[bonanova]

Curious about this place called Morty's, mentioned more than

once here at the Den, you wandered in last night to meet the

boys and have a cool one.

You must not have blended too well, because Alex saw you

coming and was striking up a conversation before you could

say Sam Adams.

Ya look like a smart lad who likes a challenge, he began.

I've got a pair of dice here, and I'm sure you know the chances

of rolling a 7 are 1/6, but to roll a 12 they are only 1/36. Well,

then; even up, I'm willing to bet you'll roll a 12 before you roll

two 7's in a row. Keep in mind two dice are thrown each roll.

It sounded like an even bet; after all, 1/6 x 1/6 = 1/36.

Do you take the bet?

Edited February 8, 2009 by bonanova
Added text in red to remind that each roll is of two dice

[Guest]

Good start.

But consider that the probability of getting a 2nd roll is 35/36 - a pretty good chance.

Thanks for the hint.

The probability for rolling a consecutive 7 on the 3rd or 4th roll is (29/36) * (1/6) * (1/6) = 2.24% and (35/36) * (29/36) * (1/6) * (1/6) = 2.18% respectively. The probability of rolling a 12 on the 3rd or 4th roll is (35/36) * (29/36) * (1/36) = 2.18% and (35/36) * (29/36) * (35/36) * (1/36) = 2.12% respectively. Therefore, except for the first roll (where I have a 1/36 chance of losing), my chances are always better by a factor of 36/35. Thus, I take the bet.

The probabilities of 35/36 and 29/36 imply that neither a 12 OR a 12 and a consecutive 7 can be rolled at certain times, otherwise the game ends.

[Prime]

It is a close call. Better than most games you find in Las Vegas. Calculating odds is a bit tricky. And there is a possibility of a draw (when you roll your second 7 just as Alex rolls his 12). Regrettably, I do not see my way to calculating the odds without resorting to geometric series. And, I know, Bonanova is going to reprimand me for that. Nonetheless, I am going to stick my head out...

To estimate the odds, I’m going to calculate the probability of 12 appearing before two 7s in a row are rolled. Then the probability of two 7s to be rolled before 12 ever appears. After that, we can compare the two numbers and decide whether to take Alex on his offer.

The probability NOT to roll 12 n times in a row is (35/36)ⁿ

Not so easy with two 7s in a row, or is it? For getting two 7s on a given roll seems to depend on what was rolled last time.

Suppose, the probability of not having two 7s in a row after n rolls (n >= 2) was P_n. Let’s figure out the probability of not having two 7s in a row after n+1 rolls. If the n-th roll was not 7 (at the probability 5/6), then no matter what we roll next, there won’t be two 7s in a row. If the last roll was 7 (at the probability 1/6), then there is 5/6 probability of still not having two 7s after n+1st roll. Thus the probability of not having two 7s in a row after n+1 rolls is P_n*(5/6)*1 + P_n*(1/6)*(5/6) = P_n*(35/36). Surprisingly, each next roll acts as an independent occurrence for 2 7s in a row.

The probability not to roll two 7s on the first two rolls is 35/36. And the probability of not having two 7s in a row after n rolls is (35/36)^n-1

Now let’s build geometric series. First for Alex’s win (12 appears first):

That is a sum of probabilities for each consecutive roll to have 12, while neither 12 appeared before, nor two 7s in a row up to and including that roll.

1st roll: 1/36 that 12 appears and wins.

2nd roll: 35/36 that 12 did not appear before; 35/36 that the first and second roll did not make two 7s; and 1/36 to roll 12 on the second roll. So (1/36)*(35/36)².

3rd roll: (35/36)² (no 12 on previous 2 rolls), times (35/36)² (no two consecutive 7s in the 3 rolls), times 1/36 to roll 12 on the third roll.

Thus the infinite geometric series for 12 to appear first is:

1/36 + (1/36)*(35/36)² + (1/36)*(35/36)⁴ + (1/36)*(35/36)⁶ + ….

