h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Hi folks!

I have a mathematical problem I hope you guys can help me out with.

For Christmas, I got a little cricket dice game called 'Howzat!' Gosho describes it rather well at A948765, but I'll give you a brief description of it here. I'll try to explain it in non-cricketing terms for anyone who doesn't know the game.

I have two dice, one for batting and one for bowling. Each has six sides. The batting die has the numbers 1, 2, 3, 4 and 6 on five of the sides, and 'Howzat!' on the sixth. The bowling die has four sides which represent different ways of getting out, a side saying 'No Ball' and one saying 'Not Out'.

I play the game by rolling the batting die. If it gives me a number, I score that number of runs. If I get 'Howzat!', I roll the bowling die, which has the following results:

* One of the four sides giving a batsman out, in which case I have used one of the ten 'lives' I begin the game with and score no points.
* The 'Not Out', which means that no points are scored, but I don't lose a life.
* The 'No Ball' which means I score a free point - I get another roll of the dice, and I score a point as well.

The 'No Ball' is important, because in a 'Twenty20' game I only get 120 rolls of the dice in each innings. A No Ball won't count towards these 120 balls. In this type of game, one team's turn is over when either either 120 balls have been bowled (ie dice rolled) or I use up all ten of my lives.

The reason I'm explaining all this is that I played a game last night (thread at F8869429?thread=6188725 and scoreboard at A45362603). The game was very tight, and I wondered about the mathematics of the game. I have a few questions I hope someone can answer for me:

1. What would be an 'average' score for a team to get in a game?
2. In cricket, we often hear commentators mention a 'par' score for a pitch, so they might say that a team should expect to score 250-270 to be competitive. What would be an equivalent range for Howzat!, and what criteria would you use to define it?
3. What should I consider to be a very poor score (one where the next team would be able to reach the same total easily) and a very good score (where the second team would have a very hard job of matching that total)?
4. In what percentage of games should I expect to reach 120 balls without losing all my lives?
5. What are the odds of one of my players being out for 0, scoring 50, and scoring 100 without being out?

Are there any other interesting probabilities you could work out from the game?

Bit of a challenge, I know, but my maths runs out of steam just below this level

There are six sides on the bowling die, yes?

I would do it by calculating for 120 balls, but adjust the other probabilities as if there was no chance of no balls (because you know that all of those will be replayed anyway). Then add the no balls on to the average score at the end.

Average score per scoring ball = (1 + 2 + 3 + 4 + 6) / 5 = 16/5
Average chance of scoring ball = 5*6/6 + 5/6 = 30/35 = 6/7
Average score per ball = 96/35
Score over 120 balls = 329 and a bit

Average chance no ball = 1/6 * 1/6 = 1/36
Average number of no balls is recursive, year. 3 + 1/3 from the standard 120. Another 1/10 from that extra 3 + 1/3. No point going on from there there.

Assuming team survives all 120 balls, average score would be 332 and a bit.

Chance of being out on a ball:
1/6 * 4/6 = 4/36 = 1/9.

Chance of being out 10 times on 120 balls:
I forget how to do this. It needs the nCr button on your calculator.

That's great

If there's a 1/9 chance of being out on any given ball, is it fair to say we'd have lost our ten lives around the 90th ball? If so, the average score would be (332/120)*90=249, yes?

Well that's if I got it right.

Which was 1-in-6 chance of having to roll the bowling die * a 4-in-6 chance of being out on the bowling die. Is that right: 4 sides of the six sided die mean you're out?

"If there's a 1/9 chance of being out on any given ball, is it fair to say we'd have lost our ten lives around the 90th ball?"

Maybe. I was never very good at stats. You're meant to do something with the binomial expansion at that point I think.

Yes, that's right. Four sides are out, one is no ball, the other is not out.

'You're meant to do something with the binomial expansion at that point I think.'

Where did you get 30/35 from for the chance of scoring? Isn't it just 5/6 (if you're ignoring no-balls, and adding them on later)?

