Probability: At a carnival, a player must spin a spinner with numbers 1-10....

In the probability game, a player must spin a spinner with numbers 1-10 on it (equal chances). Based on the result of the first spinner, they have to guess if the next spinner number is going to be higher or lower. If they guess correctly they get 1 ticket. What is the probability of the player winning?

What are your thoughts? What have you tried? Please share with us any and all work you've done on this problem, even if you know it's wrong. Thank you.

Example
The Prob of of getting a 6 for an example is 1/10

The player could choose higher or lower assuming which would be 1/2 (assuming it's a flip of the coin choice). But the player is more likely to choose lower as the number is 6 ****This is my first issue*****


There is a 5/10 chance of the next spinner being lower (1,2,3,4,5) and a 4/10 chance of it being higher (7,8,9,10). Since it depends on if the person chooses higher or lower, the probability of the second spinner changes (if they choose higher their chances would be less). How would I theoretically calculate the chance of winning then?

***** Since it is Lower OR Higher, I tried adding the probabilities, but the probability would then be the same for all numbers which is not true"

It seems like you came to the same conclusion I did, that the wording of the problem is somewhat faulty. The way I see it, we can either assume that the person decides randomly (e.g. flips a coin) whether to bet on "higher" or "lower"; Or we can assume that they play with strategy and always choose the more likely outcome. However, I doubt that random-choice is what we're meant to assume, as you're correct that the probabilities of winning in such a scenario are the same no matter what number comes up. Let's assume they play logically, and go case-by-case:

First spin is a 1 --> If they bet "lower" they always lose. If they bet "higher" they can only lose if it spins a 1 again, so "higher" is logical bet. The chance of winning is then 9/10 = 90%.

First spin is a 2 --> Betting "lower" means they can only win on a 1, whereas the "higher" bet wins on 3-10, so "higher" is the logical bet. That gives 8/10 = 80% chance of winning.

Now you do the rest. Personally, I'd recommend asking the teacher for clarification as to the meaning of that ambiguous wording.

SHAQ said:

In the probability game, a player must spin a spinner with numbers 1-10 on it (equal chances). Based on the result of the first spinner, they have to guess if the next spinner number is going to be higher or lower. If they guess correctly they get 1 ticket. What is the probability of the player winning?

Click to expand...

The player must follow this strategy: if the first number is less than five chose higher, if greater than five the lower. Always doing that give a probability greater than one-half. You can calculate each outcome

Example: If the first number is 3, chose higher then the probability of wining is 0.7 .
etc

SHAQ said:

In the probability game, a player must spin a spinner with numbers 1-10 on it (equal chances). Based on the result of the first spinner, they have to guess if the next spinner number is going to be higher or lower. If they guess correctly they get 1 ticket. What is the probability of the player winning?

Click to expand...

If you use the best strategy method then....

P(winning | you spin was a 1) = .9
P(winning | you spin was a 2) = .8
P(winning | you spin was a 3) = .7
P(winning | you spin was a 4) = .6
P(winning | you spin was a 5) = .5
P(winning | you spin was a 6) = .5
P(winning | you spin was a 7) = .6
P(winning | you spin was a 8) = .7
P(winning | you spin was a 9) = .8
P(winning | you spin was a 10) = .9

Adding up the right hand numbers yields 7

So the p(winning) = 7/10 = .7

That makes more sense to assume that they choose the one with the highest probability. Thank you