Where the common ratio is (35/36)² and the scale factor is 1/36. Applying the formula for the (infinite) geometric progression, we find the sum is (1/36)*(1296/71) = 36/71. Or approximately 50.70%.

Now the same for 7s win:

2nd roll: 1/36 to roll two 7s, times (35/36)² for the opponent not to roll 12.

3rd roll: 1/36 to roll two 7s, times (35/36) not having them in the first two rolls, times (35/36)³ for the opponent not rolling 12 on the 3 tries.

4th roll: (1/36)*(35/36)²*(35/36)⁴.

Thus the infinite geometric series is:

(1/36)*(35/36)² + (1/36)*(35/36)⁴ + (1/36)*(35/36)⁶ + ….

Where the common ratio is (35/36)² and the scale factor (1/36)*(35/36)².

Evaluating we find the probability is (1/36)*(35/36)² * (1296/71) = 35²/(36*71). Or approximately 47.92%.

The remaining approximately 1.38% must go for the draw.

So the odds are roughly 51:48 in favor of Alex (12).

I would take the bet at a small stake. The odds are very close. And knowing from the previous Bonanova’s posts that Alex gets sloppy at times with calculating the odds for his bets, taking this bet at a slight disadvantage could make Alex a profitable gambling opponent in future.

[Guest]

The remaining approximately 1.38% must go for the draw.

O.o

"Approximate exact odds" notwithstanding, how can there possibly be a draw?

[Prime]

O.o

"Approximate exact odds" notwithstanding, how can there possibly be a draw?

The OP does not state who goes first. So we have to make some reasonable assumption. Mine is: if I got two 7s on my n-th roll and Alex rolled his first 12 on his n-th roll -- that's a draw. Otherwise, you have to make two separate calculations, as the winning chances will dependend on who moves first. For example, if your first two rolls were destined to score 7 each, and Alexe's fate held 12 on his second roll; then if Alex goes first -- he wins, you go first -- you win.

Edited February 10, 2009 by Prime

[Guest]

The OP does not state who goes first. So we have to make some reasonable assumption. Mine is: if I got two 7s on my n-th roll and Alex rolled his first 12 on his n-th roll -- that's a draw. Otherwise, you have to make to separate calculations, as the winning chances will dependend on who moves first. For example, if your first two rolls were destined to score 7 each, and Alexes fate held 12 on his second roll; then if Alex goes first -- he wins, you go first -- you win.

You are the only person with a dice as I understand it? Alex does not roll at all. He is betting that YOU will roll a 12 before YOU roll 2 7s

[Prime]

You are the only person with a dice as I understand it? Alex does not roll at all. He is betting that YOU will roll a 12 before YOU roll 2 7s

Oh my! I missed that. Well, since there is no draw, the probability for 12 to show first is 1331/2556, or just over 52%. (Can be calculated as in my original post).

Still, I would take the bet.

[Guest]

Given that you can't possibly win after just 1 roll, don't take the bet.

The probability of rolling two consecutive7s is 1 in 36 and not just the 1 in 12 as stated above. If the rolls of seven did not have to be consecutive, then the 1 in 12 would be accurate, but that is not the question. I don't have the answer, but I wanted to point this out.

[bonanova]

1. The first dice roll is 12 [Alex wins], 7, or x; where x is something else.
   
2. If it's 7, you roll again and get 12 [Alex wins], 7[you win], or x.
   
3. After any x, you're back to square 1.
   

Voila! p[Alex wins] = _____.

[HoustonHokie]

1. The first dice roll is 12 [Alex wins], 7, or x; where x is something else.
   
2. If it's 7, you roll again and get 12 [Alex wins], 7[you win], or x.
   
3. After any x, you're back to square 1.
   

Voila! p[Alex wins] = _____.

