# Dice problem that I can't fully get my head around

1. Oct 19, 2006

### waterchan

There's a probability question I saw on the web that looks innocent enough, unless you pay attention to the wording:

I have two dice.

If i roll them until i get a double, which is the most likely roll to get a double on?

The first roll is the most likely.

Heres why

First Roll:

1/6 Chance to get Double. 5/6 Chance to not get double

Total Chance: 1/6

Second Roll
5/6 Chance to Reach This Stage
1/6 Chance to get double

Total Chance = 5/36

Third Roll
25/36 Chance to Reach This Stage
1/6 Chance to get double

Total Chance = 25/216

You get the idea. This carries on.

The answer makes sense to me, but probably due to the wording of the question, I am still slightly inclined to think that the answer is "equally likely". Note that the person will stop rolling the dice when he hits a double.

What do you think?

2. Oct 19, 2006

### matt grime

I think that you've just claimed that 1/6 = 5/36.

3. Oct 19, 2006

### waterchan

I don't understand it completely myself, but let me try to provide a shorter version of this answer.

Question: I have two dice. If i roll them until i get a double, which is the most likely roll to get a double on?

It is explained in the answer that, since he will stop rolling when he gets a double, the most likely roll to get a double on is the first roll. It's less likely to get a double on the next rolls because it becomes progressively unlikely that he will reach subsequent stages.

Something wrong with this?

4. Oct 19, 2006

### matt grime

Well, now you're contradicting yourself. What is it? Equally likely on each throw or more likely on some (the first, obviously)?

5. Oct 19, 2006

### waterchan

I'm not totally sure. If it were a bet, I would put my money on the choice that the answer is the first throw. But I'm not completely convinced.

6. Oct 19, 2006

### Simon 6

You've come up with a good one!

I'm not sure that the apparent paradox has been understood.

Incidentally, the same calculations apply to a single dice. It is more likely it will take one throw to roll a given number. The probability starts at 1/6 and becomes progressively smaller.

This may seem like a contradiction, but it isn't.

The following are the approximate odds for how many throws it should take to roll a particluar number on a single dice (or any double number on a pair of dice).

1 throw = 1/6 chance
2 throws = 1/7 chance
3 throws = 1/8 chance
4 throws = 1/10 chance
5 throws = 1/12 chance
6 throws = 1/15 chance

The trick is recognising that these odds apply to the rolls collectively, not individually.

Say you decide to throw a dice until you get a six. Each time you throw a six you record the number of throws it took. You repeat this over and over.

With a large enough sample, the pattern will emerge. The winner will be "1 throw". In second place "2 throws". In third place "3 throws" - and so on.

However, if you take each throw of the dice individually, the odds are always the same - 1/6. If you fail to get a six on the first throw, you now have a 1/6 chance of achieving it on the second.

Simon

Last edited: Oct 20, 2006
7. Oct 25, 2006

### Simon 6

Incidentally, the problem is even more counter-intuitive when applied to larger number of possible outcomes.

Consider this example:

Given an unbiased roulette wheel with a single zero:

"How many consective spins are you most likely to get without the ball landing on zero?"

In other words, how many consecutive spins of any number in the 1-36 range are most likely to separate two zero spins?

Most people would say 36. That is in fact the average.

The Mode number of spins is 0. You are more likely get two consecutive zero spins than acheive any number of non-zero spins in between.

I suspect some may dispute this.

Simon

Last edited: Oct 26, 2006
8. Oct 26, 2006

### haki

waterchan from what I read I think you are thinking like...

supose you are at a roulette, and you want to bet on red or black. Then you look at a roulette history and see that the last 5 numbers were black! Now you assume that it is more likely that the next number will not be black since last 5 were black! This is ofcourse bull ****. Each event is independant soo there is 50/50 chance and not more than 50 % that the next number will be the opposite of black. Forgive me for assigning a color to a number it is mathematically not correct.

I guess the question you should be asking is after how many throws is it possible to assume that number 2 has fallen.

9. Oct 26, 2006

### DaveC426913

Lert's see what happens if we actualy run the experiment.

Start rolling. If your hypothesis is such that each scenario is equally likely, then after 100 sessions of rolling, you will have an even distribution of hits.

100 scenarios:

rolled 1 times to get doubles - ~1 occurrence
rolled 2 times to get doubles - ~1 occurrence
rolled 3 times to get doubles - ~1 occurrence
rolled 4 times to get doubles - ~1 occurrence
...
rolled 100 times to get doubles - ~1 occurrence

This last one means that in at least one scenario, you had to roll 100 times before getting doubles! Would you say that this scenario is every bit as likely as getting doubles on the first throw?

Last edited: Oct 26, 2006
10. Oct 30, 2006

### Ozducati

The outcome of a throw of a die, or a pair of dice for that matter, is not influenced by previous or subsequent throws. Therefore the chance of getting a double on a certain throw of a pair of dice is the same as for every other throw.

11. Oct 30, 2006

### Simon 6

But the question is: "how many rolls of a pair of dice is most likely required to achieve a double?"

They are not all equally likely. If they were, then the prospect of having to roll 1000 times before getting a double would be equally probable as having to roll 6 times - as I think Dave was also pointing out.