The binomial formula I think you're on about is normally used to calculate the probability of getting r things in n attempts (where the probability of r is given by p)... the formula is:

P(X) = (nCr) * p^r * (1-p)^(n-r)

where n is the number of "rolls" - so 120 in this case - and r is the number of "outs", with p being the probability of getting "out".

So with n=120, r=10, p=1/9, that gives us a probability of everyone in the team getting out within 120 balls of P = 0.0786 ... which seems a bit small to me, but it could be right I suppose - these things always seem smaller than I expect them to be.

I'm not sure whether there is an equivalent formula for working out how many balls you would expect everyone in the team to be out in; I can't find it at the moment.

Hmm, replying to myself...

... anyway, the number in the previous calculation appears to be low because what it's actually calculating is the probability of EXACTLY 10 occurences with a probability of 1/9 in 120 tries; it doesn't take into account anything above that (which is obviously impossible in the game, but the calculation doesn't know that).

Which means, actually, that there is a way of figuring it out - you can turn the question on its head and ask what the probability of not getting more than 9 occurences with a probability of 1/9 in 120 tries. Effectively this is adding the formula in the previous posting together for r from 0 to 9, and then subtracting the answer from 1. For 120 tries, this gives me a probability of at least 10 occurences (i.e. everyone was out) of 0.870. (If you increase it to 180 tries, say, then the probability increases to 0.996; if you reduce it to 60 then the probability is 0.125 - been playing with Excel a bit too much I think...!)

Still can't find an answer to your other question yet, but I suppose if you were to look at the number of tries required to have a probability of 0.5 for at least 10 occurences then this would give you some idea - from my calculations the smallest number of tries required for this is 87 - so 90 isn't too far off.

When a no-ball happens (1/36 chance), we get another go. So, we know that we still have the same number of results that aren't no balls. I therefore took it out and scaled up the rest of the probabilities to total 1.

You could simply do it as 16/6 * (120 + the expected extra balls). Gives the same result.

The reason I did it the first way was because the number of extra balls recurses, so I thought I'd cut out a bit of inaccuracy. Turns out it's negligible.

This is marvellous

So the average score, based on teams surviving 87 balls, would be 240. The Cavaliers and the World XI both did pretty well, then!

What I'd really like is a formula that can tell me where any score would appear in the rankings, if that makes sense. So if team A were to score 280, for example, the formula could tell me that the next team to bat would score 281 or more 25% of the time, or whatever. And I could perhaps use it the other way, to say that 300 would be a great total because only 10% of innings would beat it, or that 120 would be awful because 90% would score more.

And the odds of a batsman scoring a duck must be almost the same as the chance of getting out on any ball, 1/9, yes? Although that's not quite right; in the event that I rolled a 'Howzat!' and then got a 'Not Out' or 'No Ball', the batsman would still 'face' the next ball on 0. This is more complex, I think:

Odds of being out on any ball: 1/9
Odds of being still on 0 on the next ball but not being out: 1/6 * 1/3

So would the odds of a batsman being out for 0 be: 1/9 + 1/(9*18) + 1/(9*18*18)... which equals about 0.167, or roughly 1/6?

And how about the odds of a century or fifty?

I'd love to work out how you guys figure out stuff like this. I find it fascinating.

I did it myself to start off with - using the formula above, but then I discovered Excel had a function called BINOMDIST that did it all itself.

Your formula for getting a duck is correct, but the answer you got isn't the same as what I just got - 0.117 (1/9 + 1/162 + ...), which seems more reasonable. It should be close to 1/9.

I can currently rank scores based upon the average score per ball worked out earlier combined with my average number of balls faced. So to answer your specific questions a score of 280 would be achieved, on average, in 105 balls, which would be achieved 25.8% of the time (so pretty close to your estimate). 300 would be scored in about 113 balls (about 18% of the time); 120 in about 45 balls (about 2% of the time). I could send you the Excel spreadsheet I've made if you'd like to see it.

You'd perhaps have to work out the standard deviation on the score per ball to get some sort of better rank for scores.

If you could email me your spreadsheet, that would be marvellous