### Coin Tossing Games

You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?

### Win or Lose?

A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with?

A counter is placed in the bottom right hand corner of a grid. You toss a coin and move the star according to the following rules: ... What is the probability that you end up in the top left-hand corner of the grid?

# Last One Standing

##### Stage: 4 Challenge Level:

Many wonderful solutions have been received and here are a few selected ones. Enjoy!

Anon from Horrington Primary experimented with the simulator and arrived at an answer:

The mean average of this question was 10 for me on the simulator, (I did 156) so I think someone would get heads more than 6 times.

Jacques and Tahmid from Wilson's School ran more values through the simulator and got the answer below:
number of spins till none left standing
number of people       try one       try two       try three      average
4                                3                3                3                3
16                              5                5                 5              6.67
64                              5                11               6               7.34
256                           8                7                 11             8.67
1024                         17               13               12             14
If half of the crowd got tails each time though (the probability of spinning a head being 1/2), then the results would be so:
number of people               number of spins till none standing
4                                        range from 3-4
16                                      range from 5-6
64                                      range from 7-8
256                                    range from 9-10
1024                                  range from  11-12
(margin of difference is for the last one standing)
The results we got practically weren't far from those in theory. Had we done more tests, the results from the practical tests wouldve been closer to those in theory from our results, we'd need somewhere between 256 to 1024 people in the assembly to expect someone to get over 10 heads in a row, though as you can see from the results table, with 256 people, someone still managed to flip the coin eleven times.

Vatsal from Wilson's School gave a clear demonstration of the process:

On average after every turn half of the amount of people who were there will flip again so in a way the amount of people remaining will keep on halfing. So this is how it will go about.
250
125
62
31
15
7
3
1

Patrick from Otterbourne and Shaun from Wilson's School gave a concise solution to the problem.

If 250 people start tossing coins, then you can expect about 4 to get 6 heads in a row. The chances of getting 6 straight heads are 1/64. The last one standing can expect about 8 straight heads. The chances of 8 consecutive heads are 1/256. The chances of 10 heads are 1/1024. You need 1024 people to expect someone to get 10 heads.
If there are 2 jackpot winners, then there are about 28 million tickets are sold each week.
The chances of 3 people sharing their birthday is 1/365 squared, not cubed. This equates to 1/133225 (ignoring leap years).

Derren Brown would take about 1024 tries.

Lindon from Wilson's School gave an answer to the lottery question:

Lottery question If the probability of winning is 1 in 14 million so for 2 people to win the jackpot about 28 million tickets would need to be bought.

Jack from Wilson's School made a good attempt to the rest of the problem
For the birthday problem:
The papers worked out the probability of all three children arriving on a specific day, which is:
1 in 365 is the chance 1 would land on October 7th. 2 must be 365 squared which is 1 in 133,225. So 3 must be 365 cubed which is 1 in 48,627,125, as the papers reported.
But they reported it as the probability of three children being born on the same day, when it is in fact the probability of the three children being born on a specific day
The correct probability is just $1\text{ in } 365^2 = 1 \text{ in }133,225$ , as the first child can pick any day they want, and then the other two have to pick that specific day. The papers calculated the probability of 4 children
So with more than a million families with three children in the UK, there are probably more families where this has occured.

Well done to everyone!