It looks like you're suggesting the problem can be boiled down to what happens in the next two rolls of the dice. In two rolls of the dice, Alex has two chances to win (2*1/36), but you only have one (1*1/36). Which would suggest that, overall, Alex is twice as likely to win as you are (p[Alex] = 0.67; p[you] = 0.33). Gotta get a cup of coffee and see if that makes sense when I'm done with it, but it sounds reasonable right now...

[Prime]

1. The first dice roll is 12 [Alex wins], 7, or x; where x is something else.
   
2. If it's 7, you roll again and get 12 [Alex wins], 7[you win], or x.
   
3. After any x, you're back to square 1.
   

Voila! p[Alex wins] = _____.

That's a nice recursive solution!

On the other hand, my infinite series turned out wrong. I am stumped by calculating probability of rolling/not rolling two 7s in a given number of rolls. There seems no way to make a roll independent.

Now to the actual calculation of your solution

Probability of Alex's win is P.

Probability of rolling a number other than 12, or 7: 29/36.

Probability of rolling 7: 1/6.

Probability to get 12 on a single roll: 1/36

Alex's win probability equation:

P = (29/36)P + (1/6)*(29/36)P + 1/36 + (1/6)*(1/36); [x on the first roll + 7 on the first and x on the second + 12 on the first + 7 on the first and 12 on the second]

P = (203/216)P + 7/216; P = 7/13

Just to check, we can calculate the probability Q for two 7s to win:

Q = (29/36)Q + (1/6)(29/36)Q + (1/6)*(1/6);

Q = (203/216)Q + 1/36; Q = 6/13

And so P+Q=1

Still, I would take the bet.

[HoustonHokie]

It looks like you're suggesting the problem can be boiled down to what happens in the next two rolls of the dice. In two rolls of the dice, Alex has two chances to win (2*1/36), but you only have one (1*1/36). Which would suggest that, overall, Alex is twice as likely to win as you are (p[Alex] = 0.67; p[you] = 0.33). Gotta get a cup of coffee and see if that makes sense when I'm done with it, but it sounds reasonable right now...

Because I made that last post prior to my morning coffee and because I was torn between my logic & Prime's, I decided to do a little simulation to see if I could understand what happens. Using a random number generator in a spreadsheet, I simulated 10000 pairs of rolls of the dice and evaluated whether or not there was a winner in those pairs of rolls. Based on my last post, I expected Alex to win 2/36*10000 = 556 times and me to win 1/36*10000 = 278 times. And that was pretty much what I got. Every time I updated the calcs for the spreadsheet, the numbers regenerated and I got new values for wins, but, in general, Alex was twice as winning as I was. But I was left with a doubt - what about those times when nobody won and a 7 showed up on the second roll? Surely, I was more likely to win on the 3rd roll than Alex was in those cases! Of course, he would still win on other situations that led to a third roll. So, I tried another tactic.

Starting with the same 10000 games, I looked at each roll of the dice individually. After the first roll, Alex should have won 10000/36 or 278 times. I couldn't win in the first roll, but of the remaining 9722 games where the action continues after the first roll, a 7 should appear 10000/6 = 1667 times. So, in round 2, Alex should win 9722/36 times = 270, and I should win 1667/6 times = 278. Now, after (278+270+278 = 826) winners, there are 9174 games still ongoing. And, of those, 9722/6 - 278 = 1342 have a 7 showing. I carried this operation out for 110 rolls, which is the point at which the rounding didn't allow anyone to win any of the 14 remaining games. I found that Alex won 5372 times and I won 4614 times, which is approximately the 7/13 vs 6/13 relationship suggested by Prime (and blew my 1/3, 2/3 thing completely out of the water - like I said, no coffee!). To make sure the rounding didn't affect things too much, I removed it and found that Alex won 5385 times and I won 4615 times, which matches the 7/13 vs 6/13 relationship exactly.