The most probable Mode number of rolls, over any other individual number of rolls, is 1. That is the number of rolls that will occur the most often, given a decent sample.

Getting a double straight away is more likely than any other specific number of rolls to acheive it.

Clearly, the answer is as contraversial the Monty Hall conundrum!

Simon

12. Oct 30, 2006

### Moo Of Doom

The crux of the problem is that you are asked to find the most likely number of throws needed to get a double, not the expected number of throws.

Rolling just once to get a double is far more likely than having to roll 1000 times to get a double, and it turns out that it is more likely than any individual number of throws. Of course we know that it us more likely we need to roll more than one times (probability 5/6) to get a double than exactly one time (probability 1/6).

The expected mode of the number of throws is 1. The expected mean of the number of throws is 6.

13. Apr 21, 2007

### Dan 1st

Hmm .... interesting variety of answers to this question.

But I believe the answer can be found logically.
The dice throws . They're consecutive, yes, but they are not "linked" , they don't "multiply" each other or anything like that. Each one , no matter be it roll number 1 , 24 or 300 , is a dice roll , they all happen in the same way, starting off from scratch each time. Therefore, each individual roll has the same chance of being a double.

It may seem unlikely that roll number 1000 is the first roll to get a double , but what you are doing is much nearer calculating the "average" dice roll that gives a double.
Just because averages might say " 1 in every X throws is a double " doesn't mean you can't get 999 throws without getting a double . Each of those 999 throws was just as likely as the " most likely " first throw to get a double. Just because they didn't doesn't mean they are any less likely to.

Another thing - it doesn't actually change anything if the person stops throwing the dice or not after he gets a double. Let's imagine he rolled the dice 3 times to get doubles. The first double was on the 3rd rool, and so was the second. The 3rd double was on the fourth roll. From what some people have been saying, it is more likely you will get a double ( in this case ) on the third roll , because the thrid roll scored a double twice but the fourth roll scored a double once. But - since there was no fourth roll on the first two throws, since he stops throwing dice when he gets a double, then to say you are more likely to get a double on the 3rd roll is not true.
If he doesn't stop throwing the dice after he gets a double it's back to the beginning - no matter how many dice rolls the person has done; each seperate roll has the same chance of being a double.

It might sound weird , but each throw does have equal chance to being a double. Since each throw is individual ( even if it's the 200th throw or the 2nd ) , why should one throw have , say, 1/6 chance of being a double , and another throw have less chance ? the chance of each throw being double is not influenced by the chance the last one had ; each throw has the same chance.

Just because you might never reach 1000 rolls due to getting a double before doesn't mean it's any less likely, as you would have stopped throwing and therefore can't say the 1000th roll is less likely because you never did a thousandth roll . And if you never reached that, it's quite normal only because 1/6 chance is still not a negelectable chance, like for example 1/6000 ; so one of the rolls fell into that 1/6 chance and was a double. That doesn't necessarily have to happen before the 1000 ( or any other high number ... )

In fact I doubt it's 1 - since there are 36 possible outcomes of a dice roll ; of which 6 are doubles - then there is 1/6 chance for your throw to be a double. So why would it not be more likely to take 6 throws ?

If you think of that last bit , you might find it "more likely" to get a double on any of the first 6 rolls ; not just the first one.

...... I hope that's not too confusing , because I think I'm starting to confuse myself here , but hopefully somebody might be able to make some sense out of something there ....

14. Apr 22, 2007

### Simon 6

Yes Dan, you are getting thoroughly confused!

Naturally each individual throw has the same 1/6 chance of being a double. It also has the same 5/6 chance of not being a double. That is the point. The question is really about the most likely number of consecutive non-doubles to occur before a double is thrown.

Perhaps it becomes clearer if asked like this:

What are the probabilties of:

a) Throwing a double immediately.
b) Throwing a non-double exactly once, followed by a double.
c) Throwing a non-double exactly twice, followed by a double.
d) Throwing a non-double exactly three times, followed by a double.
e) Throwing a non-double exactly four times, followed by a double.
f) Throwing a non-double exactly five times, followed by a double.
g) Throwing a non-double exactly 999 times, followed by a double.

a) 1/6 = 16.66%
b) 5/6 x 1/6 = 13.88%
c) 5/6 x 5/6 x 1/6 = 11.57%
d) 5/5 x 5/6 x 5/6 x 1/6 = 9.6%
e) 5/5 x 5/6 x 5/6 x 5/6 x 1/6 = 8%
f) 5/5 x 5/6 x 5/6 x 5/6 x x 5/6 x 1/6 = 6.7%
g) 5/6 (to the power of 999) x 1/6 = less than 0.0000000000000000001%

I'm sure you can see why the probabilties become progressively smaller.

Simon

15. Apr 22, 2007

### -Job-

Or, in other words, the probability of rolling a double on the Nth throw is smaller than the probability of exactly rolling the preceding sequence of n-1 numbers (between 1 and 11).
Since the probability of rolling a given sequence of n-1 numbers decreases as n increases, then so does the probability of rolling a double on the Nth throw.

Last edited: Apr 22, 2007
16. Apr 23, 2007

### Dan 1st

ah I get it haha !
thanks for the explanation, it makes it a lot easier to understand , I know see where I was wrong .
As they say, you learn through your mistakes ; I might try and baffle some friends with this one now ....