Although in each group game the probability of him picking the winner is 1/3 (although Paul wasn't (as far as I know) given the opportunity to pick a draw, the possibility of there being a draw still remained), and each knockout game the probability is 2/1, that doesn't mean that the probability of Paul guessing all the games correctly so far is 1/3 x 1/3 x 1/3 x 1/2 x 1/2 x 1/2, because the possibility of Paul picking the winner of one game relies on him getting the previous guess correct. So, for example, if he had guessed the result of the 1st knockout game wrongly, he wouldn't have got the chance to guess the winner of the Quarter-Final. So the options of what he could have done are (what Paul actually did is in bold):

**Guessed first game correctly**

Guessed first game wrongly (Germany Lose)

Guessed first game wrongly (Germany Draw)

**Guessed 2 correctly**

Guessed first correctly, second wrongly (Germany Win)

Guessed first correctly, second wrongly (Germany Draw)

**Guessed 3 correctly**

Guessed 2 games correctly, third wrongly (Germany Lose)

Guessed 2 games correctly, third wrongly (Germany Draw)

**Guessed 4 correctly**

Guessed 3 correctly, fourth wrongly

**Guessed 5 correctly**

Guessed 4 correctly, fifth wrongly

**Guessed 6 correctly**

Guessed 5 correctly, sixth wrongly

So he has had 6 outcomes from a possible 15. At first, I thought that 6/15, or 2/5, were the odds, but if at any point Paul has got all his guesses right, he will guess the forthcoming game, so any previous outcome involving him getting them all right can be discarded. This leaves:

Guessed first game wrongly (Germany Lose)

Guessed first game wrongly (Germany Draw)

Guessed first correctly, second wrongly (Germany Win)

Guessed first correctly, second wrongly (Germany Draw)

Guessed 2 correctly, third wrongly (Germany Lose)

Guessed 2 correctly, third wrongly (Germany Draw)

Guessed 3 correctly, fourth wrongly

Guessed 4 correctly, fifth wrongly

Guessed 5 correctly, sixth wrongly

**Guessed all six games correctly**

Which gives us 1 in 10. Am I barking up the wrong tree here?

Never in the field of human endeavour has so much been considered about such little consequence, though reports of death threats and calamari specials mean it is hardly of little consequence to Paul!

ReplyDeleteI base my calculation on his getting it wright or wrong. Right meaning he guesses the correct team to win, wrong meaning he either guesses the wrong team or the game ends in a draw.

I also leave out any consequence of Germany being favoured or not as he (presumably) is not au fait with the betting odds.

Each decision then being an either/or (an even money chance in betting terms or 50:50 in chance terms) the likelihood of his getting six such selections correct in a sequence are 1 in 64.

This is based on 1 winning one by getting it right and making 2, then 2 winning 2 getting it right making four and so on through 6 sequences. If he was betting on it he would double his money each time.

It is the same by the law of averages that in order to get six right in a row he would need to start 64 times, going back to one whenever he got it wrong.

That is my take on it. Must go back to the real world now but it has been a pleasure :-)

Rob

For the record I completely agree that if he is guessing on a game by game basis, the odds are 1 in 64. We're coming at it on different angles though - I'm assuming if he gets it wrong, he gets no more guesses. I have to admit I don't know if that would have been the case, but by the time I thought of that I was so far into the calculations I was determined to finish it.

ReplyDeleteCheers for the debate though, even if Twitter is a little limiting in what you can put!