops
= (1+3+4+5+6)/5
= 19/5
= 3.8
money bob expected after n rolls
= 3.8(n-1) + 2
# the 2 is the last roll.

nick's average in a roll
= (1+2+3+4+6)/5
= 16/5
= 3.2
money nick expected after n roll
= 3.2(n-1) + 5
# the 5 is the last roll
(all this assuming the last roll is included in X and Y)

clearly bob's average is higher by 0.6. but if both rolled the same number of times and it's under 6, nick's at advantage. at round 6, both get the same expectation.

so one of them does have an advantage over the other, if both roll 2 fair dices and that would let us assuming both will stop at the same round.
bob's get the better deal (in the long run).

Answer 2

Bob has the advantage. His sum will always include a 2 while Nick's will always include a 5. The odds are that over the long run, the number of throws each is required to take is 6, since probability of rolling any particular number from 1 -6 is 1/6.

The probability is the other 5 rolls will have similar sums for each player. Thus in the long run, Bob should come out ahead by $3.00.

Answer 3

Neither one really has an advantage. If it's a fair die, the odds of rolling a 2 are 1/6. and the odds of rolling a 5 are 1/6. The odds of rolling are the same.
However, the number Nick is rolling for is higher (5); so if each guy rolls the die the same number of times and gets the same number each time, until getting their "correct" number, Nick will have the advantage because 5 is greater than 2.

Answer 4

Bob has the advantage because all of the 5's count for him. Where Nick can count all of the 2's.

Bob might roll a 5 twice before he gets his 2. Where Nick might roll a 2 twice before he gets his 5. Those two scenerios are equal in probability, but Bob's total would be 12 and Nicks would be 9.


If you mean sum as in the number of rolls, then they are equal.

Answer 5

Nick. He at least gets five dollars and Bob only gets at least two. Bob could roll a 1, 1 and a 2 but he would only get 4 dollars. Nick could roll a 1, 1 and a 5 and he would get 7 dollars.

Answer 6

They're the same because the chances of bob rolling 1 number (2) out of 6 numbers are the same as nick rolling 1 number (5) out of 6 numbers...

1/6 = 1/6

Answer 7

i think my argument should not be nicely gained. I say the chance is 50% enable a ??, enable b ?? and randomly go with the values for a and b. As already talked about, for a ? 0, P( a < b²) = a million, it is trivial. in simple terms slightly a lot less trivial is the concept P(a < 0 ) = a million/2 and to that end P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what takes position even as a > 0 For a > 0, even as it truly is elementary to exhibit there's a non 0 chance for a finite b, the decrease, the chance is 0. a < b² is an similar as declaring 0 < a < b², bear in mind we are in simple terms searching at a > 0. If this a finite period on an unlimited line. The chance that a is an ingredient of this period is 0. P( a < b² | a > 0) = 0 As such we've a finished chance P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 bear in mind, it really is because of countless gadgets. whatever variety of period you draw on paper or on a computing gadget you'll stumble on a finite chance that seems to mind-set a million. yet it really is because of the finite random type turbines on the computing gadget and if we had this question requested with finite values there's a a answer more suitable than 50%. i do not recommend to be condescending, yet please clarify why utilizing the Gaussian to approximate a uniform distribution is a sturdy idea? are not countless numbers relaxing. Cantor even as mad operating with them! :)

Answer 8

Neither one has an advantage. If it's a fair die, the odds of rolling a 2 are 1/6. and the odds of rolling a 5 are 1/6. The odds are the same being they pay eachother an equal amount(x=y).

I'm not sure what you mean by sum of rolls(if you mean how many the answer is above)

Answer 9

good question ... I initially thought that Bob had the advantage since his last roll is a "5" .....

but then realized that "all 2's count" for him while
"all 5's count" for Nick ...

not sure yet

I just wrote a crude little program to simulate this situation about 100,000 times each player, and the preliminary results tend to favor the guy who stops when a "5" is rolled, but it's AWFULLLY close ...


after running 500,000 roll-length-sum sequences for each player, I believe that the "average" sum of digits = 21 and it's an equal expectation of payoff.

ssumsum5= 105180908
ssumsum2= 105032917

a clean 1,000,000 roll-length sequence sums each yields:
ssumsum5= 209876751
ssumsum2= 210062492

hard to call it as anything other than "even"

interestingly enough (tho not unexpected) changing the "2 and 5" to "1 and 6" makes no difference ... stilll an even-game and still have probable "roll-sum" of 21

/*********************************/

FOR m = 1 TO 50000
sumsum5 = 0
sumsum2 = 0


RANDOMIZE TIMER

FOR n = 1 TO 200
sum5 = 0
sum2 = 0


WHILE die <> 5
GOSUB roll
sum5 = sum5 + die
' PRINT "die="; die; "sum5="; sum5
WEND

'PRINT "..............................."
WHILE die <> 2
GOSUB roll
sum2 = sum2 + die
' PRINT "die="; die; "sum2="; sum2
WEND


sumsum5 = sumsum5 + sum5
sumsum2 = sumsum2 + sum2

NEXT n

ssumsum5# = ssumsum5# + sumsum5
ssumsum2# = ssumsum2# + sumsum2

NEXT m

PRINT "sumsum5="; sumsum5
PRINT "sumsum2="; sumsum2

PRINT "ssumsum5="; ssumsum5#
PRINT "ssumsum2="; ssumsum2#



INPUT asd


roll:
die = 1 + INT(6 * RND)
RETURN

Answer 10

Yeah, Nick.