Four tricky 'uns

[mathare]

First off let me just state I don't know the answers to these. They are from a competition in the latest issue of Inside Edge. And I thought as a forum of bright gamblers we must be able to crack them :)

1. A coin is tossed 1,000 times. The first toss comes down heads. Heads lead 1-0. We now keep score after every toss. Which is the most likely number of changes of lead?
a) 1
b) 21
c) 7
d) 3
e) 12

2. A set of 28 dominoes is shuffled face down and formed into a 4x7 rectangle. I wish to bet that at least one of the four corner dominoes is a double (there are seven doubles from double blank to double six). What's the nearest correct price to offer me?
a) 4/5
b) 7/4
c) 4/9
d) 5/4
e) 8/13

3. In a football match, there's a 60% chance of a goal in the first 45 minutes play, a 6% chance of a goal in first-half injury time, a 70% chance of a goal in the second 45 minutes play and a 12% chance of a goal in second-half injury time. What's the probability that the final score will be 0-0?
a) 10.92%
b) 9.46%
c) 8.14%
d) 10.37%
e) 9.93%

4. You have a piece of rice paper one thousandth of an inch think and large enough for the purposes of this problem. You tear it in half and place on piece neatly on top of the other (tear 1). You then tear the two pieces in half and neatly stak the four peieces (tear 2). You continue this process for 50 tears and stackings. Approximately how high will the resultant stack be?
a) 86 feet
b) 13,108 miles
c) 462 miles
d) 18 million miles
e) 723 yards

There is a 5th question that basically says how do you know you are right but I suspect the answer to that may lie in the column from which this was ripped.

I have answers to 3 and 4 but I won't put them up yet to give people a chance to consider their own. Some indication of how you got your answer would be a good idea as even Vegy could bang down four guesses :)

By the way I am unlikely to enter this competition even if I can get the forum to answer the questions as the "valuable prize" on offer may be connected to the poor-looking website the column's author runs.

[Merkin]

I think the answer to question 4 is 18 million miles

I doubled up on the thousand of an inch by fifty times.
Then converted to inches by dividing by 100.
Convert the inches into feet by dividing by 12
then the feet into miles (1 mile = 5 280 feet)
17769884 miles.
Is near enough to 18 million miles.

[sportingprofit]

3 - thats e)9.93%

FH 100% - 60% = 40% or 0.4
FH IT 100% - 6% = 94% or 0.94
SH 100% - 70% = 30% or 0.3
SH IT 100% - 12% = 88% or 0.88

Times these together = 9.93%

[sparkyminer]

3 - thats e)9.93%

FH 100% - 60% = 40% or 0.4
FH IT 100% - 6% = 94% or 0.94
SH 100% - 70% = 30% or 0.3
SH IT 100% - 12% = 88% or 0.88

Times these together = 9.93%
That's what I got SP.

[mathare]

We agree on 3 and 4 then :)

I don't know where to start on 1. Once heads has that 1-0 lead you need two consecutive tails for the lead to change. But if the next toss comes down heads you need 3 tails to get a change in the lead. I just can't make the problem generic/logical enough to solve.

With 2 I thought that if you turn up the first corner domino it has a 21/28 chance of not being a double. The other corners are then 20/27, 19/26 and 18/25 for not being doubles which gives a 29.2% chance of none of them being doubles. That leaves 70.8% for it to be at least one double. Which is almost halfway between 8/13 and 4/5 so I think I must have mucked up.

[GlosRFC]

I think you're partly there with no 1, Matt. The common fallacy is that the lead would continue to change but, statistically, it's much more likely that the lead will never change, or perhaps change just the once. If you tossed a coin 20 times, there's only a 6% probability that the lead will change 10 times. However there's a 35% probability that the lead will never change. And the longer you play the greater the disparity so, for 1,000 tosses as per the example, then the likelihood is that the lead won't change 99.99% of the time and change 0.01% hence 1 is the probable answer. If it hasn't changed after the first couple of tosses then it's extremely unlikely to change after another 998.

[GlosRFC]

Okay, I think I can work 1 out in a more mathematical fashion. We start with the score at 0-0 (the fact that it's heads that comes up first is immaterial) so the probability of getting either a head or a tail is 1/2 or 50%. So now heads is 1-0 in front therefore we now need to obtain two tails in a row in order for the lead to change. The odds on completing this is 1/2 * 1/2 or 25%.

Assuming that we're successful, then we now need to throw 3 heads in succession for the lead to change back. The odds are now 1/2 * 1/2 * 1/2 or 12.5%. Next we'd need to throw 4 tails in order to change the lead so the odds are 1/2 * 1/2 * 1/2 * 1/2 or 6.25%.

Now if we return to the start where we need to throw 2 tails and, on one of the throws, a head appears we're obviously unsuccesful in changing the lead. Heads will still be in front, either 2-0 or 2-1 and we'll either need to throw 3 tails (a probability of 12.5%) or we'll still need to throw 2 tails (a probability of 25%) just to get the score to 2-3.

So we can see that there's a pattern emerging in that, for each successful attempt to alter the lead, the probability of the lead changing again is reduced by half! That also holds true for each unsuccesful attempt. Because of this 50% reduction in the likelihood of successfully completing the coup, I'd be inclined to stick to my original assessment that 1 is the likely answer of the lead changing but it's not beyond the realms of possibility that 3 might be right too. Statistically, the 7 option is extremely unlikely 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 0.78% so that miniscule probability discounts 12 and 21 as contenders.

[GlosRFC]

I think I'm now pretty certain why the lead is only ever likely to change once and not 3, 7, 12, 21 or any other number of times.

To get the lead to change at least once from 1-0 to 1-2 requires the successful toss of 2 heads in a row - a probability of (1/2 * 1/2) or 25%. In order to get the lead to change a second time from 1-2 to 3-2 requires throwing 2 heads in a row so that's another (1/2 * 1/2) - combine the likelihood of both occuring (i.e. throwing 2 tails followed by 2 heads) and the probability is (1/2 * 1/2) * (1/2 * 1/2) or 6.25%.

To get it to change for a third time from 3-2 to 3-4 by throwing another 2 tails in a row requires another 25% success so, even right at the start of the game we're looking at a probability of (1/2 * 1/2) * (1/2 * 1/2) * (1/2 * 1/2) or 2 tails followed by 2 heads followed by 2 tails in other words a 1.56% chance of success!

It gets even worse if you throw 2 heads in the first two tosses. Now, for the lead to change 3 times, we need to get the score from 2-0 to 2-3, then 4-3 and then 4-5 so we need to throw 3 tails, followed by 2 heads, and then 2 tails. The probability of this happening is (1/2 * 1/2 * 1/2) * (1/2 * 1/2) * (1/2 * 1/2) or 0.78%.

For a 3-0 start by heads then the probability of the lead changing 3 times is going to be even less likely at 0.39%. Once again we can see how the probability of being successful, and therefore the probability of the lead changing in the game, is reduced by half at each stage.

At this point my brain is throbbing so I'll try to approach Q2 tomorrow :D

[GlosRFC]

I think you're also pretty close with Q2, Matt but you've ignored the corners. Yes, there's a 21/28 probability (75%) that any of the dominoes won't be a double but you can only select from 4 of the 28 (14%) possible positions. So the probability that it won't be a double is more like 10.5%

Similarly, if unsuccessful, then there's a probability that 20/27 of the dominoes can be a double but now you can only select from 3 of the 27 remaining positions. Same for 19/26 probability from 2/26 positions and 18/25 probability from 1/25 positions.

I'll let you do the maths :D