I checked something else, too, just to satisfy my curiosity. It's obvious that Alex's probability of winning on any one roll is constant at 1/36, but I wondered about my probability of winning on a particular roll. So I took the ratio of wins for me to games played (both without rounding) and came up with the following:

Roll 1: p(me) = 0

Roll 2: p(me) = 0.028571

Roll 3: p(me) = 0.024390

Roll 4: p(me) = 0.025018

Roll 5: p(me) = 0.024924

Roll 6: p(me) = 0.024938

Roll 7 (& higher): p(me) = 0.024936

So it looks to me like things are in flux through the first 6 rolls, but even out from 7 on. That suggests some level of dependency at the beginning which turns into independence by the 7th roll, but I haven't figured out why just yet.

[Prime]

Because I made that last post prior to my morning coffee and because I was torn between my logic & Prime's, I decided to do a little simulation to see if I could understand what happens. Using a random number generator in a spreadsheet, I simulated 10000 pairs of rolls of the dice and evaluated whether or not there was a winner in those pairs of rolls. Based on my last post, I expected Alex to win 2/36*10000 = 556 times and me to win 1/36*10000 = 278 times. And that was pretty much what I got. Every time I updated the calcs for the spreadsheet, the numbers regenerated and I got new values for wins, but, in general, Alex was twice as winning as I was. But I was left with a doubt - what about those times when nobody won and a 7 showed up on the second roll? Surely, I was more likely to win on the 3rd roll than Alex was in those cases! Of course, he would still win on other situations that led to a third roll. So, I tried another tactic.

Starting with the same 10000 games, I looked at each roll of the dice individually. After the first roll, Alex should have won 10000/36 or 278 times. I couldn't win in the first roll, but of the remaining 9722 games where the action continues after the first roll, a 7 should appear 10000/6 = 1667 times. So, in round 2, Alex should win 9722/36 times = 270, and I should win 1667/6 times = 278. Now, after (278+270+278 = 826) winners, there are 9174 games still ongoing. And, of those, 9722/6 - 278 = 1342 have a 7 showing. I carried this operation out for 110 rolls, which is the point at which the rounding didn't allow anyone to win any of the 14 remaining games. I found that Alex won 5372 times and I won 4614 times, which is approximately the 7/13 vs 6/13 relationship suggested by Prime (and blew my 1/3, 2/3 thing completely out of the water - like I said, no coffee!). To make sure the rounding didn't affect things too much, I removed it and found that Alex won 5385 times and I won 4615 times, which matches the 7/13 vs 6/13 relationship exactly.

I checked something else, too, just to satisfy my curiosity. It's obvious that Alex's probability of winning on any one roll is constant at 1/36, but I wondered about my probability of winning on a particular roll. So I took the ratio of wins for me to games played (both without rounding) and came up with the following:

Roll 1: p(me) = 0

Roll 2: p(me) = 0.028571

Roll 3: p(me) = 0.024390

Roll 4: p(me) = 0.025018

Roll 5: p(me) = 0.024924

Roll 6: p(me) = 0.024938

Roll 7 (& higher): p(me) = 0.024936

So it looks to me like things are in flux through the first 6 rolls, but even out from 7 on. That suggests some level of dependency at the beginning which turns into independence by the 7th roll, but I haven't figured out why just yet.

The first two turns account for about 8.26% of entire probability. There is 5.48% chance (71/1296) for 12 to win within the first two rolls, and about 2.78% chance (1/36) for 7s to win on the second roll. The remaining 91.74% are split more evenly, but still slightly in favor of 12.

The probability to roll 12 on any turn is 1/36, regardless of what happened before.

The probability to get second 7 on any turn past the first two turns is less than 1/36. Here is why:

To get a second 7 on a turn, the previous roll must have been 7. Consider all possible sequences of dice rolls. Exactly 1/6 of them will end with "7". But within the framework of the game, we must discard those sequences ending with 7, where a turn before that also was 7. Because if that was the case, then the game would be over on the last turn. Thus the proportion of strings ending with 7 is diminished, it resides between 1/6 and 5/36. That times the 1/6 probability to roll another 7, yields less than 1/36 chance.