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# Does The Probability Of Someone Rolling A Seven With Two Six Sided Dice Increase As More Attempts Are Made?

SterlingTime | 06:20 Mon 22nd Jul 2024 | Gaming

Regading the "Bank" app game from the Apple app store. This is a game played with two six sided dice. Essentially, you are hoping to avoid rolling a seven as players in the circle take turns rolling both dice. Does the probability of someone rolling a seven increase as more non-seven rolls are made? In other words, do the chance of someone rolling a seven increase as more people take their turn rolling the two dice?

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No....the dice has no memory

No, the odds of a seven are the same every throw.  This is an urban myth that persists

Like already said, dice don't have memories.

Similarly, imagine flipping a two sided coin and you get ten heads on the trot ..... the odds of it being a tails next are still 50/50 .... the coin has no memory of what has previously happened and the next flip of the coin is completely independent to whatever has happened before.

It's exactly the same for the dice.

The odds of throwing a seven (1/6) always remain the same. The likelihood of throwing a seven increases the more consecutive non-sevens are thrown.  IMV

"The likelihood of throwing a seven increases the more consecutive non-sevens are thrown.  IMV"

Why, dave?

The odds represent the likelihood. They are just an arithmetical representaton of the chances.

The likelihood of (say) ten consecutive sevens are far lower than of just one. But the likelihood of each one is the same.

NJ, "The likelihood of (say) ten consecutive sevens are far lower than of just one. But the likelihood of each one is the same."

Eh?

Likelihood is imv not a technical term - it is inevitable (?) that a seven is thrown eventually so the longer it isn't the more likely it will be.

Is this a serious question ?

Try writing a small programme to come up with random results and check the obvious for yourself.

So you're talking about cumulative probability, davebro.

I see the confusion here, as above the dice have no memory. For every throw the chance of throwing a 7 is the same. Now IF we said from the outset that a 7 would be thrown within n throws then the probability of a 7 increases depending on n. So if we said n is 5 then we have 5 goes to throw a 7 then the chance of a 5 is 1/6+1/6+1/6+1/6+1/6= 5/6 , ie 83% so in the circle it's a near certainty that someone will throw a 6 but each individual throw is 1/6.

"NJ, "The likelihood of (say) ten consecutive sevens are far lower than of just one. But the likelihood of each one is the same."

Eh?"

Sorry if I was not clear, Corby.

Keeping it simple, the likelihood of throwing a total of seven with two dice is 1 in 6. (There are six ways that the result will be seven out of the 36 possible combinations).  The likelihood of throwing two sevens in a row is 1/6 * 1/6, so one in 36. But on each occasion that two dice are thrown, the likelihood of throwing seven remains 1/6, no matter what the previous results.

The likelihood that a specific outcome will result n times consecutively decreases as n increases. But the likelihood of seeing that outcome on each occasion remains the same.

"it is inevitable (?) that a seven is thrown eventually..."

No it isn't. Seven might never be thrown, no matter how many attempts are made.

The chances of something other than seven being thrown at each attempt is 5/6. As with consecutive sevens, the chances of consecutive "no sevens" decreases as more attempts are made. With 10 throws the probability of no seven being thrown is 0.161 (where 1 is a certainty that it will happen and 0 is a certainty that it won't). Increase that to 20 throws and the probability decreases to 0.026. Increase to 30 throws and the probability of no seven decreases to 0.004 (i.e. there are only four chances in a thousand that you will not throw a seven in 30 attempts). But, although the probability of "no seven" will become oncreasingly smaller as more attempts are made it will never reach zero.

//So if we said n is 5 then we have 5 goes to throw a 7 then the chance of a 5 is 1/6+1/6+1/6+1/6+1/6= 5/6 ,//

The maths is incorrect there. In that example the chance of NOT getting a 7 on a throw is 5/6.  The chance of not getting a 7 in two goes is 5/6 x 5/6= 25/26. You multiply probbailities  rather than add them

Typo-25/36

NJ, I mis-read your previous post as meaning the chances of each example as being the same but you meant each of the ten rolls.

09:30, yep apologies, you are correct I was ass about face. So we should be saying the chance of no 7 gets smaller and smaller. So  (5/6)**n would tend to zero never actually reaching it. Thanks.

13:16, yep apologies, you are correct I was ass about face. So we should be saying the chance of no 7 gets smaller and smaller. So  (5/6)**n would tend to zero never actually reaching it. Thanks.

Regading the "Bank" app game from the Apple app store. This is a game played with two six sided dice. Essentially, you are hoping to avoid rolling a seven as players in the circle take turns rolling both dice. Does the probability of someone rolling a seven increase as more non-seven rolls are made? In other words, do the chance of someone rolling a seven increase as more people take their turn rolling the two dice?

If we were talking about two fair dice, the answer would be "No".  But as we're talking about the "Bank" app game from the Apple app store, who knows?

"But as we're talking about the "Bank" app game from the Apple app store, who knows?"

Good point. "Virtual" dice (like virtual roulette wheels and virtual packs of playing cards) may not conform to probability theory as the real things might.

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